CBSE Class 12 Mathematics Indefinite and Definite Integrals Assignment Set 02

Assignments

(i) Integration by substitution

LEVEL I

Question 1. \(\displaystyle \int \frac{\sec^2(\log x)}{x} \, dx\)
Answer: Let \(t = \log x\), then \(dt = \frac{1}{x} \, dx\)
\(= \int \sec^2 t \, dt\)
\(= \tan t + C\)
\(= \tan(\log x) + C\)
In simple words: Substitute \(\log x\) as \(t\) since its derivative \(\frac{1}{x}\) is present in the integrand.

Exam Tip: Recognize standard derivative forms like \(\frac{1}{x}\) paired with functions of \(\log x\) to apply direct substitution.

Question 2. \(\displaystyle \int \frac{e^{m \tan^{-1} x}}{1 + x^2} \, dx\)
Answer: Let \(t = \tan^{-1} x\), then \(dt = \frac{1}{1 + x^2} \, dx\)
\(= \int e^{mt} \, dt\)
\(= \frac{e^{mt}}{m} + C\)
\(= \frac{e^{m \tan^{-1} x}}{m} + C\)
In simple words: Substitute the inverse trigonometric exponent \(\tan^{-1} x\) as \(t\) because its derivative \(\frac{1}{1 + x^2}\) multiplies the differential.

Exam Tip: Standard substitution instantly reduces exponential functions containing inverse trigonometric powers into basic exponential integrals.

Question 3. \(\displaystyle \int \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \, dx\)
Answer: Let \(t = \sin^{-1} x\), then \(dt = \frac{1}{\sqrt{1 - x^2}} \, dx\)
\(= \int e^t \, dt\)
\(= e^t + C\)
\(= e^{\sin^{-1} x} + C\)
In simple words: Put \(t = \sin^{-1} x\), transforming the integral into the simple form \(\int e^t \, dt\).

Exam Tip: Always look out for standard derivative pairs such as \(\frac{1}{\sqrt{1 - x^2}}\) alongside \(\sin^{-1} x\).

LEVEL II

Question 1. \(\displaystyle \int \frac{1}{x + \sqrt{x}} \, dx\)
Answer: Rewrite the denominator by factoring out \(\sqrt{x}\):
\(= \int \frac{1}{\sqrt{x}(\sqrt{x} + 1)} \, dx\)
Let \(t = \sqrt{x} + 1\), then \(dt = \frac{1}{2\sqrt{x}} \, dx \implies \frac{1}{\sqrt{x}} \, dx = 2 \, dt\)
\(= \int \frac{2}{t} \, dt\)
\(= 2 \log |t| + C\)
\(= 2 \log |\sqrt{x} + 1| + C\)
In simple words: Factor \(\sqrt{x}\) out from the denominator so that substitution simplifies the integrand into a logarithmic standard form.

Exam Tip: Expressions containing sums of powers of \(x\) with fractional indices can often be integrated cleanly by substituting the lower fractional power or sum.

Question 2. \(\displaystyle \int \frac{1}{x(x^6 - 1)} \, dx\)
Answer: Multiply numerator and denominator by \(x^5\):
\(= \int \frac{x^5}{x^6(x^6 - 1)} \, dx\)
Let \(t = x^6\), then \(dt = 6x^5 \, dx \implies x^5 \, dx = \frac{1}{6} \, dt\)
\(= \frac{1}{6} \int \frac{1}{t(t - 1)} \, dt\)
Using partial fractions: \(\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac{1}{t}\)
\(= \frac{1}{6} \int \left( \frac{1}{t - 1} - \frac{1}{t} \right) dt\)
\(= \frac{1}{6} (\log |t - 1| - \log |t|) + C\)
\(= \frac{1}{6} \log \left| \frac{x^6 - 1}{x^6} \right| + C\)
In simple words: Introduce \(x^5\) in the numerator to match the derivative of \(x^6\), then apply partial fractions.

Exam Tip: Multiplying top and bottom by \(x^{n-1}\) is a standard technique when dealing with integrals containing terms of the form \(x^n(x^n - a)\).

Question 3. \(\displaystyle \int \frac{1}{e^x - 1} \, dx\)
Answer: Multiply numerator and denominator by \(e^{-x}\):
\(= \int \frac{e^{-x}}{1 - e^{-x}} \, dx\)
Let \(t = 1 - e^{-x}\), then \(dt = e^{-x} \, dx\)
\(= \int \frac{1}{t} \, dt\)
\(= \log |t| + C\)
\(= \log |1 - e^{-x}| + C\)
Alternatively, write as \(\int \frac{e^{-x}}{1 - e^{-x}} \, dx = -\log |1 - e^{-x}| + C\) or \(\log |e^x - 1| - x + C\).
In simple words: Multiply top and bottom by \(e^{-x}\) to make the numerator the exact derivative of the denominator.

Exam Tip: For exponential fractions, expressing terms in negative powers of \(e\) simplifies substitution.

LEVEL III

Question 1. \(\displaystyle \int \frac{\tan x}{\sin x \cos x} \, dx\)
Answer: Simplify the integrand:
\(\frac{\tan x}{\sin x \cos x} = \frac{\frac{\sin x}{\cos x}}{\sin x \cos x} = \frac{1}{\cos^2 x} = \sec^2 x\)
\(= \int \sec^2 x \, dx\)
\(= \tan x + C\)
In simple words: Convert tangent into sine and cosine to cancel terms, reducing the problem to the standard integral of \(\sec^2 x\).

Exam Tip: Always simplify trigonometric expressions using fundamental identities before attempting complex substitutions.

Question 2. \(\displaystyle \int \frac{\tan x}{\sec x + \cos x} \, dx\)
Answer: Express in terms of sine and cosine:
\(= \int \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \cos x} \, dx = \int \frac{\frac{\sin x}{\cos x}}{\frac{1 + \cos^2 x}{\cos x}} \, dx = \int \frac{\sin x}{1 + \cos^2 x} \, dx\)
Let \(t = \cos x\), then \(dt = -\sin x \, dx \implies \sin x \, dx = -dt\)
\(= -\int \frac{1}{1 + t^2} \, dt\)
\(= -\tan^{-1} t + C\)
\(= -\tan^{-1}(\cos x) + C\)
In simple words: Convert all trigonometric functions to sine and cosine, simplify the complex fraction, and substitute \(\cos x\).

Exam Tip: Converting trigonometric expressions to sine and cosine is a reliable fallback strategy when substitution is not immediately obvious.

Question 3. \(\displaystyle \int \frac{1}{\sin x \cos^3 x} \, dx\)
Answer: Multiply and divide numerator by \(\cos^2 x\) or use the identity \(\sin^2 x + \cos^2 x = 1\):
\(= \int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x} \, dx = \int \left( \frac{\sin^2 x}{\sin x \cos^3 x} + \frac{\cos^2 x}{\sin x \cos^3 x} \right) dx\)
\(= \int \left( \frac{\sin x}{\cos^3 x} + \frac{1}{\sin x \cos x} \right) dx = \int (\tan x \sec^2 x + \sec x \csc x) \, dx\)
Alternatively, divide numerator and denominator by \(\cos^4 x\):
\(= \int \frac{\sec^4 x}{\tan x} \, dx = \int \frac{(1 + \tan^2 x)\sec^2 x}{\tan x} \, dx\)
Let \(t = \tan x\), then \(dt = \sec^2 x \, dx\)
\(= \int \frac{1 + t^2}{t} \, dt = \int \left( \frac{1}{t} + t \right) dt\)
\(= \log |t| + \frac{t^2}{2} + C\)
\(= \log |\tan x| + \frac{\tan^2 x}{2} + C\)
In simple words: Divide the numerator and denominator by \(\cos^4 x\) to convert the expression entirely into powers of \(\tan x\) and \(\sec^2 x\).

Exam Tip: When integrating inverse powers of sines and cosines in denominators, dividing through by \(\cos^k x\) is a powerful standard method.

(ii) Application of trigonometric function in integrals

LEVEL I

Question 1. \(\displaystyle \int \sin^3 x \, dx\)
Answer: Using the triple angle identity \(\sin 3x = 3\sin x - 4\sin^3 x \implies \sin^3 x = \frac{3\sin x - \sin 3x}{4}\):
\(= \int \frac{3\sin x - \sin 3x}{4} \, dx\)
\(= \frac{1}{4} \left( -3\cos x + \frac{\cos 3x}{3} \right) + C\)
Alternatively, using substitution: \(\int \sin^2 x \sin x \, dx = \int (1 - \cos^2 x)\sin x \, dx\) (let \(t = \cos x\)).
In simple words: Express \(\sin^3 x\) using trigonometric multiple-angle identities or substitute \(\cos x\).

Exam Tip: Odd powers of sine or cosine can be integrated by saving one factor to form the differential and converting the rest using Pythagorean identities.

Question 2. \(\displaystyle \int \cos^2 3x \, dx\)
Answer: Using the half-angle formula \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\):
\(= \int \frac{1 + \cos 6x}{2} \, dx\)
\(= \frac{1}{2} \left( x + \frac{\sin 6x}{6} \right) + C\)
\(= \frac{x}{2} + \frac{\sin 6x}{12} + C\)
In simple words: Reduce the power using the trigonometric half-angle identity before integrating term by term.

Exam Tip: Even powers of sine and cosine should always be reduced to linear multiples of angles using double-angle or half-angle formulas.

Question 3. \(\displaystyle \int \cos x \cos 2x \cos 3x \, dx\)
Answer: Use the product-to-sum formula \(2\cos A \cos B = \cos(A + B) + \cos(A - B)\):
Combine \(\cos 3x\) and \(\cos x\) first:
\(2\cos 3x \cos x = \cos 4x + \cos 2x\)
Now multiply by \(\cos 2x\):
\(= \frac{1}{2} \int (\cos 4x + \cos 2x)\cos 2x \, dx\)
\(= \frac{1}{2} \int (\cos 4x \cos 2x + \cos^2 2x) \, dx\)
Using \(\cos 4x \cos 2x = \frac{1}{2}(\cos 6x + \cos 2x)\) and \(\cos^2 2x = \frac{1 + \cos 4x}{2}\):
\(= \frac{1}{2} \int \left[ \frac{1}{2}(\cos 6x + \cos 2x) + \frac{1 + \cos 4x}{2} \right] dx\)
\(= \frac{1}{4} \int (1 + \cos 2x + \cos 4x + \cos 6x) \, dx\)
\(= \frac{1}{4} \left( x + \frac{\sin 2x}{2} + \frac{\sin 4x}{4} + \frac{\sin 6x}{6} \right) + C\)
In simple words: Convert the product of cosines into a sum of cosines using standard product-to-sum formulas, then integrate each term.

Exam Tip: Pair the largest and smallest angles first when applying product-to-sum formulas to simplify subsequent steps.

LEVEL II

Question 1. \(\displaystyle \int \sec^4 x \tan x \, dx\)
Answer: Rewrite the integral as:
\(= \int \sec^3 x (\sec x \tan x) \, dx\)
Let \(t = \sec x\), then \(dt = \sec x \tan x \, dx\)
\(= \int t^3 \, dt\)
\(= \frac{t^4}{4} + C\)
\(= \frac{\sec^4 x}{4} + C\)
In simple words: Separate \(\sec x \tan x\) to serve as the derivative of \(\sec x\), then apply substitution.

Exam Tip: When powers of secant and tangent appear together, isolate \(\sec x \tan x\) if secant has an odd power, or isolate \(\sec^2 x\) if tangent has an odd power.

Question 2. \(\displaystyle \int \frac{\sin 4x}{\sin x} \, dx\)
Answer: Use multiple angle formulas: \(\sin 4x = 2\sin 2x \cos 2x = 4\sin x \cos x \cos 2x\)
\(= \int \frac{4\sin x \cos x \cos 2x}{\sin x} \, dx\)
\(= 4 \int \cos x \cos 2x \, dx\)
Using \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\):
\(= 2 \int (\cos 3x + \cos x) \, dx\)
\(= 2 \left( \frac{\sin 3x}{3} + \sin x \right) + C\)
In simple words: Expand \(\sin 4x\) using multiple-angle identities to cancel \(\sin x\) in the denominator.

Exam Tip: Always expand higher multiple angles in the numerator to see if they share factors with the denominator.

LEVEL III

Question 1. \(\displaystyle \int \cos^5 x \, dx\)
Answer: Isolate one cosine factor:
\(= \int \cos^4 x \cos x \, dx = \int (\cos^2 x)^2 \cos x \, dx = \int (1 - \sin^2 x)^2 \cos x \, dx\)
Let \(t = \sin x\), then \(dt = \cos x \, dx\)
\(= \int (1 - t^2)^2 \, dt = \int (1 - 2t^2 + t^4) \, dt\)
\(= t - \frac{2t^3}{3} + \frac{t^5}{5} + C\)
\(= \sin x - \frac{2\sin^3 x}{3} + \frac{\sin^5 x}{5} + C\)
In simple words: Split off a single cosine factor, convert the remaining even power into sine terms using Pythagorean identities, and substitute \(t = \sin x\).

Exam Tip: For odd powers of sine or cosine, reserve one trigonometric function for the differential \(dt\) and express the rest in terms of the complementary function.

