CBSE Class 12 Mathematics Integrals Assignment Set 05

TOPIC 6: INDEFINITE & DEFINITE INTEGRALS

SCHEMATIC DIAGRAM & IMPORTANT RESULTS

Assignments

(i) Integration by substitution - LEVEL I

Question 1. \(\displaystyle\int \frac{\sec^2(\log x)}{x} \, dx\)
Answer: Let \(\log x = t\)
Differentiating both sides with respect to \(x\), we get \(\frac{1}{x} \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int \sec^2(t) \, dt = \tan t + C\)
Replacing \(t\) with \(\log x\):
\(= \tan(\log x) + C\)
In simple words: Substitute the inner logarithmic function with a single variable to simplify the integral into a standard trigonometric form.

Exam Tip: Whenever the derivative of the inner function is present as a factor in the integrand, use the substitution method.

Question 2. \(\displaystyle\int \frac{e^{m \tan^{-1} x}}{1 + x^2} \, dx\)
Answer: Let \(\tan^{-1} x = t\)
Differentiating both sides, we get \(\frac{1}{1 + x^2} \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int e^{m t} \, dt = \frac{e^{m t}}{m} + C\)
Replacing \(t\) with \(\tan^{-1} x\):
\(= \frac{1}{m} e^{m \tan^{-1} x} + C\)
In simple words: Substitute \(\tan^{-1} x\) as \(t\) because its derivative \(\frac{1}{1 + x^2}\) is already multiplying the term.

Exam Tip: Memorize standard inverse trigonometric derivatives to quickly spot substitution candidates.

Question 3. \(\displaystyle\int \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \, dx\)
Answer: Let \(\sin^{-1} x = t\)
Differentiating both sides, we get \(\frac{1}{\sqrt{1 - x^2}} \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int e^t \, dt = e^t + C\)
Replacing \(t\) with \(\sin^{-1} x\):
\(= e^{\sin^{-1} x} + C\)
In simple words: The derivative of \(\sin^{-1} x\) is \(\frac{1}{\sqrt{1 - x^2}}\), which makes substitution straightforward.

Exam Tip: Exponential functions with composite powers often simplify completely when the exponent is substituted.

(i) Integration by substitution - LEVEL II

Question 1. \(\displaystyle\int \frac{1}{x + \sqrt{x}} \, dx\)
Answer: Rewrite the denominator by factoring out \(\sqrt{x}\):
\(\displaystyle\int \frac{1}{\sqrt{x}(\sqrt{x} + 1)} \, dx\)
Let \(\sqrt{x} + 1 = t\)
Differentiating both sides, we get \(\frac{1}{2\sqrt{x}} \, dx = dt \implies \frac{1}{\sqrt{x}} \, dx = 2 \, dt\)
Substituting into the integral:
\(\displaystyle\int \frac{2}{t} \, dt = 2 \log |t| + C\)
Replacing \(t\) with \(\sqrt{x} + 1\):
\(= 2 \log |\sqrt{x} + 1| + C\)
In simple words: Factor \(\sqrt{x}\) out from the denominator so that the differential of \(\sqrt{x}\) appears in the expression.

Exam Tip: Expressions containing sums of powers of \(x\) and roots can often be integrated by substituting the root term.

Question 2. \(\displaystyle\int \frac{1}{x(x^6 - 1)} \, dx\)
Answer: Multiply numerator and denominator by \(x^5\):
\(\displaystyle\int \frac{x^5}{x^6(x^6 - 1)} \, dx\)
Let \(x^6 = t\)
Differentiating both sides, we get \(6x^5 \, dx = dt \implies x^5 \, dx = \frac{1}{6} \, dt\)
Substituting into the integral:
\(\frac{1}{6} \displaystyle\int \frac{1}{t(t - 1)} \, dt\)
Using partial fractions: \(\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac{1}{t}\)
\(= \frac{1}{6} \displaystyle\left( \int \frac{1}{t - 1} \, dt - \int \frac{1}{t} \, dt \right)\)
\(= \frac{1}{6} \left( \log |t - 1| - \log |t| \right) + C\)
\(= \frac{1}{6} \log \left| \frac{t - 1}{t} \right| + C\)
Replacing \(t\) with \(x^6\):
\(= \frac{1}{6} \log \left| \frac{x^6 - 1}{x^6} \right| + C\)
In simple words: Multiply top and bottom by \(x^5\) to create the derivative of \(x^6\) in the numerator, then apply partial fractions.

Exam Tip: This is a standard algebraic manipulation technique for rational functions involving high powers of \(x\).

Question 3. \(\displaystyle\int \frac{1}{e^x - 1} \, dx\)
Answer: Multiply numerator and denominator by \(e^{-x}\):
\(\displaystyle\int \frac{e^{-x}}{1 - e^{-x}} \, dx\)
Let \(1 - e^{-x} = t\)
Differentiating both sides, we get \(e^{-x} \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int \frac{1}{t} \, dt = \log |t| + C\)
Replacing \(t\) with \(1 - e^{-x}\):
\(= \log |1 - e^{-x}| + C\)
In simple words: Adjust the fraction by multiplying by \(e^{-x}\) so that the numerator becomes the exact derivative of the denominator.

Exam Tip: For exponential denominators, factoring out the highest power of \(e^x\) or multiplying by \(e^{-x}\) is the standard approach.

(i) Integration by substitution - LEVEL III

Question 1. \(\displaystyle\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx\)
Answer: Rewrite the denominator by dividing numerator and denominator by \(\cos^2 x\):
\(\displaystyle\int \frac{\sqrt{\tan x}}{\frac{\sin x \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} \, dx = \int \frac{\sqrt{\tan x}}{\tan x} \cdot \sec^2 x \, dx\)
\(= \displaystyle\int \frac{\sec^2 x}{\sqrt{\tan x}} \, dx\)
Let \(\tan x = t\)
Differentiating both sides, we get \(\sec^2 x \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt = 2t^{1/2} + C\)
Replacing \(t\) with \(\tan x\):
\(= 2\sqrt{\tan x} + C\)
In simple words: Convert the product of sine and cosine in the denominator into tangent and secant squared by dividing through by \(\cos^2 x\).

Exam Tip: Look for combinations of sine and cosine in denominators that simplify when divided by \(\cos^2 x\).

Question 2. \(\displaystyle\int \frac{\sqrt{\tan x}}{\sec x + \cos x} \, dx\)
Answer: Convert all trigonometric terms into sine and cosine:
\(\displaystyle\int \frac{\sqrt{\frac{\sin x}{\cos x}}}{\frac{1}{\cos x} + \cos x} \, dx = \int \frac{\sqrt{\frac{\sin x}{\cos x}}}{\frac{1 + \cos^2 x}{\cos x}} \, dx\)
\(= \displaystyle\int \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \cdot \frac{\cos x}{1 + \cos^2 x} \, dx = \int \frac{\sqrt{\sin x} \cos^{1/2} x}{1 + \cos^2 x} \, dx\) (This form can be approached via substitution after proper algebraic setup).
Alternatively, rewriting: let \(\tan x = t^2\), then \(\sec^2 x \, dx = 2t \, dt\), \(dx = \frac{2t \, dt}{1 + t^4}\).
With \(\sin x = \frac{t^2}{\sqrt{1+t^4}}\) and \(\cos x = \frac{1}{\sqrt{1+t^4}}\), the integral simplifies to a rational algebraic function in \(t\).
Performing the algebraic reduction yields:
\(= \sqrt{2} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2}\sqrt{\tan x}}\right) + C\) (Standard reduction).
In simple words: Expressing terms in sine and cosine and using strategic substitutions transforms complex trigonometric expressions into solvable polynomials.

Exam Tip: Trigonometric integrals with fractional powers of tangent often simplify by substituting \(\tan x = t^2\).

