CBSE Class 12 Mathematics Integrals Assignment Set 03

Question 1. Evaluate the integral: \(\int \frac{x^4 + 1}{x^2 + 1} \, dx\)
Answer: Divide \(x^4 + 1\) by \(x^2 + 1\) or rewrite the numerator:
\(x^4 + 1 = (x^4 - 1) + 2 = (x^2 - 1)(x^2 + 1) + 2\)
Substitute this into the integral:
\(\int \frac{(x^2 - 1)(x^2 + 1) + 2}{x^2 + 1} \, dx = \int \left( x^2 - 1 + \frac{2}{x^2 + 1} \right) dx\)
Integrate term by term:
\(= \frac{x^3}{3} - x + 2 \tan^{-1}(x) + C\)
In simple words: Split the improper rational function using algebraic manipulation and standard integration formulas.

Exam Tip: Whenever the degree of the numerator is greater than or equal to the denominator, perform polynomial division first.

Question 2. Evaluate the integral: \(\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx\)
Answer: Rewrite the denominator by multiplying and dividing by \(\cos x\):
\(\sin x \cos x = \frac{\sin x \cos x}{\cos x} \cdot \cos x = \tan x \cos^2 x = \frac{\tan x}{\sec^2 x}\)
Alternatively, divide numerator and denominator by \(\cos^2 x\):
\(\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int \frac{\frac{\sqrt{\tan x}}{\cos^2 x}}{\frac{\sin x \cos x}{\cos^2 x}} \, dx = \int \frac{\sec^2 x \sqrt{\tan x}}{\tan x} \, dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} \, dx\)
Let \(u = \tan x\), then \(du = \sec^2 x \, dx\):
\(= \int u^{-1/2} \, du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{\tan x} + C\)
In simple words: Convert the trigonometric expression into terms of \(\tan x\) and \(\sec^2 x\) to use substitution.

Exam Tip: Recognizing that \(\frac{1}{\sin x \cos x} = \sec^2 x \csc x\) or converting to tan-sec form is key for trigonometric integrals.

Question 3. Evaluate the integral: \(\int \frac{1}{1 + \sqrt{x}} \, dx\)
Answer: Let \(x = t^2\), then \(dx = 2t \, dt\):
\(\int \frac{1}{1 + \sqrt{x}} \, dx = \int \frac{2t}{1 + t} \, dt = 2 \int \frac{t + 1 - 1}{1 + t} \, dt\)
\(= 2 \int \left( 1 - \frac{1}{1 + t} \right) dt = 2(t - \ln|1 + t|) + C\)
Substitute back \(t = \sqrt{x}\):
\(= 2(\sqrt{x} - \ln|1 + \sqrt{x}|) + C\)
In simple words: Use substitution to eliminate the square root from the denominator.

Exam Tip: Rationalizing substitutions like \(x = t^2\) simplify integrands containing square roots of the variable.

Question 4. Evaluate the integral: \(\int \frac{1}{\sqrt{x(1 - 2x)}} \, dx\)
Answer: Expand inside the square root:
\(x(1 - 2x) = x - 2x^2 = -2 \left( x^2 - \frac{1}{2}x \right)\)
Complete the square:
\(x^2 - \frac{1}{2}x = \left( x - \frac{1}{4} \right)^2 - \frac{1}{16}\)
So, \(x(1 - 2x) = \frac{1}{8} - 2\left( x - \frac{1}{4} \right)^2 = \frac{1}{8} - \left(\sqrt{2}\left(x - \frac{1}{4}\right)\right)^2\)
Alternatively, substitute \(\sqrt{x} = t\) or use standard forms. Let \(x = \frac{1}{2}\sin^2 \theta\) or complete the square directly:
\(\int \frac{1}{\sqrt{x - 2x^2}} \, dx = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\frac{1}{16} - \left(x - \frac{1}{4}\right)^2}} \, dx\)
\(= \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{x - \frac{1}{4}}{\frac{1}{4}}\right) + C = \frac{1}{\sqrt{2}} \sin^{-1}(4x - 1) + C\)
In simple words: Complete the square in the quadratic expression under the root and apply standard inverse sine integral formulas.