Question 2. \(\displaystyle \int \sin^2 x \cos^3 x \, dx\)
Answer: Isolate one cosine factor:
\(= \int \sin^2 x \cos^2 x (\cos x \, dx) = \int \sin^2 x (1 - \sin^2 x) \cos x \, dx\)
Let \(t = \sin x\), then \(dt = \cos x \, dx\)
\(= \int t^2 (1 - t^2) \, dt = \int (t^2 - t^4) \, dt\)
\(= \frac{t^3}{3} - \frac{t^5}{5} + C\)
\(= \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C\)
In simple words: Since cosine has an odd power, save one cosine for \(dt\) and convert the remaining even powers of cosine into sine.

Exam Tip: When both functions are present, look at their powers: substitute for the function whose power is odd.

(iii) Integration using standard results

LEVEL I

Question 1. \(\displaystyle \int \frac{dx}{\sqrt{4x^2 - 9}}\)
Answer: Rewrite the denominator:
\(= \int \frac{dx}{\sqrt{(2x)^2 - 3^2}}\)
Using the standard formula \(\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + C\) and dividing by the coefficient of \(x\) (which is 2):
\(= \frac{1}{2} \log |2x + \sqrt{4x^2 - 9}| + C\)
In simple words: Express the integral in standard form and remember to divide by the linear coefficient of \(x\).

Exam Tip: Always divide the final logarithmic result by the coefficient of \(x\) when applying standard formulas.

Question 2. \(\displaystyle \int \frac{1}{x^2 + 2x + 10} \, dx\)
Answer: Complete the square in the denominator:
\(x^2 + 2x + 10 = (x^2 + 2x + 1) + 9 = (x + 1)^2 + 3^2\)
\(= \int \frac{dx}{(x + 1)^2 + 3^2}\)
Using the standard formula \(\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\):
\(= \frac{1}{3} \tan^{-1}\left(\frac{x + 1}{3}\right) + C\)
In simple words: Complete the square for the quadratic denominator to convert it into the standard form \(\frac{1}{x^2 + a^2}\).

Exam Tip: Completing the square is the standard method for integrating reciprocal quadratic expressions without linear roots in the denominator.

Question 3. \(\displaystyle \int \frac{dx}{9x^2 + 12x + 13}\)
Answer: Complete the square in the denominator:
\(9x^2 + 12x + 13 = (3x)^2 + 2(3x)(2) + 2^2 - 4 + 13 = (3x + 2)^2 + 9 = (3x + 2)^2 + 3^2\)
\(= \int \frac{dx}{(3x + 2)^2 + 3^2}\)
Using the standard formula and dividing by the coefficient of \(x\) (which is 3):
\(= \frac{1}{3} \cdot \frac{1}{3} \tan^{-1}\left(\frac{3x + 2}{3}\right) + C\)
\(= \frac{1}{9} \tan^{-1}\left(\frac{3x + 2}{3}\right) + C\)
In simple words: Complete the square, apply the inverse tangent formula, and divide by the derivative coefficient of the linear term \(3x + 2\).

Exam Tip: When the coefficient of \(x^2\) is not 1, factor it out or complete the square carefully before applying standard formulas.

LEVEL II

Question 1. \(\displaystyle \int \frac{x}{x^4 + x^2 + 1} \, dx\)
Answer: Let \(t = x^2\), then \(dt = 2x \, dx \implies x \, dx = \frac{1}{2} \, dt\)
\(= \frac{1}{2} \int \frac{dt}{t^2 + t + 1}\)
Complete the square in the denominator:
\(t^2 + t + 1 = \left( t + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2\)
\(= \frac{1}{2} \int \frac{dt}{(t + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}\)
Using the standard formula \(\frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right)\):
\(= \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C\)
\(= \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2x^2 + 1}{\sqrt{3}}\right) + C\)
In simple words: Substitute \(t = x^2\) to transform the integral into a standard quadratic denominator form.

Exam Tip: Whenever an integral contains powers of \(x^2\) and \(x\) in the numerator, substituting \(t = x^2\) simplifies the polynomial degree.

Question 2. \(\displaystyle \int \frac{\cos x}{\sin^2 x + 4\sin x + 5} \, dx\)
Answer: Let \(t = \sin x\), then \(dt = \cos x \, dx\)
\(= \int \frac{dt}{t^2 + 4t + 5}\)
Complete the square in the denominator:
\(t^2 + 4t + 5 = (t + 2)^2 + 1^2\)
\(= \int \frac{dt}{(t + 2)^2 + 1^2}\)
\(= \tan^{-1}(t + 2) + C\)
\(= \tan^{-1}(\sin x + 2) + C\)
In simple words: Substitute \(t = \sin x\) since \(\cos x \, dx\) is the numerator, then complete the square.

Exam Tip: Spot the derivative of sine or cosine in the numerator to immediately apply substitution before completing the square.

Question 3. \(\displaystyle \int \frac{dx}{\sqrt{7 - 6x - x^2}}\)
Answer: Complete the square inside the square root:
\(7 - 6x - x^2 = 7 - (x^2 + 6x + 9 - 9) = 7 + 9 - (x + 3)^2 = 16 - (x + 3)^2 = 4^2 - (x + 3)^2\)
\(= \int \frac{dx}{\sqrt{4^2 - (x + 3)^2}}\)
Using the standard formula \(\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C\):
\(= \sin^{-1}\left(\frac{x + 3}{4}\right) + C\)
In simple words: Complete the square inside the radical and apply the standard inverse sine integration formula.

Exam Tip: Be careful with negative signs when completing the square for quadratic expressions inside square roots.

LEVEL III

Question 1. \(\displaystyle \int \frac{2x}{\sqrt{1 - x^2 - x^4}} \, dx\)
Answer: Let \(t = x^2\), then \(dt = 2x \, dx\)
\(= \int \frac{dt}{\sqrt{1 - t - t^2}}\)
Complete the square in the denominator:
\(1 - t - t^2 = 1 - \left( t^2 + t + \frac{1}{4} - \frac{1}{4} \right) = \frac{5}{4} - \left( t + \frac{1}{2} \right)^2 = \left( \frac{\sqrt{5}}{2} \right)^2 - \left( t + \frac{1}{2} \right)^2\)
\(= \int \frac{dt}{\sqrt{(\frac{\sqrt{5}}{2})^2 - (t + \frac{1}{2})^2}}\)
Using the standard formula \(\sin^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{5}}{2}}\right)\):
\(= \sin^{-1}\left(\frac{2x^2 + 1}{\sqrt{5}}\right) + C\)
In simple words: Substitute \(t = x^2\), complete the square, and use the inverse sine standard integral.

Exam Tip: Recognizing \(2x \, dx\) as the differential of \(x^2\) simplifies fourth-degree polynomials under radicals.

Question 2. \(\displaystyle \int \frac{x^2 + x + 1}{x^2 - x + 1} \, dx\)
Answer: Since the degree of the numerator equals the degree of the denominator, perform polynomial division or rewrite the numerator:
Numerator = \((x^2 - x + 1) + 2x\)
\(= \int \frac{(x^2 - x + 1) + 2x}{x^2 - x + 1} \, dx = \int \left( 1 + \frac{2x}{x^2 - x + 1} \right) dx\)
For the second term, adjust the numerator to match the derivative of the denominator \(\frac{d}{dx}(x^2 - x + 1) = 2x - 1\):
\(\frac{2x}{x^2 - x + 1} = \frac{(2x - 1) + 1}{x^2 - x + 1}\)
\(= \int 1 \, dx + \int \frac{2x - 1}{x^2 - x + 1} \, dx + \int \frac{1}{(x - \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \, dx\)
\(= x + \log |x^2 - x + 1| + \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right) + C\)
In simple words: Divide the polynomials since degrees are equal, then split the remainder integral into a logarithmic part and an inverse tangent part.

Exam Tip: When the numerator degree matches or exceeds the denominator degree, always divide first before applying standard integration techniques.

Question 3. \(\displaystyle \int \frac{x + 2}{\sqrt{x^2 + 5x + 6}} \, dx\)
Answer: Express the numerator in terms of the derivative of the quadratic expression inside the radical: \(\frac{d}{dx}(x^2 + 5x + 6) = 2x + 5\)
Let \(x + 2 = A(2x + 5) + B\)
Comparing coefficients: \(2A = 1 \implies A = \frac{1}{2}\), and \(5A + B = 2 \implies \frac{5}{2} + B = 2 \implies B = -\frac{1}{2}\)
\(= \int \frac{\frac{1}{2}(2x + 5) - \frac{1}{2}}{\sqrt{x^2 + 5x + 6}} \, dx\)
\(= \frac{1}{2} \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} \, dx - \frac{1}{2} \int \frac{dx}{\sqrt{(x + \frac{5}{2})^2 - (\frac{1}{2})^2}}\)
\(= \frac{1}{2} \cdot 2\sqrt{x^2 + 5x + 6} - \frac{1}{2} \log \left| x + \frac{5}{2} + \sqrt{x^2 + 5x + 6} \right| + C\)
\(= \sqrt{x^2 + 5x + 6} - \frac{1}{2} \log \left| x + \frac{5}{2} + \sqrt{x^2 + 5x + 6} \right| + C\)
In simple words: Express the numerator as a linear combination of the derivative of the radicand and a constant, then integrate each part separately.

Exam Tip: Integrals of the form \(\int \frac{px + q}{\sqrt{ax^2 + bx + c}} \, dx\) are always solved by splitting the numerator using the derivative of the quadratic expression.

Question 4. \(\displaystyle \int \sqrt{\frac{1 - x}{1 + x}} \, dx\)
Answer: Rationalize the numerator by multiplying top and bottom by \(\sqrt{1 - x}\):
\(= \int \frac{1 - x}{\sqrt{1 - x^2}} \, dx = \int \frac{1}{\sqrt{1 - x^2}} \, dx - \int \frac{x}{\sqrt{1 - x^2}} \, dx\)
For the first integral: \(\sin^{-1} x\)
For the second integral, let \(t = 1 - x^2\), then \(dt = -2x \, dx\):
\(= -\int -\frac{x}{\sqrt{1 - x^2}} \, dx = \sqrt{1 - x^2}\)
Combining both: \(= \sin^{-1} x + \sqrt{1 - x^2} + C\)
In simple words: Rationalize the fraction inside the square root to split it into a standard inverse sine integral and a simple substitution integral.

Exam Tip: Rationalization is an effective first step when dealing with square roots of linear fractional expressions.

Question 5. \(\displaystyle \int \frac{6x + 7}{\sqrt{(x - 5)(x - 4)}} \, dx\) 
Answer: Expand the denominator: \((x - 5)(x - 4) = x^2 - 9x + 20\)
Derivative of the quadratic expression: \(\frac{d}{dx}(x^2 - 9x + 20) = 2x - 9\)
Express the numerator \(6x + 7\) in the form \(A(2x - 9) + B\):
\(6x + 7 = 2Ax + (-9A + B)\)
Equating coefficients: \(2A = 6 \implies A = 3\), and \(-9(3) + B = 7 \implies -27 + B = 7 \implies B = 34\)
\(= \int \frac{3(2x - 9) + 34}{\sqrt{x^2 - 9x + 20}} \, dx\)
\(= 3 \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} \, dx + 34 \int \frac{dx}{\sqrt{(x - \frac{9}{2})^2 - (\frac{1}{2})^2}}\)
\(= 6\sqrt{x^2 - 9x + 20} + 34 \log \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C\)
In simple words: Multiply out the factors in the denominator, then split the numerator using the derivative of the quadratic expression.

Exam Tip: Standard splitting formulas apply directly once the denominator is expanded into standard quadratic form.

(iv) Integration using Partial Fraction

LEVEL I

Question 1. \(\displaystyle \int \frac{2x + 1}{(x + 1)(x - 1)} \, dx\)
Answer: Express using partial fractions:
\(\frac{2x + 1}{(x + 1)(x - 1)} = \frac{A}{x + 1} + \frac{B}{x - 1}\)
\(2x + 1 = A(x - 1) + B(x + 1)\)
Put \(x = 1 \implies 3 = 2B \implies B = \frac{3}{2}\)
Put \(x = -1 \implies -1 = -2A \implies A = \frac{1}{2}\)
\(= \int \left( \frac{1/2}{x + 1} + \frac{3/2}{x - 1} \right) dx\)
\(= \frac{1}{2} \log |x + 1| + \frac{3}{2} \log |x - 1| + C\)
In simple words: Decompose the rational function into partial fractions and integrate each logarithmic term.

Exam Tip: Use the cover-up method for linear non-repeated factors to quickly find constants \(A\) and \(B\).

Question 2. \(\displaystyle \int \frac{x^2}{(x - 1)(x - 2)(x - 3)} \, dx\)
Answer: Decompose into partial fractions:
\(\frac{x^2}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}\)
\(x^2 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)\)
Put \(x = 1 \implies 1 = A(-1)(-2) \implies A = \frac{1}{2}\)
Put \(x = 2 \implies 4 = B(1)(-1) \implies B = -4\)
Put \(x = 3 \implies 9 = C(2)(1) \implies C = \frac{9}{2}\)
\(= \int \left( \frac{1/2}{x - 1} - \frac{4}{x - 2} + \frac{9/2}{x - 3} \right) dx\)
\(= \frac{1}{2} \log |x - 1| - 4 \log |x - 2| + \frac{9}{2} \log |x - 3| + C\)
In simple words: Split the expression into three partial fractions corresponding to each linear factor in the denominator.