Question 3. \(\displaystyle\int \frac{1}{\sin x \cos^3 x} \, dx\)
Answer: Multiply and divide by \(\cos^x\) or rewrite using \(\sec^2 x\):
\(\displaystyle\int \frac{\sec^2 x}{\sin x / \cos x} \cdot \sec^2 x \, dx\) - alternative method:
Divide numerator and denominator by \(\cos^4 x\):
\(\displaystyle\int \frac{\sec^4 x}{\tan x} \, dx = \int \frac{(1 + \tan^2 x) \sec^2 x}{\tan x} \, dx\)
Let \(\tan x = t \implies \sec^2 x \, dx = dt\)
\(\displaystyle\int \frac{1 + t^2}{t} \, dt = \int \left( \frac{1}{t} + t \right) dt\)
\(= \log |t| + \frac{t^2}{2} + C\)
Replacing \(t\) with \(\tan x\):
\(= \log |\tan x| + \frac{1}{2} \tan^2 x + C\)
In simple words: Divide numerator and denominator by \(\cos^4 x\) to introduce \(\sec^2 x\) and express everything in terms of \(\tan x\).

Exam Tip: Dividing by appropriate powers of cosine is a powerful tool for integrating products of sine and cosine in denominators.

(ii) Application of trigonometric function in integrals - LEVEL I

Question 1. \(\displaystyle\int \sin^3 x \, dx\)
Answer: Use the identity \(\sin^3 x = \frac{3 \sin x - \sin 3x}{4}\):
\(\displaystyle\int \frac{3 \sin x - \sin 3x}{4} \, dx\)
\(= \frac{1}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} \right] + C\)
\(= -\frac{3}{4} \cos x + \frac{1}{12} \cos 3x + C\)
In simple words: Use triple-angle or half-angle trigonometric identities to lower the power of trigonometric functions before integrating.

Exam Tip: Odd powers of sine or cosine can be integrated by isolating one factor and converting the rest using Pythagorean identities, or by using multiple-angle formulas.

Question 2. \(\displaystyle\int \cos^2 3x \, dx\)
Answer: Use the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\):
\(\displaystyle\int \frac{1 + \cos 6x}{2} \, dx\)
\(= \frac{1}{2} \left[ x + \frac{\sin 6x}{6} \right] + C\)
\(= \frac{x}{2} + \frac{\sin 6x}{12} + C\)
In simple words: Use the double-angle formula to convert squared cosine into a linear term in cosine.

Exam Tip: Even powers of sine and cosine should always be reduced using power-reducing formulas.

Question 3. \(\displaystyle\int \cos x \cos 2x \cos 3x \, dx\)
Answer: Combine two terms using the product formula \(2 \cos A \cos B = \cos(A + B) + \cos(A - B)\):
Consider \(\cos 3x \cos x = \frac{1}{2} [\cos 4x + \cos 2x]\)
Now integrate \(\frac{1}{2} \int (\cos 4x + \cos 2x) \cos 3x \, dx\)
\(= \frac{1}{2} \int (\cos 4x \cos 3x + \cos 2x \cos 3x) \, dx\)
Using \(2 \cos A \cos B = \cos(A + B) + \cos(A - B)\) again:
\(= \frac{1}{4} \int (\cos 7x + \cos x + \cos 5x + \cos x) \, dx\)
\(= \frac{1}{4} \int (\cos 7x + \cos 5x + 2\cos x) \, dx\)
\(= \frac{1}{4} \left[ \frac{\sin 7x}{7} + \frac{\sin 5x}{5} + 2\sin x \right] + C\)
In simple words: Convert products of cosines into sums using standard factor formulas before integrating term by term.

Exam Tip: Product-to-sum formulas are essential when integrating products of multiple trigonometric functions.

(ii) Application of trigonometric function in integrals - LEVEL II

Question 1. \(\displaystyle\int \sec^4 x \tan x \, dx\)
Answer: Rewrite the integrand as:
\(\displaystyle\int \sec^3 x (\sec x \tan x) \, dx\)
Let \(\sec x = t\)
Differentiating both sides, we get \(\sec x \tan x \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int t^3 \, dt = \frac{t^4}{4} + C\)
Replacing \(t\) with \(\sec x\):
\(= \frac{\sec^4 x}{4} + C\)
In simple words: Separate a \(\sec x \tan x\) factor to serve as the derivative of \(\sec x\), then use substitution.

Exam Tip: When an odd power of tangent is multiplied by an even power of secant, reserve \(\sec x \tan x\) for \(dt\).

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

Exam Tip: Always check if multiple-angle expansion can simplify denominators in trigonometric integrals.

(ii) Application of trigonometric function in integrals - LEVEL III

Question 1. \(\displaystyle\int \cos^5 x \, dx\)
Answer: Isolate one cosine factor: \(\displaystyle\int \cos^4 x (\cos x \, dx)\)
Rewrite \(\cos^4 x = (1 - \sin^2 x)^2\):
\(\displaystyle\int (1 - \sin^2 x)^2 \cos x \, dx\)
Let \(\sin x = t \implies \cos x \, dx = dt\)
\(\displaystyle\int (1 - t^2)^2 \, dt = \int (1 - 2t^2 + t^4) \, dt\)
\(= t - \frac{2t^3}{3} + \frac{t^5}{5} + C\)
Replacing \(t\) with \(\sin x\):
\(= \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C\)
In simple words: Save one cosine factor for \(dt\) and express the remaining even power of cosine in terms of sine using the Pythagorean identity.

Exam Tip: For odd powers of sine or cosine, saving one factor for substitution is a foolproof technique.

Question 2. \(\displaystyle\int \sin^2 x \cos^3 x \, dx\)
Answer: Isolate one cosine factor: \(\displaystyle\int \sin^2 x \cos^2 x (\cos x \, dx)\)
Replace \(\cos^2 x = 1 - \sin^2 x\):
\(\displaystyle\int \sin^2 x (1 - \sin^2 x) \cos x \, dx\)
Let \(\sin x = t \implies \cos x \, dx = dt\)
\(\displaystyle\int t^2 (1 - t^2) \, dt = \int (t^2 - t^4) \, dt\)
\(= \frac{t^3}{3} - \frac{t^5}{5} + C\)
Replacing \(t\) with \(\sin x\):
\(= \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C\)
In simple words: Since the power of cosine is odd, save one cosine for substitution and convert the remaining cosine terms into sines.

Exam Tip: When both sine and cosine appear with powers, pick the function with the odd power to substitute.

(iii) Integration using standard results - LEVEL I

Question 1. \(\displaystyle\int \frac{dx}{\sqrt{4x^2 - 9}}\)
Answer: Rewrite the denominator: \(\sqrt{(2x)^2 - 3^2}\)
Using the standard formula \(\displaystyle\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + C\):
\(= \frac{1}{2} \log |2x + \sqrt{4x^2 - 9}| + C\) (accounting for the coefficient 2 of \(x\)).
In simple words: Match the integral to the standard logarithmic inverse-hyperbolic form and divide by the coefficient of \(x\).

Exam Tip: Remember to divide by the derivative of the linear inner function \(2x\).

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

Exam Tip: Completing the square is the primary technique for integrating rational functions with irreducible quadratic denominators.

Question 3. \(\displaystyle\int \frac{dx}{9x^2 + 12x + 13}\)
Answer: Complete the square in the denominator:
\(9x^2 + 12x + 13 = (3x)^2 + 2(3x)(2) + 2^2 - 4 + 13 = (3x + 2)^2 + 9 = (3x + 2)^2 + 3^2\)
Using the standard formula for \(\frac{1}{x^2 + a^2}\):
\(= \frac{1}{3} \cdot \frac{1}{3} \tan^{-1}\left(\frac{3x + 2}{3}\right) + C\)
\(= \frac{1}{9} \tan^{-1}\left(\frac{3x + 2}{3}\right) + C\)
In simple words: Group the terms to form a perfect square and apply the standard arctangent formula, dividing by the coefficient of \(x\).

Exam Tip: Always factor out the coefficient of \(x^2\) or set up the square carefully when the coefficient is not 1.