Exam Tip: Substitute the roots of each linear factor directly to find the partial fraction coefficients efficiently.

Question 3. \(\displaystyle \int \frac{3x - 2}{(x + 1)^2(x + 3)} \, dx\)
Answer: Decompose into partial fractions with a repeated linear factor:
\(\frac{3x - 2}{(x + 1)^2(x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + 3}\)
\(3x - 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)^2\)
Put \(x = -1 \implies -5 = B(2) \implies B = -\frac{5}{2}\)
Put \(x = -3 \implies -11 = C(-2)^2 \implies 4C = -11 \implies C = -\frac{11}{4}\)
Equating coefficients of \(x^2\) on both sides (\(A + C = 0\)): \(A = -C = \frac{11}{4}\)
\(= \int \left( \frac{11/4}{x + 1} - \frac{5/2}{(x + 1)^2} - \frac{11/4}{x + 3} \right) dx\)
\(= \frac{11}{4} \log |x + 1| + \frac{5}{2(x + 1)} - \frac{11}{4} \log |x + 3| + C\)
In simple words: Repeated linear factors require terms of powers up to the multiplicity in the partial fraction expansion.

Exam Tip: For repeated linear factors, finding coefficients by substituting roots and comparing highest degree coefficients is standard practice.

LEVEL II

Question 1. \(\displaystyle \int \frac{x^2 + 2x + 8}{(x - 1)(x - 2)} \, dx\)
Answer: Since the degree of the numerator equals the denominator degree, perform polynomial division first:
\(x^2 + 2x + 8 = (x^2 - 3x + 2) + 5x + 6\)
Wait, using polynomial long division: \(\frac{x^2 + 2x + 8}{x^2 - 3x + 2} = 1 + \frac{5x + 6}{x^2 - 3x + 2}\)
Now split the remainder into partial fractions:
\(\frac{5x + 6}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}\)
\(5x + 6 = A(x - 2) + B(x - 1)\)
Put \(x = 1 \implies 11 = A(-1) \implies A = -11\)
Put \(x = 2 \implies 16 = B(1) \implies B = 16\)
\(= \int \left( 1 - \frac{11}{x - 1} + \frac{16}{x - 2} \right) dx\)
\(= x - 11 \log |x - 1| + 16 \log |x - 2| + C\)
In simple words: Divide first since degrees are equal, then apply partial fractions to the proper rational remainder.

Exam Tip: Never forget to perform long division when the numerator degree is greater than or equal to the denominator degree.

Question 2. \(\displaystyle \int \frac{x^2 + x + 1}{x^2(x + 2)} \, dx\)
Answer: Decompose into partial fractions:
\(\frac{x^2 + x + 1}{x^2(x + 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 2}\)
\(x^2 + x + 1 = Ax(x + 2) + B(x + 2) + Cx^2\)
Put \(x = 0 \implies 1 = 2B \implies B = \frac{1}{2}\)
Put \(x = -2 \implies 4 - 2 + 1 = C(-2)^2 \implies 3 = 4C \implies C = \frac{3}{4}\)
Equate coefficients of \(x^2\) (\(A + C = 1\)): \(A = 1 - \frac{3}{4} = \frac{1}{4}\)
\(= \int \left( \frac{1/4}{x} + \frac{1/2}{x^2} + \frac{3/4}{x + 2} \right) dx\)
\(= \frac{1}{4} \log |x| - \frac{1}{2x} + \frac{3}{4} \log |x + 2| + C\)
In simple words: Set up partial fractions accounting for the repeated root \(x^2\) and the distinct linear root \(x + 2\).

Exam Tip: A squared term in the denominator requires both a linear term and a squared term in the partial fraction setup.

Question 3. \(\displaystyle \int \frac{x^2 + 1}{(x - 1)^2(x + 3)} \, dx\)
Answer: Decompose into partial fractions:
\(\frac{x^2 + 1}{(x - 1)^2(x + 3)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 3}\)
\(x^2 + 1 = A(x - 1)(x + 3) + B(x + 3) + C(x - 1)^2\)
Put \(x = 1 \implies 2 = B(4) \implies B = \frac{1}{2}\)
Put \(x = -3 \implies 9 + 1 = C(-4)^2 \implies 10 = 16C \implies C = \frac{5}{8}\)
Equate coefficients of \(x^2\) (\(A + C = 1\)): \(A = 1 - \frac{5}{8} = \frac{3}{8}\)
\(= \int \left( \frac{3/8}{x - 1} + \frac{1/2}{(x - 1)^2} + \frac{5/8}{x + 3} \right) dx\)
\(= \frac{3}{8} \log |x - 1| - \frac{1}{2(x - 1)} + \frac{5}{8} \log |x + 3| + C\)
In simple words: Use standard partial fraction expansion for repeated and non-repeated linear factors, then integrate directly.

Exam Tip: Substituting roots simplifies coefficient finding for linear factors in partial fraction decompositions.

LEVEL III

Question 1. \(\displaystyle \int \frac{8}{(x + 2)(x^2 + 4)} \, dx\)
Answer: Decompose into partial fractions involving a quadratic factor:
\(\frac{8}{(x + 2)(x^2 + 4)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 4}\)
\(8 = A(x^2 + 4) + (Bx + C)(x + 2)\)
Put \(x = -2 \implies 8 = A(4 + 4) \implies 8A = 8 \implies A = 1\)
Equate coefficients of \(x^2\) (\(A + B = 0\)): \(B = -1\)
Equate constant terms (\(4A + 2C = 8\)): \(4(1) + 2C = 8 \implies 2C = 4 \implies C = 2\)
\(= \int \left( \frac{1}{x + 2} + \frac{-x + 2}{x^2 + 4} \right) dx\)
\(= \int \frac{dx}{x + 2} - \frac{1}{2} \int \frac{2x}{x^2 + 4} \, dx + 2 \int \frac{dx}{x^2 + 2^2}\)
\(= \log |x + 2| - \frac{1}{2} \log |x^2 + 4| + \frac{2}{2} \tan^{-1}\left(\frac{x}{2}\right) + C\)
\(= \log |x + 2| - \frac{1}{2} \log |x^2 + 4| + \tan^{-1}\left(\frac{x}{2}\right) + C\)
In simple words: Quadratic irreducible factors in the denominator require a linear numerator term \(Bx + C\) in partial fractions.

Exam Tip: Always include a linear term \(Bx + C\) over irreducible quadratic factors in partial fraction decompositions.

Question 2. \(\displaystyle \int \frac{1}{\sin x + \sin 2x} \, dx\)
Answer: Expand \(\sin 2x = 2\sin x \cos x\):
\(= \int \frac{dx}{\sin x(1 + 2\cos x)}\)
Multiply numerator and denominator by \(\sin x\):
\(= \int \frac{\sin x}{\sin^2 x(1 + 2\cos x)} \, dx = \int \frac{\sin x}{(1 - \cos^2 x)(1 + 2\cos x)} \, dx\)
Let \(t = \cos x\), then \(dt = -\sin x \, dx\)
\(= -\int \frac{dt}{(1 - t^2)(1 + 2t)} = \int \frac{dt}{(t - 1)(t + 1)(2t + 1)}\)
Using partial fractions: \(\frac{1}{(t - 1)(t + 1)(2t + 1)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{2t + 1}\)
Solving for coefficients: \(A = \frac{1}{6}, B = -\frac{1}{2}, C = \frac{4}{3}\)
\(= \frac{1}{6} \log |t - 1| - \frac{1}{2} \log |t + 1| + \frac{2}{3} \log |2t + 1| + C\)
Substitute back \(t = \cos x\):
\(= \frac{1}{6} \log |\cos x - 1| - \frac{1}{2} \log |\cos x + 1| + \frac{2}{3} \log |2\cos x + 1| + C\)
In simple words: Convert into algebraic form by substituting \(t = \cos x\), then apply partial fractions.

Exam Tip: Multiplying top and bottom by \(\sin x\) prepares the integral for substitution followed by partial fraction decomposition.

Question 3. \(\displaystyle \int \frac{1}{1 + x^3} \, dx\)
Answer: Factor the denominator using sum of cubes: \(1 + x^3 = (1 + x)(1 - x + x^2)\)
Decompose into partial fractions:
\(\frac{1}{(1 + x)(1 - x + x^2)} = \frac{A}{1 + x} + \frac{Bx + C}{1 - x + x^2}\)
\(1 = A(1 - x + x^2) + (Bx + C)(1 + x)\)
Put \(x = -1 \implies 1 = A(1 - (-1) + (-1)^2) = 3A \implies A = \frac{1}{3}\)
Equate coefficients of \(x^2\) (\(A + B = 0\)): \(B = -\frac{1}{3}\)
Equate constant terms (\(A + C = 1\)): \(C = 1 - \frac{1}{3} = \frac{2}{3}\)
\(= \int \left( \frac{1/3}{1 + x} + \frac{-\frac{1}{3}x + \frac{2}{3}}{1 - x + x^2} \right) dx\)
For the second term, adjust numerator to derivative of denominator \(\frac{d}{dx}(1 - x + x^2) = 2x - 1\):
\(-\frac{1}{3}x + \frac{2}{3} = -\frac{1}{6}(2x - 1) + \frac{1}{2}\)
\(= \frac{1}{3} \log |1 + x| - \frac{1}{6} \log |1 - x + x^2| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right) + C\)
In simple words: Factor the sum of cubes into linear and quadratic factors, then apply partial fractions.

Exam Tip: Integrals with cubic denominators factor nicely into linear and irreducible quadratic terms suitable for partial fractions.

(v) Integration by Parts

LEVEL I

Question 1. \(\displaystyle \int x \sec^2 x \, dx\)
Answer: Using integration by parts formula \(\int u v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx\):
Let \(u = x\) and \(v = \sec^2 x\)
\(= x \int \sec^2 x \, dx - \int \left( \frac{d}{dx}(x) \int \sec^2 x \, dx \right) dx\)
\(= x \tan x - \int 1 \cdot \tan x \, dx\)
\(= x \tan x - \log |\sec x| + C\)
In simple words: Apply integration by parts taking \(x\) as the first function and \(\sec^2 x\) as the second function.

Exam Tip: Use the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) to choose your first function \(u\).

Question 2. \(\displaystyle \int \log x \, dx\)
Answer: Write as \(\int (\log x) \cdot 1 \, dx\):
Let \(u = \log x\) and \(v = 1\)
\(= \log x \int 1 \, dx - \int \left( \frac{1}{x} \int 1 \, dx \right) dx\)
\(= x \log x - \int \frac{1}{x} \cdot x \, dx\)
\(= x \log x - \int 1 \, dx\)
\(= x \log x - x + C\)
In simple words: Treat logarithmic integrals as a product with \(1\) and apply integration by parts.

Exam Tip: Standard integrals of logarithmic and inverse trigonometric functions are solved by multiplying by 1 and applying integration by parts.

Question 3. \(\displaystyle \int e^x (\tan x + \log \sec x) \, dx\)
Answer: Using the standard result \(\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C\):
Let \(f(x) = \log \sec x\), then \(f'(x) = \frac{1}{\sec x} \sec x \tan x = \tan x\)
\(= e^x \log \sec x + C\)
In simple words: Recognize the standard exponential form \(e^x [f(x) + f'(x)]\) to write the answer immediately.

Exam Tip: Whenever \(e^x\) multiplies a sum of two terms, check if one term is the derivative of the other.

LEVEL II

Question 1. \(\displaystyle \int \sin^{-1} x \, dx\)
Answer: Write as \(\int (\sin^{-1} x) \cdot 1 \, dx\):
Let \(u = \sin^{-1} x\) and \(v = 1\)
\(= \sin^{-1} x \int 1 \, dx - \int \left( \frac{1}{\sqrt{1 - x^2}} \cdot x \right) dx\)
\(= x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} \, dx\)
For the second integral, substitute \(t = 1 - x^2 \implies dt = -2x \, dx\):
\(= x \sin^{-1} x + \sqrt{1 - x^2} + C\)
In simple words: Multiply by 1 and use integration by parts, followed by a simple substitution.

Exam Tip: Inverse trigonometric integrals are always integrated by parts with 1 as the second function.

Question 2. \(\displaystyle \int x^2 \sin^{-1} x \, dx\)
Answer: Let \(u = \sin^{-1} x\) and \(v = x^2\):
\(= \sin^{-1} x \left( \frac{x^3}{3} \right) - \int \frac{1}{\sqrt{1 - x^2}} \cdot \frac{x^3}{3} \, dx\)
\(= \frac{x^3}{3} \sin^{-1} x + \frac{1}{3} \int \frac{x^3}{\sqrt{1 - x^2}} \, dx\)
For the remaining integral, let \(t = 1 - x^2 \implies x^2 = 1 - t\) and \(x \, dx = -\frac{1}{2} \, dt\):
\(= \frac{x^3}{3} \sin^{-1} x + \frac{1}{3} \int \frac{(1 - t)(-\frac{1}{2} dt)}{\sqrt{t}}\)
After evaluating, we get:
\(= \frac{x^3}{3} \sin^{-1} x + \frac{1}{9}(2 + x^2)\sqrt{1 - x^2} + C\)
In simple words: Apply integration by parts, then use substitution to evaluate the remaining algebraic radical integral.

Exam Tip: Algebraic powers multiplied by inverse trigonometric functions are tackled effectively by taking the inverse function as \(u\).

Question 3. \(\displaystyle \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx\)
Answer: Let \(t = \sin^{-1} x \implies x = \sin t\) and \(dx = \cos t \, dt\):
\(= \int \frac{\sin t \cdot t}{\cos t} \cos t \, dt = \int t \sin t \, dt\)
Using integration by parts on \(t \sin t\):
\(= t(-\cos t) - \int 1 \cdot (-\cos t) \, dt\)
\(= -t \cos t + \sin t + C\)
Substitute back \(t = \sin^{-1} x\) and \(\cos t = \sqrt{1 - x^2}\):
\(= -\sin^{-1} x \cdot \sqrt{1 - x^2} + x + C\)
In simple words: Substitute \(t = \sin^{-1} x\) first to convert the integral into a simple algebraic-trigonometric product.

Exam Tip: Substitution can simplify complex integration by parts problems significantly.

Question 4. \(\displaystyle \int \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) dx\)
Answer: Substitute \(x = \tan \theta \implies dx = \sec^2 \theta \, d\theta\):
The inverse trigonometric expression simplifies: \(\cos^{-1}\left(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta\)
\(= \int 2\theta \sec^2 \theta \, d\theta = 2 \int \theta \sec^2 \theta \, d\theta\)
Using integration by parts:
\(= 2 \left( \theta \tan \theta - \int 1 \cdot \tan \theta \, d\theta \right)\)
\(= 2(\theta \tan \theta - \log |\sec \theta|) + C\)
Substitute back \(\theta = \tan^{-1} x\) and \(\sec \theta = \sqrt{1 + x^2}\):
\(= 2x \tan^{-1} x - \log(1 + x^2) + C\)
In simple words: Simplify the inverse trigonometric function using standard substitutions before integrating by parts.

Exam Tip: Always simplify inverse trigonometric functions using substitutions like \(x = \tan \theta\) before attempting integration.

Question 5. \(\displaystyle \int \sec^3 x \, dx\)
Answer: Write as \(\int \sec x \cdot \sec^2 x \, dx\) and apply integration by parts:
Let \(u = \sec x\) and \(v = \sec^2 x\)
\(= \sec x \tan x - \int (\sec x \tan x) \tan x \, dx\)
\(= \sec x \tan x - \int \sec x \tan^2 x \, dx\)
\(= \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx\)
\(= \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx\)
Let \(I = \int \sec^3 x \, dx\), then:
\(I = \sec x \tan x - I + \log |\sec x + \tan x|\)
\(2I = \sec x \tan x + \log |\sec x + \tan x|\)
\(I = \frac{1}{2} [\sec x \tan x + \log |\sec x + \tan x|] + C\)
In simple words: Use integration by parts and bring the recurring integral term back to the left-hand side to solve for it.

Exam Tip: This is a classic cyclic integration by parts problem where the original integral reappears on the right side.

LEVEL III

Question 1. \(\displaystyle \int \cos(\log x) \, dx\)
Answer: Let \(x = e^t \implies dx = e^t \, dt\):
\(= \int \cos t \cdot e^t \, dt = \int e^t \cos t \, dt\)
Using the standard exponential-trigonometric integration formula \(\int e^a t \cos(bt) \, dt = \frac{e^{at}}{a^2 + b^2}[a \cos(bt) + b \sin(bt)]\):
\(= \frac{e^t}{2}(\cos t + \sin t) + C\)
Substitute back \(t = \log x \implies e^t = x\):
\(= \frac{x}{2} [\cos(\log x) + \sin(\log x)] + C\)
In simple words: Substitute \(x = e^t\) to convert the logarithmic argument into a standard exponential-trigonometric integral.

Exam Tip: Substituting \(x = e^t\) effectively eliminates logarithms from trigonometric arguments.

Question 2. \(\displaystyle \int \frac{e^x(1 + x)}{(2 + x)^2} \, dx\)
Answer: Adjust the numerator to match the denominator term \(2 + x\):
\(\frac{1 + x}{(2 + x)^2} = \frac{(2 + x) - 1}{(2 + x)^2} = \frac{1}{2 + x} - \frac{1}{(2 + x)^2}\)
\(= \int e^x \left[ \frac{1}{2 + x} - \frac{1}{(2 + x)^2} \right] dx\)
Let \(f(x) = \frac{1}{2 + x}\), then \(f'(x) = -\frac{1}{(2 + x)^2}\)
Using \(\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C\):
\(= \frac{e^x}{2 + x} + C\)
In simple words: Express the rational fraction as the sum of a function and its derivative, then apply the standard exponential formula.

Exam Tip: Adjusting numerators to match denominators in exponential fractions helps reveal the hidden \(e^x [f(x) + f'(x)]\) form.

Question 3. \(\displaystyle \int \frac{\log x}{(1 + \log x)^2} \, dx\)
Answer: Let \(x = e^t \implies dx = e^t \, dt\):
\(= \int \frac{t}{(1 + t)^2} e^t \, dt = \int e^t \left[ \frac{(1 + t) - 1}{(1 + t)^2} \right] dt = \int e^t \left[ \frac{1}{1 + t} - \frac{1}{(1 + t)^2} \right] dt\)
Using \(\int e^t [f(t) + f'(t)] \, dt = e^t f(t) + C\):
\(= \frac{e^t}{1 + t} + C\)
Substitute back \(t = \log x \implies e^t = x\):
\(= \frac{x}{1 + \log x} + C\)
In simple words: Substitute \(x = e^t\) to convert logarithms into algebraic variables, then apply the exponential derivative formula.

Exam Tip: Substituting \(x = e^t\) simplifies logarithmic integrals into standard exponential forms.

Question 4. \(\displaystyle \int e^x \left( \frac{2 + \sin x}{1 + \cos 2x} \right) dx\)
Answer: Use the identity \(1 + \cos 2x = 2\cos^2 x\):
\(= \int e^x \left( \frac{2 + \sin x}{2\cos^2 x} \right) dx = \int e^x \left( \frac{1}{\cos^2 x} + \frac{\sin x}{2\cos^2 x} \right) dx\)
\(= \int e^x \left[ \frac{1}{2} \tan x + \frac{1}{2} \sec^2 x \right] dx\)
Let \(f(x) = \frac{1}{2} \tan x\), then \(f'(x) = \frac{1}{2} \sec^2 x\)
Using \(\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C\):
\(= \frac{1}{2} e^x \tan x + C\)
In simple words: Expand denominator using double-angle identities, split into terms, and identify the function-derivative pair for exponential integrals.

Exam Tip: Trigonometric denominator identities like \(1 + \cos 2x = 2\cos^2 x\) are crucial for converting terms into secants and tangents.

Question 5. \(\displaystyle \int e^{2x} \cos 3x \, dx\)
Answer: Using the standard exponential-trigonometric integration formula \(\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2}[a \cos(bx) + b \sin(bx)] + C\):
Here \(a = 2\) and \(b = 3\):
\(= \frac{e^{2x}}{2^2 + 3^2} [2 \cos 3x + 3 \sin 3x] + C\)
\(= \frac{e^{2x}}{13} [2 \cos 3x + 3 \sin 3x] + C\)
In simple words: Apply the direct formula for exponential products with trigonometric functions.

Exam Tip: Memorize the standard result for exponential-trigonometric integrals to save time during exams.

(vi) Some Special Integrals

LEVEL I

Question 1. \(\displaystyle \int \sqrt{4 + x^2} \, dx\)
Answer: Using the standard formula \(\int \sqrt{a^2 + x^2} \, dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log|x + \sqrt{a^2 + x^2}| + C\):
\(= \frac{x}{2}\sqrt{4 + x^2} + \frac{4}{2}\log|x + \sqrt{4 + x^2}| + C\)
\(= \frac{x}{2}\sqrt{4 + x^2} + 2\log|x + \sqrt{4 + x^2}| + C\)
In simple words: Apply the standard formula for square roots of the sum of squares directly.

Exam Tip: Memorize standard square root formulas to write down answers instantly.

Question 2. \(\displaystyle \int \sqrt{1 - 4x^2} \, dx\)
Answer: Rewrite as \(\int \sqrt{1^2 - (2x)^2} \, dx\)
Using the standard formula \(\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C\) and dividing by the coefficient of \(x\) (which is 2):
\(= \frac{1}{2} \left[ \frac{2x}{2}\sqrt{1 - 4x^2} + \frac{1}{2}\sin^{-1}(2x) \right] + C\)
\(= \frac{x}{2}\sqrt{1 - 4x^2} + \frac{1}{4}\sin^{-1}(2x) + C\)
In simple words: Apply the standard inverse sine square root formula and divide by the coefficient of \(x\).

Exam Tip: Always divide by the coefficient of \(x\) when applying standard integration formulas to linear inner functions.

LEVEL II

Question 1. \(\displaystyle \int \sqrt{x^2 + 4x + 6} \, dx\)
Answer: Complete the square inside the radical:
\(x^2 + 4x + 6 = (x + 2)^2 + 2 = (x + 2)^2 + (\sqrt{2})^2\)
\(= \int \sqrt{(x + 2)^2 + (\sqrt{2})^2} \, dx\)
Using the standard formula for \(\sqrt{x^2 + a^2}\):
\(= \frac{x + 2}{2}\sqrt{x^2 + 4x + 6} + \frac{2}{2}\log|x + 2 + \sqrt{x^2 + 4x + 6}| + C\)
\(= \frac{x + 2}{2}\sqrt{x^2 + 4x + 6} + \log|x + 2 + \sqrt{x^2 + 4x + 6}| + C\)
In simple words: Complete the square inside the square root and apply the standard sum-of-squares formula.

Exam Tip: Completing the square transforms quadratic square root integrals into standard forms.

Question 2. \(\displaystyle \int \sqrt{1 - 4x - x^2} \, dx\)
Answer: Complete the square inside the radical:
\(1 - 4x - x^2 = 1 - (x^2 + 4x + 4 - 4) = 5 - (x + 2)^2 = (\sqrt{5})^2 - (x + 2)^2\)
\(= \int \sqrt{(\sqrt{5})^2 - (x + 2)^2} \, dx\)
Using the standard formula for \(\sqrt{a^2 - x^2}\):
\(= \frac{x + 2}{2}\sqrt{1 - 4x - x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x + 2}{\sqrt{5}}\right) + C\)
In simple words: Complete the square inside the radical and apply the standard inverse sine square root formula.

Exam Tip: Be careful with signs when completing the square for negative quadratic coefficients inside square roots.

LEVEL III

Question 1. \(\displaystyle \int (x + 1)\sqrt{1 - x - x^2} \, dx\)
Answer: Derivative of the radicand: \(\frac{d}{dx}(1 - x - x^2) = -1 - 2x\)
Express the numerator factor \(x + 1\) as \(A(-1 - 2x) + B\):
\(x + 1 = -2Ax + (-A + B)\)
Equating coefficients: \(-2A = 1 \implies A = -\frac{1}{2}\), and \(-A + B = 1 \implies \frac{1}{2} + B = 1 \implies B = \frac{1}{2}\)
\(= \int \left[ -\frac{1}{2}(-1 - 2x) + \frac{1}{2} \right] \sqrt{1 - x - x^2} \, dx\)
\(= -\frac{1}{2} \int (-1 - 2x)\sqrt{1 - x - x^2} \, dx + \frac{1}{2} \int \sqrt{(\sqrt{5}/2)^2 - (x + 1/2)^2} \, dx\)
Evaluating the first part using substitution: \(-\frac{1}{3}(1 - x - x^2)^{3/2}\)
Evaluating the second part using standard square root formulas:
\(= -\frac{1}{3}(1 - x - x^2)^{3/2} + \frac{1}{2}\left[ \frac{x + 1/2}{2}\sqrt{1 - x - x^2} + \frac{5}{8}\sin^{-1}\left(\frac{2x + 1}{\sqrt{5}}\right) \right] + C\)
In simple words: Express the linear factor as a derivative of the radicand plus a constant, then integrate each part.

Exam Tip: Integrals of the form \(\int (px + q)\sqrt{ax^2 + bx + c} \, dx\) are solved by splitting using the derivative of the quadratic expression.

Question 2. \(\displaystyle \int (x - 5)\sqrt{x^2 + x} \, dx\)
Answer: Derivative of the radicand: \(\frac{d}{dx}(x^2 + x) = 2x + 1\)
Express \(x - 5\) as \(A(2x + 1) + B\):
\(x - 5 = 2Ax + (A + B)\)
Equating coefficients: \(2A = 1 \implies A = \frac{1}{2}\), and \(A + B = -5 \implies \frac{1}{2} + B = -5 \implies B = -\frac{11}{2}\)
\(= \int \left[ \frac{1}{2}(2x + 1) - \frac{11}{2} \right] \sqrt{x^2 + x} \, dx\)
\(= \frac{1}{2} \int (2x + 1)\sqrt{x^2 + x} \, dx - \frac{11}{2} \int \sqrt{(x + 1/2)^2 - (1/2)^2} \, dx\)
\(= \frac{1}{3}(x^2 + x)^{3/2} - \frac{11}{2} \left[ \frac{x + 1/2}{2}\sqrt{x^2 + x} - \frac{1}{8}\log\left|x + \frac{1}{2} + \sqrt{x^2 + x}\right| \right] + C\)
In simple words: Split the linear multiplier using the derivative of the quadratic expression inside the square root.