(iii) Integration using standard results - LEVEL II

Question 1. \(\displaystyle\int \frac{x}{x^4 + x^2 + 1} \, dx\)
Answer: Let \(x^2 = t \implies 2x \, dx = dt \implies x \, dx = \frac{1}{2} \, dt\)
Substituting into the integral:
\(\frac{1}{2} \displaystyle\int \frac{1}{t^2 + t + 1} \, dt\)
Complete the square for \(t^2 + t + 1 = \left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2\)
\(= \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C\)
\(= \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) + C\)
Replacing \(t\) with \(x^2\):
\(= \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2x^2 + 1}{\sqrt{3}}\right) + C\)
In simple words: Substitute \(x^2 = t\) to reduce the degree of the denominator, then complete the square.

Exam Tip: Functions of even powers like \(x^4\) and \(x^2\) can often be treated as quadratic functions in \(x^2\).

Question 2. \(\displaystyle\int \frac{\cos x}{\sin^2 x + 4 \sin x + 5} \, dx\)
Answer: Let \(\sin x = t \implies \cos x \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int \frac{1}{t^2 + 4t + 5} \, dt\)
Complete the square in the denominator:
\(t^2 + 4t + 5 = (t + 2)^2 + 1^2\)
Using the standard arctangent formula:
\(= \frac{1}{1} \tan^{-1}\left(\frac{t + 2}{1}\right) + C\)
Replacing \(t\) with \(\sin x\):
\(= \tan^{-1}(\sin x + 2) + C\)
In simple words: Substitute \(\sin x = t\) so that \(\cos x \, dx\) becomes \(dt\), then use completing the square.

Exam Tip: Whenever the derivative of a trigonometric function is present in the numerator, substitution simplifies the problem into algebraic form.

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

Exam Tip: Be careful with negative signs when completing the square inside a square root.

(iii) Integration using standard results - LEVEL III

Question 1. \(\displaystyle\int \frac{2x}{\sqrt{1 - x^2 - x^4}} \, dx\)
Answer: Let \(x^2 = t \implies 2x \, dx = dt\)
Substituting into the integral:
\(\displaystyle\int \frac{dt}{\sqrt{1 - t - t^2}}\)
Complete the square in the denominator:
\(1 - t - t^2 = 1 - \left(t^2 + t + \frac{1}{4} - \frac{1}{4}\right) = \frac{5}{4} - \left(t + \frac{1}{2}\right)^2 = \left(\frac{\sqrt{5}}{2}\right)^2 - \left(t + \frac{1}{2}\right)^2\)
Using the standard formula \(\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C\):
\(= \sin^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{5}}{2}}\right) + C\)
\(= \sin^{-1}\left(\frac{2t + 1}{\sqrt{5}}\right) + C\)
Replacing \(t\) with \(x^2\):
\(= \sin^{-1}\left(\frac{2x^2 + 1}{\sqrt{5}}\right) + C\)
In simple words: Substitute \(x^2 = t\) to change the expression into a standard inverse sine form with completing the square.

Exam Tip: Spotting the substitution \(x^2 = t\) instantly reduces higher-degree polynomials into manageable quadratics.

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:
\(\frac{x^2 + x + 1}{x^2 - x + 1} = 1 + \frac{2x}{x^2 - x + 1}\)
Now integrate term by term:
\(\displaystyle\int 1 \, dx + \int \frac{2x}{x^2 - x + 1} \, dx = x + \int \frac{2x - 1 + 1}{x^2 - x + 1} \, dx\)
\(= x + \int \frac{2x - 1}{x^2 - x + 1} \, dx + \int \frac{1}{\left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dx\)
\(= x + \log |x^2 - x + 1| + \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right) + C\)
In simple words: Divide the improper fraction first, then adjust the numerator of the remaining fraction to match the derivative of the denominator.

Exam Tip: For rational functions where numerator degree equals or exceeds denominator degree, always divide first.

Question 3. \(\displaystyle\int \frac{x + 2}{\sqrt{x^2 + 5x + 6}} \, dx\)
Answer: Express the numerator as a multiple of the derivative of the quadratic expression inside the radical.
Let \(x + 2 = A \frac{d}{dx}(x^2 + 5x + 6) + B = A(2x + 5) + B\)
Equating coefficients: \(2A = 1 \implies A = \frac{1}{2}\), and \(5A + B = 2 \implies \frac{5}{2} + B = 2 \implies B = -\frac{1}{2}\)
Thus, \(x + 2 = \frac{1}{2}(2x + 5) - \frac{1}{2}\)
Substitute into the integral:
\(\displaystyle\int \frac{\frac{1}{2}(2x + 5) - \frac{1}{2}}{\sqrt{x^2 + 5x + 6}} \, dx = \frac{1}{2} \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} \, dx - \frac{1}{2} \int \frac{dx}{\sqrt{(x + \frac{5}{2})^2 - (\frac{1}{2})^2}}\)
\(= \frac{1}{2} \left[ 2\sqrt{x^2 + 5x + 6} \right] - \frac{1}{2} \log \left| x + \frac{5}{2} + \sqrt{x^2 + 5x + 6} \right| + C\)
\(= \sqrt{x^2 + 5x + 6} - \frac{1}{2} \log \left| x + \frac{5}{2} + \sqrt{x^2 + 5x + 6} \right| + C\)
In simple words: Split the numerator into a part equal to the derivative of the inner function and a constant part.

Exam Tip: This standard form \(\frac{px + q}{\sqrt{ax^2 + bx + c}}\) is always solved by expressing the numerator as \(A(2ax + b) + B\).

Question 4. \(\displaystyle\int \sqrt{\frac{1 - x}{1 + x}} \, dx\)
Answer: Rationalize the numerator by multiplying top and bottom by \(\sqrt{1 - x}\):
\(\displaystyle\int \frac{1 - x}{\sqrt{1 - x^2}} \, dx = \int \frac{1}{\sqrt{1 - x^2}} \, dx - \int \frac{x}{\sqrt{1 - x^2}} \, dx\)
The first integral is \(\sin^{-1} x\). For the second, substitute \(1 - x^2 = t\):
\(= \sin^{-1} x + \sqrt{1 - x^2} + C\)
In simple words: Rationalize the expression to break it into a standard inverse sine integral and a simple substitution integral.

Exam Tip: Rationalizing numerators containing square roots of binomials is a reliable simplification tactic.

Question 5. \(\displaystyle\int \frac{6x + 7}{\sqrt{(x - 5)(x - 4)}} \, dx\) [CBSE 2011]
Answer: Expand the product inside the root: \((x - 5)(x - 4) = x^2 - 9x + 20\)
Integral becomes \(\displaystyle\int \frac{6x + 7}{\sqrt{x^2 - 9x + 20}} \, dx\)
Let numerator \(6x + 7 = A \frac{d}{dx}(x^2 - 9x + 20) + B = A(2x - 9) + B\)
Equating coefficients: \(2A = 6 \implies A = 3\), and \(-9A + B = 7 \implies -27 + B = 7 \implies B = 34\)
Rewrite the integral as:
\(3 \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} \, dx + 34 \int \frac{dx}{\sqrt{(x - \frac{9}{2})^2 - (\frac{1}{2})^2}}\)
\(= 3(2\sqrt{x^2 - 9x + 20}) + 34 \log \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C\)
In simple words: Multiply out the factors in the root and use the linear-over-root splitting method.

Exam Tip: Always multiply out binomial factors under square roots before applying derivative splitting.

(iv) Integration using Partial Fraction - LEVEL I

Question 1. \(\displaystyle\int \frac{2x + 1}{(x + 1)(x - 1)} \, dx\)
Answer: Decompose into 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 \(x = -1 \implies -1 = -2A \implies A = \frac{1}{2}\)
Integrate term by term:
\(\displaystyle\int \left( \frac{1/2}{x + 1} + \frac{3/2}{x - 1} \right) dx = \frac{1}{2} \log |x + 1| + \frac{3}{2} \log |x - 1| + C\)
In simple words: Split the rational function into simpler fractions with linear denominators using the cover-up method.

Exam Tip: Partial fractions turn complex rational expressions into easy logarithmic integrals.

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 \(x = 3 \implies 9 = C(2)(1) \implies C = \frac{9}{2}\)
Integrate term by term:
\(\frac{1}{2} \log |x - 1| - 4 \log |x - 2| + \frac{9}{2} \log |x - 3| + C\)
In simple words: Break a fraction with three linear factors into three separate terms and integrate each into a logarithm.

Exam Tip: Finding constants in partial fractions by substituting roots of linear factors is very efficient.

Question 3. \(\displaystyle\int \frac{3x - 2}{(x + 1)^2(x + 3)} \, dx\)
Answer: Decompose into partial fractions with repeated linear factors:
\(\frac{3x - 2}{(x + 1)^2(x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + 3}\)
\(3x - 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)^2\)
Put \(x = -1 \implies -5 = B(2) \implies B = -\frac{5}{2}\)
Put \(x = -3 \implies -11 = C(-2)^2 \implies 4C = -11 \implies C = -\frac{11}{4}\)
Comparing coefficients of \(x^2\) (\(A + C = 0\)): \(A = -C = \frac{11}{4}\)
Integrate term by term:
\(\frac{11}{4} \log |x + 1| + \frac{5}{2(x + 1)} - \frac{11}{4} \log |x + 3| + C\)
In simple words: Repeated linear factors require terms for each power up to the multiplicity in the denominator.

Exam Tip: Remember that a squared linear factor generates both a linear term and a squared term in partial fractions.

(iv) Integration using Partial Fraction - LEVEL II

Question 1. \(\displaystyle\int \frac{x^2 + 2x + 8}{(x - 1)(x - 2)} \, dx\)
Answer: Since the degree of numerator equals the degree of denominator, divide first:
\(\frac{x^2 + 2x + 8}{x^2 - 3x + 2} = 1 + \frac{5x + 6}{x^2 - 3x + 2}\)
Decompose the proper fraction:
\(\frac{5x + 6}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}\)
\(5x + 6 = A(x - 2) + B(x - 1)\)
Put \(x = 1 \implies 11 = -A \implies A = -11\)
Put \(x = 2 \implies 16 = B\)
Integrate:
\(\int \left( 1 - \frac{11}{x - 1} + \frac{16}{x - 2} \right) dx = x - 11 \log |x - 1| + 16 \log |x - 2| + C\)
In simple words: Perform long division first when degrees are equal, then apply partial fractions to the remainder.

Exam Tip: Never skip polynomial 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 \(x = -2 \implies 4 - 2 + 1 = C(4) \implies 4C = 3 \implies C = \frac{3}{4}\)
Compare coefficients of \(x^2\) (\(A + C = 1\)): \(A = 1 - \frac{3}{4} = \frac{1}{4}\)
Integrate:
\(\int \left( \frac{1/4}{x} + \frac{1/2}{x^2} + \frac{3/4}{x + 2} \right) dx = \frac{1}{4} \log |x| - \frac{1}{2x} + \frac{3}{4} \log |x + 2| + C\)
In simple words: A squared term in the denominator requires both a single power and a squared power in the partial fraction expansion.

Exam Tip: Match every power of repeated factors correctly to avoid missing terms.

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 \(x = -3 \implies 9 + 1 = C(-4)^2 \implies 16C = 10 \implies C = \frac{5}{8}\)
Compare coefficients of \(x^2\) (\(A + C = 1\)): \(A = 1 - \frac{5}{8} = \frac{3}{8}\)
Integrate:
\(\int \left( \frac{3/8}{x - 1} + \frac{1/2}{(x - 1)^2} + \frac{5/8}{x + 3} \right) dx = \frac{3}{8} \log |x - 1| - \frac{1}{2(x - 1)} + \frac{5}{8} \log |x + 3| + C\)
In simple words: Set up partial fractions with linear and squared terms for the repeated factor, then solve for coefficients.

Exam Tip: Substituting roots simplifies coefficient finding for linear factors.

(iv) Integration using Partial Fraction - LEVEL III

Question 1. \(\displaystyle\int \frac{8}{(x + 2)(x^2 + 4)} \, dx\)
Answer: Decompose using a linear and an irreducible 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 \(x = -2 \implies 8 = A(8) \implies A = 1\)
Compare coefficients of \(x^2\) (\(A + B = 0\)): \(B = -1\)
Compare constant terms (\(4A + 2C = 8\)): \(4(1) + 2C = 8 \implies 2C = 4 \implies C = 2\)
Integrate term by term:
\(\int \left( \frac{1}{x + 2} + \frac{-x + 2}{x^2 + 4} \right) dx = \log |x + 2| - \frac{1}{2} \log |x^2 + 4| + \frac{2}{2} \tan^{-1}\left(\frac{x}{2}\right) + C\)
\(= \log |x + 2| - \frac{1}{2} \log |x^2 + 4| + \tan^{-1}\left(\frac{x}{2}\right) + C\)
In simple words: Irreducible quadratic factors require a linear numerator term \(Bx + C\) in partial fractions.

Exam Tip: Always pair quadratic factors with a linear numerator in partial fraction decompositions.

Question 2. \(\displaystyle\int \frac{dx}{\sin x + \sin 2x}\)
Answer: Rewrite \(\sin 2x = 2 \sin x \cos x\):
\(\displaystyle\int \frac{dx}{\sin x(1 + 2 \cos x)} = \int \frac{\sin x}{\sin^2 x (1 + 2 \cos x)} \, dx = \int \frac{\sin x}{(1 - \cos^2 x)(1 + 2 \cos x)} \, dx\)
Let \(\cos x = t \implies -\sin x \, dx = dt \implies \sin x \, dx = -dt\)
\(-\displaystyle\int \frac{dt}{(1 - t^2)(1 + 2t)} = \int \frac{dt}{(t - 1)(t + 1)(2t + 1)}\)
Decompose into partial fractions and integrate to find the solution in terms of logarithms.
In simple words: Convert the trigonometric expression into an algebraic rational function using substitution and partial fractions.

Exam Tip: Multiplying numerator and denominator by \(\sin x\) prepares the integral for the substitution \(\cos x = t\).

Question 3. \(\displaystyle\int \frac{1}{1 + x^3} \, dx\)
Answer: Factor the denominator: \(1 + x^3 = (1 + x)(1 - x + x^2)\)
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 \(x = -1 \implies 1 = A(1 + 1 + 1) \implies 3A = 1 \implies A = \frac{1}{3}\)
Compare coefficients of \(x^2\) (\(A + B = 0\)): \(B = -\frac{1}{3}\)
Compare constant terms (\(A + C = 1\)): \(C = 1 - \frac{1}{3} = \frac{2}{3}\)
Integrate:
\(\frac{1}{3} \log |1 + x| - \frac{1}{6} \log |1 - x + x^2| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right) + C\)
In simple words: Factor the sum of cubes and split into a linear and an irreducible quadratic partial fraction.

Exam Tip: Memorize the factorization formula for sums and differences of cubes.

(v) Integration by Parts - LEVEL I

Question 1. \(\displaystyle\int x \sec^2 x \, dx\)
Answer: Use Integration by Parts formula \(\int u v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx\):
Let \(u = x\) (algebraic) and \(v = \sec^2 x\) (trigonometric).
\(= x \int \sec^2 x \, dx - \int \left( \frac{d}{dx}(x) \int \sec^2 x \, dx \right) dx\)
\(= x \tan x - \int 1 \cdot \tan x \, dx\)
\(= x \tan x - \log |\sec x| + C\)
In simple words: Apply the ILATE rule to choose the first function \(u = x\), then integrate by parts.

Exam Tip: ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) is the standard priority order for choosing \(u\).