Exam Tip: Splitting linear terms using derivatives simplifies otherwise difficult radical integrals.

(vii) Miscellaneous Questions

LEVEL II

Question 1. \(\displaystyle \int \frac{1}{2 - 3\cos 2x} \, dx\)
Answer: Use half-angle formulas for cosine: \(\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}\)
\(= \int \frac{dx}{2 - 3\left(\frac{1 - \tan^2 x}{1 + \tan^2 x}\right)} = \int \frac{(1 + \tan^2 x) \, dx}{2(1 + \tan^2 x) - 3(1 - \tan^2 x)}\)
\(= \int \frac{\sec^2 x \, dx}{2 + 2\tan^2 x - 3 + 3\tan^2 x} = \int \frac{\sec^2 x \, dx}{5\tan^2 x - 1}\)
Let \(t = \tan x \implies dt = \sec^2 x \, dx\):
\(= \int \frac{dt}{5t^2 - 1} = \frac{1}{5} \int \frac{dt}{t^2 - (1/\sqrt{5})^2}\)
Using the standard formula \(\frac{1}{2a}\log\left|\frac{t - a}{t + a}\right|\):
\(= \frac{1}{5} \cdot \frac{\sqrt{5}}{2} \log\left|\frac{\tan x - 1/\sqrt{5}}{\tan x + 1/\sqrt{5}}\right| + C = \frac{1}{2\sqrt{5}} \log\left|\frac{\sqrt{5}\tan x - 1}{\sqrt{5}\tan x + 1}\right| + C\)
In simple words: Convert trigonometric terms using tangent half-angle substitutions, then integrate using standard logarithmic formulas.

Exam Tip: Integrals with constants and cosines in denominators are standardly solved using \(\tan x\) substitutions.

Question 2. \(\displaystyle \int \frac{1}{3 + \sin 2x} \, dx\)
Answer: Use half-angle substitution: \(\sin 2x = \frac{2\tan x}{1 + \tan^2 x}\) and \(dx = \frac{dt}{1 + t^2}\) where \(t = \tan x\):
\(= \int \frac{\sec^2 x \, dx}{3(1 + \tan^2 x) + 2\tan x} = \int \frac{dt}{3t^2 + 2t + 3}\)
Complete the square in the denominator:
\(3t^2 + 2t + 3 = 3\left( t^2 + \frac{2}{3}t + 1 \right) = 3\left[ \left(t + \frac{1}{3}\right)^2 + \frac{8}{9} \right] = 3\left[ \left(t + \frac{1}{3}\right)^2 + \left(\frac{2\sqrt{2}}{3}\right)^2 \right]\)
\(= \frac{1}{3} \int \frac{dt}{(t + 1/3)^2 + (2\sqrt{2}/3)^2}\)
\(= \frac{1}{3} \cdot \frac{3}{2\sqrt{2}} \tan^{-1}\left(\frac{t + 1/3}{2\sqrt{2}/3}\right) + C = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{3\tan x + 1}{2\sqrt{2}}\right) + C\)
In simple words: Substitute \(t = \tan x\), complete the square in the quadratic denominator, and apply the inverse tangent formula.

Exam Tip: Sine and cosine linear combinations in denominators are always solved effectively using tangent substitutions.

Question 3. \(\displaystyle \int \frac{dx}{4\sin^2 x + 5\cos^2 x}\)
Answer: Divide numerator and denominator by \(\cos^2 x\):
\(= \int \frac{\sec^2 x \, dx}{4\tan^2 x + 5}\)
Let \(t = \tan x \implies dt = \sec^2 x \, dx\):
\(= \int \frac{dt}{4t^2 + 5} = \frac{1}{4} \int \frac{dt}{t^2 + (\sqrt{5}/2)^2}\)
Using the standard formula \(\frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)\):
\(= \frac{1}{4} \cdot \frac{2}{\sqrt{5}} \tan^{-1}\left(\frac{\tan x}{\sqrt{5}/2}\right) + C = \frac{1}{2\sqrt{5}} \tan^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right) + C\)
In simple words: Divide top and bottom by \(\cos^2 x\) to form terms of \(\tan x\) and \(\sec^2 x\), then substitute.

Exam Tip: When sines and cosines appear as even powers in denominators, divide by \(\cos^2 x\) to introduce \(\tan x\).

Question 4. \(\displaystyle \int \frac{dx}{1 + 3\sin^2 x + 8\cos^2 x}\)
Answer: Divide numerator and denominator by \(\cos^2 x\):
\(= \int \frac{\sec^2 x \, dx}{\sec^2 x + 3\tan^2 x + 8}\)
Substitute \(\sec^2 x = 1 + \tan^2 x\):
\(= \int \frac{\sec^2 x \, dx}{1 + \tan^2 x + 3\tan^2 x + 8} = \int \frac{\sec^2 x \, dx}{4\tan^2 x + 9}\)
Let \(t = \tan x \implies dt = \sec^2 x \, dx\):
\(= \int \frac{dt}{4t^2 + 9} = \frac{1}{4} \int \frac{dt}{t^2 + (3/2)^2}\)
\(= \frac{1}{4} \cdot \frac{2}{3} \tan^{-1}\left(\frac{2\tan x}{3}\right) + C = \frac{1}{6} \tan^{-1}\left(\frac{2\tan x}{3}\right) + C\)
In simple words: Divide by \(\cos^2 x\), convert remaining secants to tangents, and apply standard inverse tangent integration.

Exam Tip: Converting all trigonometric terms into tangents using \(\sec^2 x = 1 + \tan^2 x\) simplifies homogeneous quadratic trigonometric integrals.

Question 5. \(\displaystyle \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} \, dx\)
Answer: Write \(\sin 2x = 2\sin x \cos x\) and express denominator in terms of \(\tan^2 x\):
Divide numerator and denominator by \(\cos^4 x\):
\(= \int \frac{\frac{2\sin x \cos x}{\cos^4 x}}{\frac{\sin^4 x + \cos^4 x}{\cos^4 x}} \, dx = \int \frac{2\tan x \sec^2 x}{\tan^4 x + 1} \, dx\)
Let \(t = \tan^2 x \implies dt = 2\tan x \sec^2 x \, dx\):
\(= \int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + C\)
\(= \tan^{-1}(\tan^2 x) + C\)
In simple words: Divide numerator and denominator by \(\cos^4 x\) and substitute \(t = \tan^2 x\).

Exam Tip: Recognizing \(2\tan x \sec^2 x\) as the derivative of \(\tan^2 x\) simplifies fourth-power trigonometric expressions.

Question 6. \(\displaystyle \int \frac{\sec x}{\sec x + 4\tan x} \, dx\)
Answer: Convert secant and tangent into sine and cosine:
\(= \int \frac{\frac{1}{\cos x}}{\frac{1}{\cos x} + \frac{4\sin x}{\cos x}} \, dx = \int \frac{dx}{1 + 4\sin x}\)
Using half-angle substitution \(t = \tan(x/2) \implies dx = \frac{2 \, dt}{1 + t^2}\) and \(\sin x = \frac{2t}{1 + t^2}\):
\(= \int \frac{\frac{2 \, dt}{1 + t^2}}{1 + 4\left(\frac{2t}{1 + t^2}\right)} = \int \frac{2 \, dt}{1 + t^2 + 8t} = 2 \int \frac{dt}{t^2 + 8t + 16 - 15} \dots\) (Wait, complete square):
\(t^2 + 8t + 1 = (t + 4)^2 - 15 = (t + 4)^2 - (\sqrt{15})^2\)
\(= 2 \int \frac{dt}{(t + 4)^2 - (\sqrt{15})^2}\)
Using standard logarithmic formula: \(= \frac{2}{2\sqrt{15}} \log\left|\frac{t + 4 - \sqrt{15}}{t + 4 + \sqrt{15}}\right| + C\)
\(= \frac{1}{\sqrt{15}} \log\left|\frac{\tan(x/2) + 4 - \sqrt{15}}{\tan(x/2) + 4 + \sqrt{15}}\right| + C\)
In simple words: Convert to sine and cosine, apply half-angle substitution, and integrate using standard logarithmic formulas.

Exam Tip: Half-angle substitutions are universal tools for linear combinations of sines and cosines in denominators.

LEVEL III

Question 1. \(\displaystyle \int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx\)
Answer: Express numerator as \(A(\text{denominator}) + B(\text{derivative of denominator})\):
Denominator \(= 3\cos x + 2\sin x\)
Derivative of denominator \(= -3\sin x + 2\cos x\)
Let \(3\sin x + 2\cos x = A(3\cos x + 2\sin x) + B(-3\sin x + 2\cos x)\)
Equating coefficients of \(\sin x\) and \(\cos x\):
For \(\sin x\): \(3 = 2A - 3B\)
For \(\cos x\): \(2 = 3A + 2B\)
Solving these simultaneous equations gives \(A = \frac{12}{13}\) and \(B = -\frac{5}{13}\)
\(= \int \left[ \frac{12}{13} + \frac{-\frac{5}{13}(-3\sin x + 2\cos x)}{3\cos x + 2\sin x} \right] dx\)
\(= \frac{12}{13} x - \frac{5}{13} \log |3\cos x + 2\sin x| + C\)
In simple words: Express the numerator as a linear combination of the denominator and its derivative.

Exam Tip: Integrals of the form \(\int \frac{a\sin x + b\cos x}{c\cos x + d\sin x} \, dx\) are always solved by splitting the numerator.

Question 2. \(\displaystyle \int \frac{1}{1 - \tan x} \, dx\)
Answer: Convert into sine and cosine:
\(= \int \frac{1}{1 - \frac{\sin x}{\cos x}} \, dx = \int \frac{\cos x}{\cos x - \sin x} \, dx\)
Rewrite numerator using the derivative trick: Write \(\cos x = \frac{1}{2}[(\cos x - \sin x) + (\cos x + \sin x)]\)
\(= \frac{1}{2} \int \frac{(\cos x - \sin x) + (\cos x + \sin x)}{\cos x - \sin x} \, dx\)
\(= \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\cos x + \sin x}{\cos x - \sin x} \, dx\)
\(= \frac{1}{2} x - \frac{1}{2} \log |\cos x - \sin x| + C\)
In simple words: Convert to sine and cosine, then split the numerator into the denominator and its derivative.

Exam Tip: This standard numerator-splitting trick solves integrals involving \(\frac{1}{1 \pm \tan x}\) immediately.

Question 3. \(\displaystyle \int \frac{x^4}{x^4 - 1} \, dx\)
Answer: Rewrite numerator by adding and subtracting 1:
\(= \int \frac{(x^4 - 1) + 1}{x^4 - 1} \, dx = \int \left( 1 + \frac{1}{x^4 - 1} \right) dx\)
Expand \(x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)\)
Decompose \(\frac{1}{x^4 - 1}\) into partial fractions:
\(\frac{1}{x^4 - 1} = \frac{1/4}{x - 1} - \frac{1/4}{x + 1} - \frac{1/2}{x^2 + 1}\)
\(= \int 1 \, dx + \frac{1}{4}\int \frac{dx}{x - 1} - \frac{1}{4}\int \frac{dx}{x + 1} - \frac{1}{2}\int \frac{dx}{x^2 + 1}\)
\(= x + \frac{1}{4} \log |x - 1| - \frac{1}{4} \log |x + 1| - \frac{1}{2} \tan^{-1} x + C\)
\(= x + \frac{1}{4} \log\left|\frac{x - 1}{x + 1}\right| - \frac{1}{2} \tan^{-1} x + C\)
In simple words: Perform polynomial division by adding and subtracting 1, then apply partial fractions.

Exam Tip: Adding and subtracting terms in the numerator avoids long division when degrees are equal.

Question 4. \(\displaystyle \int \frac{x^2 + 1}{x^4 + x^2 + 1} \, dx\)
Answer: Divide numerator and denominator by \(x^2\):
\(= \int \frac{1 + \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} \, dx = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 3} \, dx\)
Let \(t = x - \frac{1}{x} \implies dt = \left( 1 + \frac{1}{x^2} \right) dx\)
\(= \int \frac{dt}{t^2 + (\sqrt{3})^2}\)
Using standard inverse tangent formula:
\(= \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{3}}\right) + C = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}x}\right) + C\)
In simple words: Divide top and bottom by \(x^2\) and substitute \(t = x - \frac{1}{x}\).

Exam Tip: This is a standard substitution technique for symmetric reciprocal polynomials of degree 4.

Question 5. \(\displaystyle \int \frac{x^2 - 1}{x^4 + 1} \, dx\)
Answer: Divide numerator and denominator by \(x^2\):
\(= \int \frac{1 - \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx = \int \frac{1 - \frac{1}{x^2}}{(x + \frac{1}{x})^2 - 2} \, dx\)
Let \(t = x + \frac{1}{x} \implies dt = \left( 1 - \frac{1}{x^2} \right) dx\)
\(= \int \frac{dt}{t^2 - (\sqrt{2})^2}\)
Using standard logarithmic formula: \(\frac{1}{2a}\log\left|\frac{t - a}{t + a}\right|\):
\(= \frac{1}{2\sqrt{2}} \log\left|\frac{t - \sqrt{2}}{t + \sqrt{2}}\right| + C = \frac{1}{2\sqrt{2}} \log\left|\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}\right| + C = \frac{1}{2\sqrt{2}} \log\left|\frac{x^2 - \sqrt{2}x + 1}{x^2 + \sqrt{2}x + 1}\right| + C\)
In simple words: Divide by \(x^2\) and substitute \(t = x + \frac{1}{x}\).

Exam Tip: Dividing by \(x^2\) and completing squares in the denominator is standard for reciprocal quartic integrals.

Question 6. \(\displaystyle \int \sqrt{\tan x} \, dx\)
Answer: Let \(t^2 = \tan x \implies 2t \, dt = \sec^2 x \, dx = (1 + \tan^2 x) \, dx = (1 + t^4) \, dx \implies dx = \frac{2t}{1 + t^4} \, dt\)
\(= \int t \cdot \frac{2t}{1 + t^4} \, dt = 2 \int \frac{t^2}{1 + t^4} \, dt\)
Divide numerator and denominator by \(t^2\):
\(= 2 \int \frac{1}{t^2 + \frac{1}{t^2}} \, dt = \int \left( \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} + \frac{1 - \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \right) dt = \int \frac{1 + \frac{1}{t^2}}{(t - \frac{1}{t})^2 + 2} \, dt + \int \frac{1 - \frac{1}{t^2}}{(t + \frac{1}{t})^2 - 2} \, dt\)
Integrating both parts using standard inverse tangent and logarithmic formulas:
\(= \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t - 1/t}{\sqrt{2}}\right) + \frac{1}{2\sqrt{2}} \log\left|\frac{t + 1/t - \sqrt{2}}{t + 1/t + \sqrt{2}}\right| + C\)
where \(t = \sqrt{\tan x}\).
In simple words: Substitute \(t = \sqrt{\tan x}\), split into two rational integrals by algebraic manipulation, and integrate.

Exam Tip: Square roots of tangent or cotangent require substituting \(t^2 = \tan x\) followed by splitting into reciprocal integrals.

Definite Integrals

(i) Definite Integrals based upon types of indefinite integrals

LEVEL I

Question 1. \(\displaystyle \int_{0}^{1} \frac{2x + 3}{5x^2 + 1} \, dx\)
Answer: Split the integral into two parts:
\(= \int_{0}^{1} \frac{2x}{5x^2 + 1} \, dx + \int_{0}^{1} \frac{3}{5x^2 + 1} \, dx\)
For the first part, let \(t = 5x^2 + 1 \implies dt = 10x \, dx\):
\(= \frac{1}{5} \left[ \log|5x^2 + 1| \right]_{0}^{1} + \frac{3}{5} \left[ \frac{1}{1/\sqrt{5}} \tan^{-1}(\sqrt{5}x) \right]_{0}^{1}\)
\(= \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})\)
In simple words: Split the fraction into logarithmic and inverse tangent parts, then evaluate at upper and lower limits.

Exam Tip: Always split linear-over-quadratic definite integrals into derivative and constant terms.

Question 2. \(\displaystyle \int_{0}^{\pi/2} \sqrt{\sin x} \cos^5 x \, dx\)
Answer: Rewrite as \(\int_{0}^{\pi/2} \sqrt{\sin x} \cos^4 x (\cos x \, dx) = \int_{0}^{\pi/2} \sqrt{\sin x}(1 - \sin^2 x)^2 \cos x \, dx\)
Let \(t = \sin x \implies dt = \cos x \, dx\). When \(x = 0, t = 0\); when \(x = \pi/2, t = 1\):
\(= \int_{0}^{1} t^{1/2}(1 - 2t^2 + t^4) \, dt = \int_{0}^{1} (t^{1/2} - 2t^{5/2} + t^{9/2}) \, dt\)
\(= \left[ \frac{2}{3}t^{3/2} - \frac{4}{7}t^{7/2} + \frac{2}{11}t^{11/2} \right]_{0}^{1}\)
\(= \frac{2}{3} - \frac{4}{7} + \frac{2}{11} = \frac{154 - 264 + 84}{231} = \frac{16}{231}\)
In simple words: Substitute \(t = \sin x\) after saving one cosine factor, change limits, and integrate polynomials.

Exam Tip: Changing limits correctly during substitution saves you from substituting back to \(x\).

Question 3. \(\displaystyle \int_{0}^{2} x\sqrt{x + 2} \, dx\)
Answer: Let \(t = x + 2 \implies x = t - 2\) and \(dx = dt\). Limits: when \(x = 0, t = 2\); when \(x = 2, t = 4\):
\(= \int_{2}^{4} (t - 2)t^{1/2} \, dt = \int_{2}^{4} (t^{3/2} - 2t^{1/2}) \, dt\)
\(= \left[ \frac{2}{5}t^{5/2} - \frac{4}{3}t^{3/2} \right]_{2}^{4}\)
\(= \left( \frac{2}{5}(32) - \frac{4}{3}(8) \right) - \left( \frac{2}{5}(4\sqrt{2}) - \frac{4}{3}(2\sqrt{2}) \right)\)
\(= \frac{64}{5} - \frac{32}{3} - \frac{8\sqrt{2}}{5} + \frac{8\sqrt{2}}{3} = \frac{32}{15} + \frac{8\sqrt{2}}{15}\)
In simple words: Substitute \(t = x + 2\) to simplify the radical expression, update limits, and integrate.

Exam Tip: Substitution inside radicals simplifies algebraic integrations significantly.

LEVEL II

Question 1. \(\displaystyle \int_{1}^{2} \frac{5x^2}{x^2 + 4x + 3} \, dx\)
Answer: Degree of numerator equals denominator degree. Perform polynomial division:
\(\frac{5x^2}{x^2 + 4x + 3} = 5 - \frac{20x + 15}{x^2 + 4x + 3} = 5 - \frac{20x + 15}{(x + 1)(x + 3)}\)
Using partial fractions: \(\frac{20x + 15}{(x + 1)(x + 3)} = \frac{-5/2}{x + 1} + \frac{45/2}{x + 3}\)
\(= \int_{1}^{2} \left( 5 + \frac{5/2}{x + 1} - \frac{45/2}{x + 3} \right) dx\)
\(= \left[ 5x + \frac{5}{2}\log|x + 1| - \frac{45}{2}\log|x + 3| \right]_{1}^{2}\)
Evaluating at limits \(2\) and \(1\):
\(= \left( 10 + \frac{5}{2}\log 3 - \frac{45}{2}\log 5 \right) - \left( 5 + \frac{5}{2}\log 2 - \frac{45}{2}\log 4 \right)\)
\(= 5 + \frac{5}{2}\log 3 - \frac{5}{2}\log 2 - \frac{45}{2}\log 5 + 45\log 2\)
In simple words: Divide first, apply partial fractions, and evaluate definite integrals using logarithmic properties.

Exam Tip: Polynomial division is mandatory when improper fractions appear in definite integrals.

Question 2. \(\displaystyle \int_{1/2}^{2} \left( \frac{1}{x} - \frac{1}{2x^2} \right) e^{2x} \, dx\)
Answer: Let \(f(x) = \frac{1}{2x}\), then \(f'(x) = -\frac{1}{2x^2}\)
Wait, with exponential multiplier \(e^{2x}\), let us expand carefully: let \(t = 2x \implies dt = 2 \, dx\):
\(= \int_{1}^{4} \left( \frac{2}{t} - \frac{2}{t^2} \right) e^t \cdot \frac{1}{2} \, dt = \int_{1}^{4} e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) dt\)
Using \(\int e^t [f(t) + f'(t)] \, dt = e^t f(t)\):
\(= \left[ e^t \cdot \frac{1}{t} \right]_{1}^{4} = \frac{e^4}{4} - \frac{e^1}{1} = \frac{e^4}{4} - e\)
In simple words: Substitute \(t = 2x\) to transform the integral into standard exponential derivative form.

Exam Tip: Substituting the linear exponent argument simplifies exponential-rational definite integrals.

(ii) Definite integrals as a limit of sum

LEVEL I

Question 1. Evaluate \(\displaystyle \int_{0}^{2} (x + 2) \, dx\) as the limit of a sum.
Answer: Using definition \(\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)\) where \(h = \frac{b - a}{n}\):
Here \(a = 0, b = 2, f(x) = x + 2\), and \(h = \frac{2}{n}\):
\(f(rh) = rh + 2\)
\(\sum_{r=0}^{n-1} f(rh) = \sum_{r=0}^{n-1} (rh + 2) = h \frac{(n-1)n}{2} + 2n = \frac{2}{n} \frac{(n-1)n}{2} + 2n = (n - 1) + 2n = 3n - 1\)
Now take limit as \(n \to \infty\) of \(h \sum f(rh) = \frac{2}{n}(3n - 1) = 6 - \frac{2}{n}\)
\(\lim_{n \to \infty} \left( 6 - \frac{2}{n} \right) = 6\)
In simple words: Apply the limit of a sum formula by partitioning the interval into \(n\) subintervals of width \(h\).

Exam Tip: Memorize standard summation formulas like \(\sum r = \frac{n(n-1)}{2}\) for limit-of-sum problems.

Question 2. Evaluate \(\displaystyle \int_{0}^{4} (1 + x) \, dx\) definite integral as the limit of a sum.
Answer: Here \(a = 0, b = 4, f(x) = 1 + x\), and \(h = \frac{4}{n}\):
\(f(rh) = 1 + rh\)
\(\sum_{r=0}^{n-1} f(rh) = \sum_{r=0}^{n-1} (1 + rh) = n + h \frac{(n-1)n}{2} = n + \frac{4}{n} \frac{n(n-1)}{2} = n + 2(n - 1) = 3n - 2\)
Multiply by \(h = \frac{4}{n}\):
\(\frac{4}{n}(3n - 2) = 12 - \frac{8}{n}\)
Taking limit as \(n \to \infty\): \(\lim_{n \to \infty} \left( 12 - \frac{8}{n} \right) = 12\)
In simple words: Substitute limits into the limit-of-sum definition and evaluate using arithmetic progression sums.

Exam Tip: Double-check your algebraic expansion of summation terms before taking the limit.

LEVEL II

Question 1. Evaluate \(\displaystyle \int_{1}^{2} (3x^2 - 1) \, dx\) as the limit of a sum.
Answer: Here \(a = 1, b = 2, f(x) = 3x^2 - 1\), and \(h = \frac{1}{n}\):
\(f(1 + rh) = 3(1 + rh)^2 - 1 = 3(1 + 2rh + r^2h^2) - 1 = 2 + 6rh + 3r^2h^2\)
Summing from \(r = 0\) to \(n-1\):
\(\sum_{r=0}^{n-1} f(1 + rh) = 2n + 6h \frac{(n-1)n}{2} + 3h^2 \frac{(n-1)n(2n-1)}{6}\)
Substitute \(h = \frac{1}{n}\):
\(= 2n + 3\left(\frac{1}{n}\right)(n^2 - n) + \frac{1}{2}\left(\frac{1}{n^2}\right)\left(\frac{2n^3 - 3n^2 + n}{3}\right) \dots\) (simplifying terms):
Taking limit as \(n \to \infty\), the result evaluates to \(6\).
In simple words: Use sum of squares formula \(\sum r^2 = \frac{n(n-1)(2n-1)}{6}\) in the limit-of-sum definition.

Exam Tip: Memorize standard sum formulas for first \(n\) natural numbers and their squares.

Question 2. Evaluate \(\displaystyle \int_{0}^{3} (x^2 + 1) \, dx\) as the limit of a sum.
Answer: Here \(a = 0, b = 3, f(x) = x^2 + 1\), and \(h = \frac{3}{n}\):
\(f(rh) = (rh)^2 + 1 = r^2h^2 + 1\)
Summing from \(r = 0\) to \(n-1\):
\(\sum_{r=0}^{n-1} f(rh) = h^2 \frac{(n-1)n(2n-1)}{6} + n\)
Multiply by \(h = \frac{3}{n}\):
\(\frac{3}{n} \left[ \frac{9}{n^2} \frac{(n-1)n(2n-1)}{6} + n \right] = \frac{27}{2n^3}(2n^3 - 3n^2 + n) + 3 = \frac{27}{2} - \frac{81}{2n} + \frac{27}{2n^2} + 3 \dots\) wait, evaluating limit as \(n \to \infty\) gives \(\frac{27}{3} + 3 = 9 + 3 = 12\).
In simple words: Apply sum of squares formula and take the limit as \(n \to \infty\).

Exam Tip: Coefficient matching for highest powers of \(n\) gives the limit instantly.

LEVEL III

Question 1. Evaluate \(\displaystyle \int_{1}^{2} (x^2 + x + 2)dx\) as the limit of a sum.
Answer: Here \(a = 1, b = 2, f(x) = x^2 + x + 2\), and \(h = \frac{1}{n}\):
\(f(1 + rh) = (1 + rh)^2 + (1 + rh) + 2 = 4 + 3rh + r^2h^2\)
Summing and multiplying by \(h = \frac{1}{n}\), then taking limit as \(n \to \infty\):
\(= \lim_{n \to \infty} \left[ 4 + \frac{3}{2}\left(1 - \frac{1}{n}\right) + \frac{1}{3}\left(1 - \frac{1}{n}\right)\left(2 - \frac{1}{n}\right) \right]\)
\(= 4 + \frac{3}{2} + \frac{2}{3} = \frac{24 + 9 + 4}{6} = \frac{37}{6}\)
In simple words: Expand polynomial terms, apply standard sums of numbers and squares, and take the limit.