Question 2. \(\displaystyle\int \log x \, dx\)
Answer: Rewrite as \(\int (\log x) \cdot 1 \, dx\)
Let \(u = \log x\) and \(v = 1\):
\(= (\log x) \int 1 \, dx - \int \left( \frac{d}{dx}(\log x) \int 1 \, dx \right) dx\)
\(= x \log x - \int \frac{1}{x} \cdot x \, dx\)
\(= x \log x - \int 1 \, dx\)
\(= x \log x - x + C\)
In simple words: Treat the logarithmic function as multiplied by 1 and integrate by parts.

Exam Tip: A standalone logarithmic function is always integrated by parts by taking 1 as the second function.

Question 3. \(\displaystyle\int e^x (\tan x + \log \sec x) \, dx\)
Answer: Recall the standard result: \(\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C\)
Here, let \(f(x) = \log \sec x\). Then \(f'(x) = \frac{1}{\sec x} \sec x \tan x = \tan x\).
Thus, the integral evaluates directly to:
\(= e^x \log \sec x + C\)
In simple words: Recognize the exponential form \(e^x(f(x) + f'(x))\) for an immediate solution.

Exam Tip: Spotting the \(e^x[f(x) + f'(x)]\) pattern saves considerable calculation time.

(v) Integration by Parts - LEVEL II

Question 1. \(\displaystyle\int \sin^{-1} x \, dx\)
Answer: Rewrite as \(\int (\sin^{-1} x) \cdot 1 \, dx\)
Let \(u = \sin^{-1} x\) and \(v = 1\):
\(= (\sin^{-1} x)x - \int \frac{1}{\sqrt{1 - x^2}} \cdot x \, dx\)
For the remaining integral, let \(1 - x^2 = t \implies -2x \, dx = dt\):
\(= x \sin^{-1} x + \sqrt{1 - x^2} + C\)
In simple words: Treat inverse trigonometric functions as the first function and 1 as the second function in integration by parts.

Exam Tip: Inverse trigonometric functions without an algebraic factor are always integrated by taking 1 as the second function.

Question 2. \(\displaystyle\int x^2 \sin^{-1} x \, dx\)
Answer: Let \(u = \sin^{-1} x\) and \(v = x^2\):
\(= \sin^{-1} x \left(\frac{x^3}{3}\right) - \int \frac{1}{\sqrt{1 - x^2}} \cdot \frac{x^3}{3} \, dx\)
\(= \frac{x^3}{3} \sin^{-1} x + \frac{1}{3} \int \frac{-x^3}{\sqrt{1 - x^2}} \, dx\)
Substitute \(x^2 = t \implies 2x \, dx = dt\):
Completing the integration yields:
\(= \frac{x^3}{3} \sin^{-1} x + \frac{1}{9}(2 + x^2)\sqrt{1 - x^2} + C\)
In simple words: Use integration by parts with the inverse trig function as \(u\) and the polynomial as \(v\).

Exam Tip: Polynomials multiplied by inverse trig functions are standard candidates for integration by parts.

Question 3. \(\displaystyle\int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx\)
Answer: Let \(\sin^{-1} x = t \implies x = \sin t\) and \(dx = \cos t \, dt\)
Substituting into the integral:
\(\displaystyle\int \frac{\sin t \cdot t}{\cos t} \cdot \cos t \, dt = \int t \sin t \, dt\)
Using integration by parts on \(t \sin t\):
\(= t(-\cos t) - \int 1 \cdot (-\cos t) \, dt = -t \cos t + \sin t + C\)
Replacing \(t\) with \(\sin^{-1} x\) and \(\cos t\) with \(\sqrt{1 - x^2}\):
\(= -x \sin^{-1} x + \sqrt{1 - x^2} + C\)
In simple words: Substitution transforms this into a simple algebraic-trigonometric integration by parts problem.

Exam Tip: Substituting the inverse trigonometric function first can drastically simplify seemingly complex integrands.

Question 4. \(\displaystyle\int \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) dx\)
Answer: Recall the trigonometric substitution: let \(x = \tan \theta \implies dx = \sec^2 \theta \, dt\)
Then \(\frac{1 - x^2}{1 + x^2} = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta\)
Thus, \(\cos^{-1}(\cos 2\theta) = 2\theta = 2 \tan^{-1} x\)
The integral becomes:
\(\displaystyle\int 2 \tan^{-1} x \, dx = 2 \left[ x \tan^{-1} x - \int \frac{x}{1 + x^2} \, dx \right]\)
\(= 2x \tan^{-1} x - \log(1 + x^2) + C\)
In simple words: Simplify the inverse trigonometric function using standard double-angle substitutions before integrating.

Exam Tip: Always simplify inverse trigonometric expressions using substitutions before applying calculus rules.

Question 5. \(\displaystyle\int \sec^3 x \, dx\)
Answer: Write as \(\int \sec x \cdot \sec^2 x \, dx\) and apply integration by parts with \(u = \sec x\) and \(v = \sec^2 x\):
\(= \sec x \tan x - \int (\sec x \tan x) \tan x \, dx\)
\(= \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx\)
\(= \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx\)
Moving \(\int \sec^3 x \, dx\) to the left hand side:
\(2 \int \sec^3 x \, dx = \sec x \tan x + \log |\sec x + \tan x|\)
\(\int \sec^3 x \, dx = \frac{1}{2} [\sec x \tan x + \log |\sec x + \tan x|] + C\)
In simple words: This is a classic cyclic integral where the original integral reappears on the right and is solved by algebraic transposition.

Exam Tip: Memorize the reduction/cyclic result for \(\sec^3 x\) as it appears frequently in exams.

(v) Integration by Parts - LEVEL III

Question 1. \(\displaystyle\int \cos(\log x) \, dx\)
Answer: Let \(\log x = t \implies x = e^t \implies dx = e^t \, dt\)
The integral becomes \(\displaystyle\int e^t \cos t \, dt\)
Using the standard cyclic formula for \(\int e^t \cos t \, dt\):
\(= \frac{e^t}{2} (\sin t + \cos t) + C\)
Replacing \(t\) with \(\log x\) and \(e^t\) with \(x\):
\(= \frac{x}{2} [\sin(\log x) + \cos(\log x)] + C\)
In simple words: Substitute \(\log x = t\) to convert the logarithmic argument into an exponential product integral.

Exam Tip: Exponential times trigonometric integrals are cyclic and require integration by parts twice.

Question 2. \(\displaystyle\int \frac{e^x(1 + x)}{(2 + x)^2} \, dx\)
Answer: Adjust the numerator to match \(2 + x\):
\(1 + x = (2 + x) - 1\)
\(\displaystyle\int e^x \left[ \frac{2 + x}{(2 + x)^2} - \frac{1}{(2 + x)^2} \right] dx = \int e^x \left[ \frac{1}{2 + x} - \frac{1}{(2 + x)^2} \right] dx\)
Let \(f(x) = \frac{1}{2 + x}\). Then \(f'(x) = -\frac{1}{(2 + x)^2}\).
Since the integral is in the form \(\int e^x [f(x) + f'(x)] \, dx\), the solution is:
\(= \frac{e^x}{2 + x} + C\)
In simple words: Manipulate the numerator to split the fraction into a function and its exact derivative.

Exam Tip: Adjusting numerators by adding and subtracting constants helps reveal the \(e^x(f(x) + f'(x))\) form.

Question 3. \(\displaystyle\int \frac{\log x}{(1 + \log x)^2} \, dx\)
Answer: Let \(\log x = t \implies x = e^t \implies dx = e^t \, dt\)
The integral becomes \(\displaystyle\int \frac{t}{(1 + t)^2} e^t \, dt\)
Rewrite the numerator as \(t = (1 + t) - 1\):
\(\displaystyle\int e^t \left[ \frac{1}{1 + t} - \frac{1}{(1 + t)^2} \right] dt\)
Using the standard result for \(e^t[f(t) + f'(t)]\):
\(= e^t \left( \frac{1}{1 + t} \right) + C\)
Replacing \(t\) with \(\log x\) and \(e^t\) with \(x\):
\(= \frac{x}{1 + \log x} + C\)
In simple words: Substitute \(\log x = t\) to transform the logarithmic integral into an exponential form.

Exam Tip: Logarithmic substitutions often convert complicated expressions into standard exponential forms.

Question 4. \(\displaystyle\int \frac{x^2 + 2\sin x}{1 + \cos 2x} \, e^x \, dx\)
Answer: Use the identity \(1 + \cos 2x = 2 \cos^2 x\):
\(\displaystyle\int e^x \left[ \frac{x^2 + 2\sin x}{2 \cos^2 x} \right] dx = \int e^x \left[ \frac{x^2}{2 \cos^2 x} + \frac{2\sin x}{2 \cos^2 x} \right] dx\)
\(= \int e^x \left[ \frac{1}{2} x^2 \sec^2 x + \tan x \sec x \right] dx\) (Further algebraic splitting reveals the derivative structure).
In simple words: Expand trigonometric denominators and separate terms to find the function and its derivative.

Exam Tip: Double-angle formulas simplify trigonometric denominators when multiplied by exponential functions.

Question 5. \(\displaystyle\int e^{2x} \cos 3x \, dx\)
Answer: Apply integration by parts twice or use the standard formula for \(\int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C\):
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: Exponential and trigonometric products are cyclic integrals that can be solved using the direct integration formula.

Exam Tip: Memorize the direct formula for integrals of the form \(e^{ax} \sin bx\) and \(e^{ax} \cos bx\).

(vi) Some Special Integrals - LEVEL I

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

Exam Tip: Memorize the three fundamental square-root standard integrals to solve these instantly.

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

Exam Tip: Always divide the final result by the linear inner coefficient of \(x\).

(vi) Some Special Integrals - LEVEL II

Question 1. \(\displaystyle\int \sqrt{x^2 + 4x + 6} \, dx\)
Answer: Complete the square inside the radical:
\(x^2 + 4x + 6 = (x + 2)^2 + 2 = (x + 2)^2 + (\sqrt{2})^2\)
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 formula.

Exam Tip: Completing the square reduces quadratic radicals into standard recognizable forms.

Question 2. \(\displaystyle\int \sqrt{1 - 4x - x^2} \, dx\)
Answer: Complete the square inside the radical:
\(1 - 4x - x^2 = 1 - (x^2 + 4x + 4 - 4) = 1 + 4 - (x + 2)^2 = 5 - (x + 2)^2 = (\sqrt{5})^2 - (x + 2)^2\)
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 negative quadratic radical and apply the inverse sine square-root formula.

Exam Tip: Be careful with the sign distribution when completing the square with a negative coefficient of \(x^2\).

(vi) Some Special Integrals - LEVEL III

Question 1. \(\displaystyle\int (x + 1)\sqrt{1 - x - x^2} \, dx\)
Answer: Express the linear factor in terms of the derivative of the quadratic expression inside the radical: \(\frac{d}{dx}(1 - x - x^2) = -1 - 2x\)
Let \(x + 1 = A(-1 - 2x) + B = -2A x - A + B\)
Equating coefficients: \(-2A = 1 \implies A = -\frac{1}{2}\), and \(-A + B = 1 \implies \frac{1}{2} + B = 1 \implies B = \frac{1}{2}\)
Substitute into the integral:
\(\int \left[ -\frac{1}{2}(-1 - 2x) + \frac{1}{2} \right] \sqrt{1 - x - x^2} \, dx\)
\(= -\frac{1}{2} \int (-1 - 2x)\sqrt{1 - x - x^2} \, dx + \frac{1}{2} \int \sqrt{1 - x - x^2} \, dx\)
Evaluate each part using standard formulas to get the complete solution.
In simple words: Split the integral into a derivative-substitution part and a completing-the-square radical part.

Exam Tip: Linear factors multiplying square roots are always solved by expressing the linear part as a derivative of the inner quadratic.

Question 2. \(\displaystyle\int (x - 5)\sqrt{x^2 + x} \, dx\)
Answer: Let \(x - 5 = A \frac{d}{dx}(x^2 + x) + B = A(2x + 1) + B\)
Equating coefficients: \(2A = 1 \implies A = \frac{1}{2}\), and \(A + B = -5 \implies \frac{1}{2} + B = -5 \implies B = -\frac{11}{2}\)
Substitute into the integral and split into two parts:
\(\frac{1}{2} \int (2x + 1)\sqrt{x^2 + x} \, dx - \frac{11}{2} \int \sqrt{x^2 + x} \, dx\)
Evaluate the first part via substitution and the second part by completing the square.
In simple words: Use derivative splitting for the linear term to separate the integral into manageable components.

Exam Tip: Systematic coefficient matching guarantees correct splitting for algebraic-radical products.

(vii) Miscellaneous Questions - LEVEL II

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

Exam Tip: Integrals with constants and cosines in the denominator are standardly solved using the \(\tan x\) half-angle substitution.

Question 2. \(\displaystyle\int \frac{1}{3 + \sin 2x} \, dx\)
Answer: Use the half-angle substitution: \(\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}\)
\(\displaystyle\int \frac{dx}{3 + \frac{2\tan x}{1 + \tan^2 x}} = \int \frac{(1 + \tan^2 x) \, dx}{3(1 + \tan^2 x) + 2\tan x}\)
\(= \displaystyle\int \frac{\sec^2 x \, dx}{3\tan^2 x + 2\tan x + 3}\)
Let \(\tan x = t \implies \sec^2 x \, dx = dt\)
\(\displaystyle\int \frac{dt}{3t^2 + 2t + 3} = \frac{1}{3} \int \frac{dt}{t^2 + \frac{2}{3}t + 1}\)
Complete the square in the denominator:
\(t^2 + \frac{2}{3}t + 1 = \left(t + \frac{1}{3}\right)^2 + \frac{8}{9} = \left(t + \frac{1}{3}\right)^2 + \left(\frac{2\sqrt{2}}{3}\right)^2\)
Using the standard arctangent formula:
\(= \frac{1}{3} \cdot \frac{3}{2\sqrt{2}} \tan^{-1}\left(\frac{t + \frac{1}{3}}{\frac{2\sqrt{2}}{3}}\right) + C = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{3\tan x + 1}{2\sqrt{2}}\right) + C\)
In simple words: Express sine in terms of tangent half-angle, substitute \(\tan x = t\), and complete the square.

Exam Tip: Half-angle tangent substitutions transform rational sine/cosine expressions into algebraic quadratics.

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

Exam Tip: Integrals with sums of squares of sine and cosine in the denominator are always solved by dividing by \(\cos^2 x\).

Question 4. \(\displaystyle\int \frac{dx}{1 + 3 \sin^2 x + 8 \cos^2 x}\)
Answer: Divide numerator and denominator by \(\cos^2 x\):
\(\displaystyle\int \frac{\sec^2 x \, dx}{\frac{1}{\cos^2 x} + 3\frac{\sin^2 x}{\cos^2 x} + 8} = \int \frac{\sec^2 x \, dx}{\sec^2 x + 3\tan^2 x + 8}\)
Replace \(\sec^2 x = 1 + \tan^2 x\):
\(\displaystyle\int \frac{\sec^2 x \, dx}{1 + \tan^2 x + 3\tan^2 x + 8} = \int \frac{\sec^2 x \, dx}{4\tan^2 x + 9}\)
Let \(\tan x = t \implies \sec^2 x \, dx = dt\)
\(\displaystyle\int \frac{dt}{4t^2 + 9} = \frac{1}{4} \int \frac{dt}{t^2 + (\frac{3}{2})^2} = \frac{1}{4} \cdot \frac{2}{3} \tan^{-1}\left(\frac{2t}{3}\right) + C\)
\(= \frac{1}{6} \tan^{-1}\left(\frac{2\tan x}{3}\right) + C\)
In simple words: Divide by \(\cos^2 x\) and convert any remaining secant squared terms into tangents.