Exam Tip: Grouping terms containing \(\frac{1}{n}\) that vanish as \(n \to \infty\) simplifies limit evaluations.

Answer: Using limit of a sum or splitting into standard definite integrals:
\(= \left[ \frac{1}{2}e^{2x} + \frac{x^3}{3} \right]_{2}^{4} = \left( \frac{1}{2}e^8 + \frac{64}{3} \right) - \left( \frac{1}{2}e^4 + \frac{8}{3} \right)\)
\(= \frac{1}{2}(e^8 - e^4) + \frac{56}{3}\)
In simple words: Exponential and polynomial sums are most easily evaluated using fundamental theorem of calculus unless limit of sum is strictly demanded.

Exam Tip: Use standard integration for complex exponential limits unless explicitly forced to use first principles.

(iii) Properties of definite Integrals

LEVEL I

Question 1. \(\displaystyle \int_{0}^{\pi/2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx\)
Answer: Let \(I = \int_{0}^{\pi/2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx\)
Using property \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx\):
\(I = \int_{0}^{\pi/2} \frac{\sqrt{\tan(\frac{\pi}{2} - x)}}{1 + \sqrt{\tan(\frac{\pi}{2} - x)}} \, dx = \int_{0}^{\pi/2} \frac{\sqrt{\cot x}}{1 + \sqrt{\cot x}} \, dx = \int_{0}^{\pi/2} \frac{1}{\sqrt{\tan x} + 1} \, dx\)
Adding both expressions for \(I\):
\(2I = \int_{0}^{\pi/2} \frac{\sqrt{\tan x} + 1}{1 + \sqrt{\tan x}} \, dx = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\)
In simple words: Use the property \(f(a + b - x)\), add the original and transformed integrals, and solve for \(I\).

Exam Tip: This is a standard symmetry property problem where adding \(I\) and \(I\) yields a constant integrand.

Question 2. \(\displaystyle \int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} \, dx\)
Answer: Let \(I = \int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} \, dx\)
Using property \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx\) where \(a + b = 4\):
\(I = \int_{1}^{3} \frac{\sqrt{4 - (4 - x)}}{\sqrt{4 - x} + \sqrt{4 - (4 - x)}} \, dx = \int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4 - x} + \sqrt{x}} \, dx\)
Adding both integrals:
\(2I = \int_{1}^{3} \frac{\sqrt{4 - x} + \sqrt{x}}{\sqrt{x} + \sqrt{4 - x}} \, dx = \int_{1}^{3} 1 \, dx = [x]_{1}^{3} = 2\)
\(I = 1\)
In simple words: Apply the property \(f(a + b - x)\), combine both integrals, and integrate a constant.

Exam Tip: Whenever denominators are symmetric under \(a + b - x\), adding the two integral forms solves the problem instantly.

Question 3. \(\displaystyle \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx\)
Answer: Let \(I = \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx\)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\):
\(I = \int_{0}^{\pi/2} \frac{\sin^4(\frac{\pi}{2} - x)}{\sin^4(\frac{\pi}{2} - x) + \cos^4(\frac{\pi}{2} - x)} \, dx = \int_{0}^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \, dx\)
Adding both expressions:
\(2I = \int_{0}^{\pi/2} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \, dx = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\)
In simple words: Replace \(x\) with \(\frac{\pi}{2} - x\), add both forms of \(I\), and integrate.

Exam Tip: Symmetric trigonometric limits like \(0\) to \(\frac{\pi}{2}\) frequently simplify using the reflection property.

LEVEL II

Question 1. \(\displaystyle \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x} \, dx\)
Answer: Let \(I = \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x} \, dx\)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\):
\(I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} \, dx = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\cos x + \sin x} \, dx\)
Adding both equations for \(I\):
\(2I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2}}{\sin x + \cos x} \, dx = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{dx}{\sqrt{2}\sin(x + \frac{\pi}{4})}\)
Evaluating the standard integral: \(\frac{\pi}{2\sqrt{2}} \left[ \log\left|\tan\left(\frac{x}{2} + \frac{\pi}{8}\right)\right| \right]_{0}^{\pi/2} = \frac{\pi}{2\sqrt{2}} \log(\sqrt{2} + 1)\)
\(I = \frac{\pi}{4\sqrt{2}} \log(\sqrt{2} + 1)\)
In simple words: Use the property \(f(a - x)\) to eliminate the \(x\) term in the numerator.

Exam Tip: Multiplying linear \(x\) terms with trigonometric denominators is standardly resolved using the property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\).

Question 2. \(\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx\)
Answer: Let \(I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx\)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\):
\(I = \int_{0}^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx = \int_{0}^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x} \, dx\)
Adding both equations:
\(2I = \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx\)
Let \(t = \cos x \implies dt = -\sin x \, dx\). Limits: when \(x = 0, t = 1\); when \(x = \pi, t = -1\):
\(2I = \pi \int_{-1}^{1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^{1} \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{-1}^{1} = \pi \left( \frac{\pi}{4} - (-\frac{\pi}{4}) \right) = \frac{\pi^2}{2}\)
\(I = \frac{\pi^2}{4}\)
In simple words: Use the reflection property to eliminate \(x\) from the numerator, then substitute \(\cos x\).

Exam Tip: This is a classic board exam problem type. Eliminating \(x\) using the upper limit property is crucial.

Question 3. \(\displaystyle \int_{0}^{\pi} \frac{x \tan x}{\sec x \csc x} \, dx\)
Answer: Simplify integrand: \(\frac{\tan x}{\sec x \csc x} = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x \cos x}} = \sin^2 x\)
\(I = \int_{0}^{\pi} x \sin^2 x \, dx\)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\):
\(I = \int_{0}^{\pi} (\pi - x)\sin^2(\pi - x) \, dx = \int_{0}^{\pi} (\pi - x)\sin^2 x \, dx\)
Adding both equations:
\(2I = \int_{0}^{\pi} \pi \sin^2 x \, dx = \pi \int_{0}^{\pi} \frac{1 - \cos 2x}{2} \, dx\)
\(2I = \pi \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\pi} = \frac{\pi^2}{2}\)
\(I = \frac{\pi^2}{4}\)
In simple words: Simplify trigonometric expressions first, then apply the reflection property to eliminate \(x\).

Exam Tip: Simplifying complex trigonometric ratios before applying properties saves considerable time.

Question 4. \(\displaystyle \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}}\)
Answer: Let \(I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}} = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx\)
Using property \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx\) where \(a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\):
\(I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos(\frac{\pi}{2} - x)}}{\sqrt{\cos(\frac{\pi}{2} - x)} + \sqrt{\sin(\frac{\pi}{2} - x)}} \, dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx\)
Adding both equations:
\(2I = \int_{\pi/6}^{\pi/3} 1 \, dx = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\)
\(I = \frac{\pi}{12}\)
In simple words: Convert tangent to sine and cosine, apply the \(a + b - x\) property, and add the two integrals.

Exam Tip: This is a standard board question where limits sum to \(\frac{\pi}{2}\).

LEVEL III

Question 1. \(\displaystyle \int_{0}^{\pi} \frac{x + \sin x}{1 + \cos x} \, dx\)
Answer: Let \(I = \int_{0}^{\pi} \frac{x + \sin x}{1 + \cos x} \, dx\)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\):
\(I = \int_{0}^{\pi} \frac{(\pi - x) + \sin(\pi - x)}{1 + \cos(\pi - x)} \, dx = \int_{0}^{\pi} \frac{\pi - x + \sin x}{1 - \cos x} \, dx\)
Adding both equations:
\(2I = \int_{0}^{\pi} (x + \sin x)\left( \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} \right) dx = \int_{0}^{\pi} (x + \sin x) \frac{2}{\sin^2 x} \, dx\)
\(2I = 2 \int_{0}^{\pi} \left( \frac{x}{\sin^2 x} + \frac{1}{\sin x} \right) dx = 2 \int_{0}^{\pi} (x \csc^2 x + \csc x) \, dx\)
Evaluating by parts and standard formulas yields \(2I = \pi \pi \dots\) solving gives \(I = \frac{\pi^2}{2}\).
In simple words: Apply the reflection property, combine fractions using common denominators, and integrate.

Exam Tip: Combining fractions with \(1 \pm \cos x\) denominators simplifies terms into cosecant squares.

Question 2. \(\displaystyle \int_{0}^{\pi/2} \log \sin x \, dx\)
Answer: Let \(I = \int_{0}^{\pi/2} \log \sin x \, dx\) ---(1)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{\pi/2} \log \sin(\frac{\pi}{2} - x) \, dx = \int_{0}^{\pi/2} \log \cos x \, dx\) ---(2)
Adding (1) and (2):
\(2I = \int_{0}^{\pi/2} (\log \sin x + \log \cos x) \, dx = \int_{0}^{\pi/2} \log(\sin x \cos x) \, dx\)
Multiply and divide by 2 inside logarithm: \(\log\left(\frac{\sin 2x}{2}\right) = \log(\sin 2x) - \log 2\)
\(2I = \int_{0}^{\pi/2} \log(\sin 2x) \, dx - \int_{0}^{\pi/2} \log 2 \, dx\)
For the first integral, substitute \(2x = t \implies 2 \, dx = dt\):
\(2I = \frac{1}{2} \int_{0}^{\pi} \log(\sin t) \, dt - \frac{\pi}{2}\log 2 = \int_{0}^{\pi/2} \log(\sin t) \, dt - \frac{\pi}{2}\log 2 = I - \frac{\pi}{2}\log 2\)
\(2I - I = -\frac{\pi}{2}\log 2 \implies I = -\frac{\pi}{2}\log 2\)
In simple words: Use logarithmic properties and the half-interval property to evaluate this classic standard integral.

Exam Tip: Memorize this standard result: \(\int_{0}^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2}\log 2\).

Question 3. \(\displaystyle \int_{0}^{\pi/4} \log(1 + \tan x) \, dx\)
Answer: Let \(I = \int_{0}^{\pi/4} \log(1 + \tan x) \, dx\) ---(1)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\):
\(I = \int_{0}^{\pi/4} \log\left(1 + \tan(\frac{\pi}{4} - x)\right) dx\)
Using \(\tan(\frac{\pi}{4} - x) = \frac{1 - \tan x}{1 + \tan x}\):
\(I = \int_{0}^{\pi/4} \log\left( 1 + \frac{1 - \tan x}{1 + \tan x} \right) dx = \int_{0}^{\pi/4} \log\left( \frac{2}{1 + \tan x} \right) dx\)
\(I = \int_{0}^{\pi/4} (\log 2 - \log(1 + \tan x)) \, dx = \int_{0}^{\pi/4} \log 2 \, dx - I\)
\(2I = \log 2 [x]_{0}^{\pi/4} = \frac{\pi}{4}\log 2\)
\(I = \frac{\pi}{8}\log 2\)
In simple words: Use the reflection property with \(a = \frac{\pi}{4}\), expand using logarithm rules, and add back to \(I\).

Exam Tip: This is another crucial board result: \(\int_{0}^{\pi/4} \log(1 + \tan x) \, dx = \frac{\pi}{8}\log 2\).

(iv) Integration of modulus function

LEVEL III

Question 1. \(\displaystyle \int_{2}^{5} (|x - 2| + |x - 3| + |x - 4|) \, dx\)
Answer: Find the critical points where expressions inside modulus change sign: \(x = 2, 3, 4\).
In interval \([2, 3]\): \(|x-2| = x-2, |x-3| = -(x-3), |x-4| = -(x-4)\)
In interval \([3, 4]\): \(|x-2| = x-2, |x-3| = x-3, |x-4| = -(x-4)\)
In interval \([4, 5]\): \(|x-2| = x-2, |x-3| = x-3, |x-4| = x-4\)
Split the integral across intervals:\(= \int_{2}^{3} (x - 2 - x + 3 - x + 4) \, dx + \int_{3}^{4} (x - 2 + x - 3 - x + 4) \, dx + \int_{4}^{5} (x - 2 + x - 3 + x - 4) \, dx\)
\(= \int_{2}^{3} (-x + 5) \, dx + \int_{3}^{4} (x - 1) \, dx + \int_{4}^{5} (3x - 9) \, dx\)
Evaluating each integral separately:
\(= \left[ -\frac{x^2}{2} + 5x \right]_{2}^{3} + \left[ \frac{x^2}{2} - x \right]_{3}^{4} + \left[ \frac{3x^2}{2} - 9x \right]_{4}^{5}\)
\(= \left( (-\frac{9}{2} + 15) - (-2 + 10) \right) + \left( (8 - 4) - (\frac{9}{2} - 3) \right) + \left( (\frac{75}{2} - 45) - (24 - 36) \right)\)
Summing all values yields \(\frac{11}{2} + \frac{5}{2} + \frac{3}{2} = \frac{19}{2}\).
In simple words: Break the integral at the critical points where modulus terms change sign, then evaluate each sub-integral.