Exam Tip: Substituting \(\sec^2 x = 1 + \tan^2 x\) unifies all terms into powers of tangent.

Question 5. \(\displaystyle\int \frac{dx}{\sin^4 x + \cos^4 x}\)
Answer: Divide numerator and denominator by \(\cos^4 x\):
\(\displaystyle\int \frac{\sec^4 x \, dx}{\tan^4 x + 1} = \int \frac{(1 + \tan^2 x)\sec^2 x \, dx}{\tan^4 x + 1}\)
Let \(\tan x = t \implies \sec^2 x \, dx = dt\)
\(\displaystyle\int \frac{(1 + t^2) \, dt}{t^4 + 1} = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \, dt = \int \frac{1 + \frac{1}{t^2}}{(t - \frac{1}{t})^2 + 2} \, dt\)
Let \(t - \frac{1}{t} = u \implies (1 + \frac{1}{t^2}) \, dt = du\)
\(\displaystyle\int \frac{du}{u^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C\)
Replacing \(u\) and \(t\):
\(= \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x - \cot x}{\sqrt{2}}\right) + C\)
In simple words: Divide by \(\cos^4 x\), substitute \(\tan x = t\), and complete the square in the reciprocal form.

Exam Tip: Integrals of \(\sin^4 x + \cos^4 x\) in denominators are standardly solved by dividing by \(\cos^4 x\) and using the substitution \(t - \frac{1}{t}\).

Question 6. \(\displaystyle\int \frac{\sec x}{\sec x + 4 \tan x} \, dx\)
Answer: Convert secant and tangent into sine and cosine:
\(\displaystyle\int \frac{\frac{1}{\cos x}}{\frac{1}{\cos x} + \frac{4\sin x}{\cos x}} \, dx = \int \frac{1}{1 + 4\sin x} \, dx\)
Use the half-angle substitution for sine: \(\sin x = \frac{2\tan(x/2)}{1 + \tan^2(x/2)}\)
\(\displaystyle\int \frac{dx}{1 + \frac{8\tan(x/2)}{1 + \tan^2(x/2)}} = \int \frac{(1 + \tan^2(x/2)) \, dx}{1 + \tan^2(x/2) + 8\tan(x/2)}\)
Let \(\tan(x/2) = t \implies \frac{1}{2}\sec^2(x/2) \, dx = dt \implies (1 + t^2) \, dx = 2 \, dt\)
Completing the square in the denominator yields the final logarithmic/arctangent solution.
In simple words: Convert to sine and cosine, then apply half-angle tangent substitutions.

Exam Tip: Linear combinations of sine and cosine are universally solvable using half-angle tangent substitutions.

(vii) Miscellaneous Questions - LEVEL III

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

Exam Tip: Integrals of the form \(\frac{a\sin x + b\cos x}{c\cos x + d\sin x}\) are always solved using numerator-derivative splitting.

Question 2. \(\displaystyle\int \frac{1}{1 - \tan x} \, dx\)
Answer: Rewrite \(\tan x = \frac{\sin x}{\cos x}\):
\(\displaystyle\int \frac{1}{1 - \frac{\sin x}{\cos x}} \, dx = \int \frac{\cos x}{\cos x - \sin x} \, dx\)
Multiply and divide numerator by 2, and rewrite the numerator as \(\cos x - \sin x + \cos x + \sin x\):
\(= \frac{1}{2} \int \frac{(\cos x - \sin x) + (\cos x + \sin x)}{\cos x - \sin x} \, dx\)
\(= \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\cos x + \sin x}{\cos x - \sin x} \, dx\)
\(= \frac{1}{2}x - \frac{1}{2}\log|\cos x - \sin x| + C\)
In simple words: Convert tangent to sine and cosine, then split the numerator into the denominator and its derivative.

Exam Tip: Standard trigonometric trick: writing the numerator as sum/difference of denominator and its derivative.

Question 3. \(\displaystyle\int \frac{x^4}{x^4 - 1} \, dx\)
Answer: Perform polynomial division or adjust the numerator:
\(\frac{x^4}{x^4 - 1} = \frac{x^4 - 1 + 1}{x^4 - 1} = 1 + \frac{1}{x^4 - 1}\)
Decompose the proper fraction using partial fractions:
\(\frac{1}{x^4 - 1} = \frac{1}{(x^2 - 1)(x^2 + 1)} = \frac{1}{2}\left( \frac{1}{x^2 - 1} - \frac{1}{x^2 + 1} \right)\)
Integrate term by term:
\(\int \left( 1 + \frac{1}{2(x^2 - 1)} - \frac{1}{2(x^2 + 1)} \right) dx = x + \frac{1}{4}\log\left|\frac{x - 1}{x + 1}\right| - \frac{1}{2}\tan^{-1}x + C\)
In simple words: Divide the improper fraction, then apply partial fractions to the remainder.

Exam Tip: Always divide first when the degree of the numerator equals or exceeds the denominator.

Question 4. \(\displaystyle\int \frac{x^2 + 1}{x^4 + x^2 + 1} \, dx\)
Answer: Divide numerator and denominator by \(x^2\):
\(\displaystyle\int \frac{1 + \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} \, dx = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 3} \, dx\)
Let \(x - \frac{1}{x} = t \implies (1 + \frac{1}{x^2}) \, dx = dt\)
\(\displaystyle\int \frac{dt}{t^2 + (\sqrt{3})^2} = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C\)
Replacing \(t\) with \(x - \frac{1}{x}\):
\(= \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}x}\right) + C\)
In simple words: Divide by \(x^2\) and use the substitution \(x - \frac{1}{x} = t\).

Exam Tip: Symmetrical rational functions of \(x^4, x^2, 1\) are solved by dividing by \(x^2\).

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

Exam Tip: Note the sign difference in the denominator substitution compared to the previous problem.

Question 6. \(\displaystyle\int \sqrt{\tan x} \, dx\)
Answer: Let \(\tan x = t^2 \implies \sec^2 x \, dx = 2t \, dt \implies dx = \frac{2t \, dt}{1 + t^4}\)
\(\displaystyle\int t \cdot \frac{2t}{1 + t^4} \, dt = \int \frac{2t^2}{1 + t^4} \, dt = \int \frac{(t^2 + 1) + (t^2 - 1)}{t^4 + 1} \, dt\)
Split into two integrals and divide numerator and denominator by \(t^2\), then integrate using standard tangent/logarithmic formulas.
In simple words: Substitute \(\tan x = t^2\) to convert the radical into a rational algebraic function.

Exam Tip: Square roots of tangent or cotangent are standardly integrated using the substitution \(t^2 = \tan x\).

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 \(5x^2 + 1 = t \implies 10x \, dx = dt \implies 2x \, dx = \frac{1}{5} \, dt\). Limits change from 1 to 6.
For the second part, use the standard inverse tangent formula with \(a = \frac{1}{\sqrt{5}}\):
\(= \left[ \frac{1}{5} \log |5x^2 + 1| + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5}x) \right]_{0}^{1}\)
\(= \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5}) - 0\)
In simple words: Evaluate the indefinite integral first, then apply the upper and lower limits according to the Second Fundamental Theorem of Calculus.

Exam Tip: Splitting fractions with linear numerators over quadratic denominators makes definite integration straightforward.

Question 2. \(\displaystyle\int_{0}^{\pi/2} \sqrt{\sin x} \cos^5 x \, dx\)
Answer: Rewrite \(\cos^5 x = \cos^4 x \cos x = (1 - \sin^2 x)^2 \cos x\)
Let \(\sin x = t \implies \cos x \, dx = dt\). When \(x = 0, t = 0\); when \(x = \pi/2, t = 1\).
\(\displaystyle\int_{0}^{1} t^{1/2} (1 - t^2)^2 \, dt = \int_{0}^{1} t^{1/2} (1 - 2t^2 + t^4) \, dt\)
\(= \displaystyle\int_{0}^{1} \left( t^{1/2} - 2t^{5/2} + t^{9/2} \right) dt\)
\(= \left[ \frac{t^{3/2}}{3/2} - 2\frac{t^{7/2}}{7/2} + \frac{t^{11/2}}{11/2} \right]_{0}^{1}\)
\(= \frac{2}{3} - \frac{4}{7} + \frac{2}{11} = \frac{154 - 132 + 42}{231} = \frac{64}{231}\)
In simple words: Substitute \(\sin x = t\), change the limits of integration, expand, and integrate term by term.

Exam Tip: Changing limits of integration during substitution avoids having to substitute back to the original variable.

Question 3. \(\displaystyle\int_{0}^{2} x \sqrt{x + 2} \, dx\)
Answer: Let \(x + 2 = t^2 \implies dx = 2t \, dt\). Also \(x = t^2 - 2\).
Limits: when \(x = 0, t = \sqrt{2}\); when \(x = 2, t = 2\).
\(\displaystyle\int_{\sqrt{2}}^{2} (t^2 - 2) \cdot t \cdot 2t \, dt = 2 \int_{\sqrt{2}}^{2} (t^4 - 2t^2) \, dt\)
\(= 2 \left[ \frac{t^5}{5} - \frac{2t^3}{3} \right]_{\sqrt{2}}^{2}\)
Evaluating at limits gives the final numerical answer.
In simple words: Use algebraic substitution for the radical term and adjust the limits accordingly.

Exam Tip: Substituting radical expressions simplifies integrands containing square roots of linear binomials.

(i) Definite Integrals based upon types of indefinite integrals - LEVEL II

Question 1. \(\displaystyle\int_{1}^{2} \frac{5x^2}{x^2 + 4x + 3} \, dx\)
Answer: Perform polynomial division since numerator and denominator degrees are equal:
\(\frac{5x^2}{x^2 + 4x + 3} = 5 - \frac{20x + 15}{(x + 1)(x + 3)}\)
Decompose the proper fraction using partial fractions, then apply the limits from 1 to 2.
\(= \left[ 5x - \frac{5}{2}\log|x + 1| - \frac{35}{2}\log|x + 3| \right]_{1}^{2}\)
Evaluating at upper and lower limits yields the exact numerical result.
In simple words: Divide the improper fraction first, then apply partial fractions and integrate across the definite limits.

Exam Tip: Polynomial division is mandatory before partial fraction expansion in definite integrals.

Question 2. \(\displaystyle\int_{1/e}^{e} \left( \frac{1}{x} - \frac{1}{2x^2} \right) e^{2x + \frac{1}{x}} \, dx\)
Answer: Let \(2x + \frac{1}{x} = t \implies \left( 2 - \frac{1}{x^2} \right) dx = dt \implies \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx = \frac{1}{2} \, dt\)
Change limits: when \(x = 1/e, t = \frac{2}{e} + e\); when \(x = e, t = 2e + \frac{1}{e}\).
Substitute into the integral to evaluate the resulting exponential definite integral.
In simple words: Substitute the complex exponent expression to simplify the integral into a basic exponential form.

Exam Tip: Look for inner derivative relations in exponential exponents to apply clean substitutions.

(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 the definition of definite integral as limit of sum: \(\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)\) where \(h = \frac{b - a}{n}\).
Here \(a = 0, b = 2\), so \(h = \frac{2}{n}\). \(f(x) = x + 2\).
\(f(0 + rh) = \frac{2r}{n} + 2\)
\(\sum_{r=0}^{n-1} \left( \frac{2r}{n} + 2 \right) = \frac{2}{n} \frac{(n-1)n}{2} + 2n = n - 1 + 2n = 3n - 1\)
Multiplying by \(h = \frac{2}{n}\) and taking \(\lim_{n \to \infty}\):
\(\lim_{n \to \infty} \frac{2}{n} (3n - 1) = \lim_{n \to \infty} \left( 6 - \frac{2}{n} \right) = 6\)
In simple words: Replace the integral with a Riemann sum limit, evaluate the series, and take the limit as \(n \to \infty\).

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 \(a = 0, b = 4\), so \(h = \frac{4}{n}\). \(f(x) = 1 + x\).
\(f(rh) = 1 + \frac{4r}{n}\)
Summing from \(r = 0\) to \(n-1\):
\(\sum_{r=0}^{n-1} \left( 1 + \frac{4r}{n} \right) = n + \frac{4}{n} \frac{(n-1)n}{2} = n + 2(n - 1) = 3n - 2\)
Multiplying by \(h = \frac{4}{n}\) and taking limit as \(n \to \infty\):
\(\lim_{n \to \infty} \frac{4}{n} (3n - 2) = 12\)
In simple words: Set up the Riemann sum, apply arithmetic progression sums, and evaluate the limit.

Exam Tip: Double-check interval widths \(h\) and summation boundaries before taking the limit.

(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 \(I = \displaystyle\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx\) ---- (1)
Using the property \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx\):
\(I = \displaystyle\int_{0}^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\cos(\pi/2 - x)} + \sqrt{\sin(\pi/2 - x)}} \, dx = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx\) ---- (2)
Adding (1) and (2):
\(2I = \displaystyle\int_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\)
In simple words: Use the king property \(f(a + b - x)\) to add the integral to itself, simplifying the integrand to 1.

Exam Tip: This is one of the most famous definite integral properties; whenever sine and cosine appear symmetrically, add \(I\) and \(I'\).

Question 2. \(\displaystyle\int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} \, dx\)
Answer: Let \(I = \displaystyle\int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} \, dx\) ---- (1)
Using property \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx\) where \(a + b = 4\):
\(I = \displaystyle\int_{1}^{3} \frac{\sqrt{4 - (4 - x)}}{\sqrt{4 - x} + \sqrt{4 - (4 - x)}} \, dx = \int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4 - x} + \sqrt{x}} \, dx\) ---- (2)
Adding (1) and (2):
\(2I = \displaystyle\int_{1}^{3} 1 \, dx = [x]_{1}^{3} = 3 - 1 = 2\)
\(I = 1\)
In simple words: Apply the reflection property \(f(a+b-x)\) so that the sum of the two integrands equals 1.

Exam Tip: The property \(\int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx = \frac{b-a}{2}\) is a direct shortcut for objective questions.

Question 3. \(\displaystyle\int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx\)
Answer: Let \(I = \displaystyle\int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx\) ---- (1)
Using property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\):
\(I = \displaystyle\int_{0}^{\pi/2} \frac{\sin^4(\pi/2 - x)}{\sin^4(\pi/2 - x) + \cos^4(\pi/2 - x)} \, dx = \int_{0}^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \, dx\) ---- (2)
Adding (1) and (2):
\(2I = \displaystyle\int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\)
In simple words: Apply the property substituting \(\pi/2 - x\) for \(x\) and add the two identical integrals together.

Exam Tip: Symmetric trigonometric definite integrals from 0 to \(\pi/2\) almost always evaluate to \(\pi/4\) using this property.

(iv) Integration of modulus function - LEVEL III

Question 1. \(\displaystyle\int_{2}^{5} (|x - 2| + |x - 3| + |x - 4|) \, dx\)
Answer: Break the integral at the critical points where the modulus expressions change sign: \(x = 2, 3, 4\).
Intervals: \([2, 3]\), \([3, 4]\), and \([4, 5]\).
- For \(x \in [2, 3]\): \(|x-2| = x-2\), \(|x-3| = -(x-3)\), \(|x-4| = -(x-4)\)
- For \(x \in [3, 4]\): \(|x-2| = x-2\), \(|x-3| = x-3\), \(|x-4| = -(x-4)\)
- For \(x \in [4, 5]\): \(|x-2| = x-2\), \(|x-3| = x-3\), \(|x-4| = x-4\)
Integrate across each sub-interval and sum the results to get the final numerical value.
In simple words: Split the absolute value integral at the points where each inner expression equals zero.

Exam Tip: Always analyze the sign of each modulus term in every sub-interval before integrating.