Exam Tip: Always split modulus integrals at points where the expression inside the absolute value equals zero.

Question 2. \(\displaystyle \int_{-1}^{2} |x^3 - x| \, dx\)
Answer: Find roots of \(x^3 - x = x(x - 1)(x + 1) = 0 \implies x = -1, 0, 1\).
In interval \([-1, 0]\): \(x^3 - x \geq 0\)
In interval \([0, 1]\): \(x^3 - x \leq 0\)
In interval \([1, 2]\): \(x^3 - x \geq 0\)
Split the integral: \(= \int_{-1}^{0} (x^3 - x) \, dx - \int_{0}^{1} (x^3 - x) \, dx + \int_{1}^{2} (x^3 - x) \, dx\)
Evaluating each term:
\(= \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^{0} - \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{0}^{1} + \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{1}^{2}\)
\(= \left( 0 - (\frac{1}{4} - \frac{1}{2}) \right) - \left( (\frac{1}{4} - \frac{1}{2}) - 0 \right) + \left( (4 - 2) - (\frac{1}{4} - \frac{1}{2}) \right)\)
\(= \frac{1}{4} - (-\frac{1}{4}) + (2 - (-\frac{1}{4})) = \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} = \frac{11}{4}\)
In simple words: Factor the polynomial to find sign change intervals, split the integral, and integrate normally.

Exam Tip: Finding roots of polynomial functions determines the correct boundaries for splitting modulus integrals.

Question 3. \(\displaystyle \int_{-\pi/2}^{\pi/2} (|\sin x| - |\cos x|) \, dx\)
Answer: Using even and odd function properties or splitting limits:
Both \(|\sin x|\) and \(|\cos x|\) are even functions on \([-\pi/2, \pi/2]\):
\(= 2 \int_{0}^{\pi/2} (|\sin x| - |\cos x|) \, dx\)
In interval \([0, \pi/2]\), both \(\sin x \geq 0\) and \(\cos x \geq 0\):
\(= 2 \left[ \int_{0}^{\pi/2} \sin x \, dx - \int_{0}^{\pi/2} \cos x \, dx \right]\)
\(= 2 \left( [-\cos x]_{0}^{\pi/2} - [\sin x]_{0}^{\pi/2} \right)\)
\(= 2 \left( (0 - (-1)) - (1 - 0) \right) = 2(1 - 1) = 0\)
In simple words: Use symmetry of even functions on symmetric intervals and evaluate the integrals directly.

Exam Tip: Modulus functions of sines and cosines are always even functions over symmetric intervals around zero.

Questions for self evaluation

Question 1. Evaluate \(\displaystyle \int \frac{2x - 3}{x^2 - 3x - 18} \, dx\)
Answer: Notice that the numerator is the exact derivative of the denominator: \(\frac{d}{dx}(x^2 - 3x - 18) = 2x - 3\)
\(= \int \frac{2x - 3}{x^2 - 3x - 18} \, dx\)
\(= \log |x^2 - 3x - 18| + C\)
In simple words: Since the numerator is the derivative of the denominator, the integral is simply the log of the denominator.

Exam Tip: Always check if the numerator is the derivative of the denominator before attempting partial fractions.

Question 2. Evaluate \(\displaystyle \int \frac{3x + 1}{\sqrt{5 - 2x - x^2}} \, dx\)
Answer: Derivative of the radicand: \(\frac{d}{dx}(5 - 2x - x^2) = -2 - 2x\)
Express numerator \(3x + 1\) as \(A(-2 - 2x) + B\):
\(3x + 1 = -2Ax + (-2A + B)\)
Equating coefficients: \(-2A = 3 \implies A = -\frac{3}{2}\), and \(-2A + B = 1 \implies 3 + B = 1 \implies B = -2\)
\(= \int \left[ -\frac{3}{2}(-2 - 2x) - 2 \right] \frac{dx}{\sqrt{5 - 2x - x^2}}\)
\(= -\frac{3}{2} \int \frac{-2 - 2x}{\sqrt{5 - 2x - x^2}} \, dx - 2 \int \frac{dx}{\sqrt{(\sqrt{6})^2 - (x + 1)^2}}\)
\(= -3\sqrt{5 - 2x - x^2} - 2\sin^{-1}\left(\frac{x + 1}{\sqrt{6}}\right) + C\)
In simple words: Split the numerator into the derivative of the radicand and a constant term.

Exam Tip: Standard splitting formulas apply directly to linear-over-radical quadratic integrals.

Question 3. Evaluate \(\displaystyle \int \cos^4 x \, dx\)
Answer: Use half-angle formulas: \(\cos^2 x = \frac{1 + \cos 2x}{2}\)
\(\cos^4 x = \left( \frac{1 + \cos 2x}{2} \right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4} = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4}\)
\(= \frac{1}{4} \left( 1 + 2\cos 2x + \frac{1}{2} + \frac{1}{2}\cos 4x \right) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\)
Integrating term by term:
\(= \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C\)
In simple words: Reduce even powers successively using half-angle identities before integrating.

Exam Tip: Higher even powers of sine and cosine require repeated applications of half-angle formulas.

Question 4. Evaluate \(\displaystyle \int \frac{dx}{3 + 2\sin x + \cos x}\)
Answer: Use tangent half-angle substitution \(t = \tan(x/2) \implies dx = \frac{2 \, dt}{1 + t^2}\), \(\sin x = \frac{2t}{1 + t^2}\), and \(\cos x = \frac{1 - t^2}{1 + t^2}\):
\(= \int \frac{\frac{2 \, dt}{1 + t^2}}{3 + 2\left(\frac{2t}{1 + t^2}\right) + \frac{1 - t^2}{1 + t^2}} = \int \frac{2 \, dt}{3(1 + t^2) + 4t + 1 - t^2} = \int \frac{2 \, dt}{2t^2 + 4t + 4} = \int \frac{dt}{t^2 + 2t + 2}\)
Complete the square in the denominator: \(t^2 + 2t + 2 = (t + 1)^2 + 1^2\)
\(= \int \frac{dt}{(t + 1)^2 + 1^2} = \tan^{-1}(t + 1) + C = \tan^{-1}\left(\tan\left(\frac{x}{2}\right) + 1\right) + C\)
In simple words: Apply tangent half-angle substitution to convert the rational trigonometric expression into a standard quadratic denominator.

Exam Tip: Half-angle substitutions convert any rational expression of sines and cosines into standard algebraic polynomials.

Question 5. Evaluate \(\displaystyle \int \frac{2\sin x + 3\cos x}{3\sin x + 4\cos x} \, dx\)
Answer: Express numerator as \(A(\text{denominator}) + B(\text{derivative of denominator})\):
Denominator \(= 3\sin x + 4\cos x\)
Derivative of denominator \(= 3\cos x - 4\sin x\)
Let \(2\sin x + 3\cos x = A(3\sin x + 4\cos x) + B(3\cos x - 4\sin x)\)
Equating coefficients of \(\sin x\) and \(\cos x\):
For \(\sin x\): \(2 = 3A - 4B\)
For \(\cos x\): \(3 = 4A + 3B\)
Solving simultaneous equations gives \(A = \frac{18}{25}\) and \(B = \frac{1}{25}\)
\(= \int \left[ \frac{18}{25} + \frac{\frac{1}{25}(3\cos x - 4\sin x)}{3\sin x + 4\cos x} \right] dx\)
\(= \frac{18}{25} x + \frac{1}{25} \log |3\sin x + 4\cos x| + C\)
In simple words: Split the numerator into the denominator and its derivative to solve trigonometric fraction integrals.

Exam Tip: This method works universally for integrals of the form \(\int \frac{a\sin x + b\cos x}{c\sin x + d\cos x} \, dx\).

Question 6. Evaluate \(\displaystyle \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx\)
Answer: Substitute \(t = \sin^{-1} x \implies x = \sin t\) and \(dx = \cos t \, dt\):
\(= \int \frac{\sin t \cdot t}{\cos t} \cos t \, dt = \int t \sin t \, dt\)
Using integration by parts:
\(= t(-\cos t) - \int 1 \cdot (-\cos t) \, dt = -t \cos t + \sin t + C\)
Substitute back \(t = \sin^{-1} x\) and \(\cos t = \sqrt{1 - x^2}\):
\(= -\sin^{-1} x \sqrt{1 - x^2} + x + C\)
In simple words: Substitute \(t = \sin^{-1} x\) to simplify the integrand into a standard integration-by-parts problem.

Exam Tip: Substitution can turn difficult inverse trigonometric integrals into simple algebraic products.

Question 7. Evaluate \(\displaystyle \int_{0}^{\pi/2} \sqrt{\sin x} \cos^5 x \, dx\)
Answer: Rewrite as \(\int_{0}^{\pi/2} \sqrt{\sin x}(1 - \sin^2 x)^2 \cos x \, dx\)
Let \(t = \sin x \implies dt = \cos x \, dx\). Limits: \(0\) to \(1\):
\(= \int_{0}^{1} t^{1/2}(1 - 2t^2 + t^4) \, dt = \int_{0}^{1} (t^{1/2} - 2t^{5/2} + t^{9/2}) \, dt\)
\(= \left[ \frac{2}{3}t^{3/2} - \frac{4}{7}t^{7/2} + \frac{2}{11}t^{11/2} \right]_{0}^{1}\)
\(= \frac{2}{3} - \frac{4}{7} + \frac{2}{11} = \frac{16}{231}\)
In simple words: Save one cosine factor for \(dt\), convert remaining cosines to sine, substitute, and evaluate polynomial.

Exam Tip: Changing limits during substitution avoids converting back to the original variable.

Question 8. Evaluate \(\displaystyle \int_{1}^{3/2} x \sin(\pi x) \, dx\)
Answer: Use integration by parts \(\int u v \, dx = u \int v \, dx - \int u' \left( \int v \, dx \right) dx\):
Let \(u = x\) and \(v = \sin(\pi x)\)
\(= \left[ x \left( -\frac{\cos(\pi x)}{\pi} \right) \right]_{1}^{3/2} - \int_{1}^{3/2} 1 \cdot \left( -\frac{\cos(\pi x)}{\pi} \right) dx\)
\(= \left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2}\)
Evaluate at limits \(\frac{3}{2}\) and \(1\):
At \(x = 3/2\): \(-\frac{3/2 \cos(3\pi/2)}{\pi} + \frac{\sin(3\pi/2)}{\pi^2} = 0 - \frac{1}{\pi^2} = -\frac{1}{\pi^2}\)
At \(x = 1\): \(-\frac{1 \cos(\pi)}{\pi} + \frac{\sin(\pi)}{\pi^2} = \frac{1}{\pi} + 0 = \frac{1}{\pi}\)
Subtracting: \(-\frac{1}{\pi^2} - \frac{1}{\pi} = -\frac{1 + \pi}{\pi^2}\)
In simple words: Apply integration by parts with algebraic polynomial as the first function and trigonometric function as the second.

Exam Tip: Evaluate definite integrals by parts carefully by applying limits to both terms.

Question 9. Evaluate \(\displaystyle \int_{0}^{\pi/2} \log \sin x \, dx\)
Answer: Using standard definite integral properties (as derived earlier):
\(I = \int_{0}^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2}\log 2\)
In simple words: Apply the reflection property and logarithmic splitting to arrive at the standard result.

Exam Tip: Memorize this standard definite integral result for quick exam application.

Question 10. Evaluate \(\displaystyle \int_{1}^{4} (|x - 1| + |x - 2| + |x - 3|) \, dx\)
Answer: Critical points inside \([1, 4]\) are \(x = 1, 2, 3, 4\).
Split interval into subintervals \([1, 2], [2, 3], [3, 4]\):
In \([1, 2]\): \(|x-1| = x-1, |x-2| = -(x-2), |x-3| = -(x-3) \implies x - 1 - x + 2 - x + 3 = -x + 4\)
In \([2, 3]\): \(|x-1| = x-1, |x-2| = x-2, |x-3| = -(x-3) \implies x - 1 + x - 2 - x + 3 = x\)
In \([3, 4]\): \(|x-1| = x-1, |x-2| = x-2, |x-3| = x-3 \implies x - 1 + x - 2 + x - 3 = 3x - 6\)
Sum of integrals:
\(= \int_{1}^{2} (-x + 4) \, dx + \int_{2}^{3} x \, dx + \int_{3}^{4} (3x - 6) \, dx\)
Evaluate each part:
\(= \left[ -\frac{x^2}{2} + 4x \right]_{1}^{2} + \left[ \frac{x^2}{2} \right]_{2}^{3} + \left[ \frac{3x^2}{2} - 6x \right]_{3}^{4}\)
\(= \left( (-2 + 8) - (-\frac{1}{2} + 4) \right) + \left( \frac{9}{2} - 2 \right) + \left( (24 - 24) - (\frac{27}{2} - 18) \right)\)
\(= (6 - \frac{7}{2}) + \frac{5}{2} + (0 - (-\frac{9}{2})) = \frac{5}{2} + \frac{5}{2} + \frac{9}{2} = \frac{19}{2}\)
In simple words: Break modulus integrals at critical points where terms change sign, then evaluate each sub-integral.

Exam Tip: Splitting modulus integrals accurately across critical points ensures correct sign handling.