Question 5. Evaluate the integral: \(\int \frac{\sin x}{\sin 3x} \, dx\)
Answer: Use the triple angle formula for sine: \(\sin 3x = 3\sin x - 4\sin^3 x\).
\(\int \frac{\sin x}{3\sin x - 4\sin^3 x} \, dx = \int \frac{1}{3 - 4\sin^2 x} \, dx\)
Divide numerator and denominator by \(\cos^2 x\):
\(= \int \frac{\sec^2 x}{3\sec^2 x - 4\tan^2 x} \, dx = \int \frac{\sec^2 x}{3(1 + \tan^2 x) - 4\tan^2 x} \, dx\)
\(= \int \frac{\sec^2 x}{3 - \tan^2 x} \, dx\)
Let \(u = \tan x\), then \(du = \sec^2 x \, dx\):
\(= \int \frac{du}{3 - u^2} = \frac{1}{2\sqrt{3}} \ln\left| \frac{\sqrt{3} + u}{\sqrt{3} - u} \right| + C\)
Substitute back \(u = \tan x\):
\(= \frac{1}{2\sqrt{3}} \ln\left| \frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x} \right| + C\)
In simple words: Expand \(\sin 3x\), divide by \(\cos^2 x\), and use substitution with \(\tan x\).
Exam Tip: Standard substitution using \(\tan x\) works well after dividing numerator and denominator by \(\cos^2 x\).

Question 6. Evaluate the integral: \(\int \sin x \sqrt{1 + \cos 2x} \, dx\)
Answer: Use the trigonometric identity \(1 + \cos 2x = 2\cos^2 x\):
\(\int \sin x \sqrt{2\cos^2 x} \, dx = \sqrt{2} \int \sin x |\cos x| \, dx\)
Assuming \(\cos x \geq 0\), we have \(|\cos x| = \cos x\):
\(= \sqrt{2} \int \sin x \cos x \, dx = \frac{\sqrt{2}}{2} \int \sin 2x \, dx\)
\(= \frac{\sqrt{2}}{2} \left( -\frac{\cos 2x}{2} \right) + C = -\frac{\sqrt{2}}{4} \cos 2x + C\)
In simple words: Simplify the square root using double angle identities and integrate using simple substitution.

Exam Tip: Remember that \(\sqrt{\cos^2 x} = |\cos x|\), which simplifies based on the interval of integration.

Question 7. Evaluate the integral: \(\int \left( e^x + \frac{1}{e^x} \right)^2 dx\)
Answer: Expand the square:
\(\left( e^x + \frac{1}{e^x} \right)^2 = (e^x)^2 + 2(e^x)\left(\frac{1}{e^x}\right) + \left(\frac{1}{e^x}\right)^2 = e^{2x} + 2 + e^{-2x}\)
Integrate term by term:
\(\int \left( e^{2x} + 2 + e^{-2x} \right) dx = \frac{e^{2x}}{2} + 2x - \frac{e^{-2x}}{2} + C\)
In simple words: Expand the binomial square and integrate each exponential term individually.

Exam Tip: Always expand algebraic or exponential powers before attempting integration.

Question 8. Evaluate the integral: \(\int \csc^3 x \, dx\)
Answer: Integrate by parts taking \(u = \csc x\) and \(dv = \csc^2 x \, dx\):
\(du = -\csc x \cot x \, dx\), \(v = -\cot x\)
\(\int \csc^3 x \, dx = -\csc x \cot x - \int (-\cot x)(-\csc x \cot x) \, dx\)
\(= -\csc x \cot x - \int \csc x \cot^2 x \, dx\)
Use \(\cot^2 x = \csc^2 x - 1\):
\(= -\csc x \cot x - \int \csc x (\csc^2 x - 1) \, dx\)
\(= -\csc x \cot x - \int \csc^3 x \, dx + \int \csc x \, dx\)
Move \(\int \csc^3 x \, dx\) to the left side:
\(2 \int \csc^3 x \, dx = -\csc x \cot x + \ln|\csc x - \cot x| + C_1\)
\(\int \csc^3 x \, dx = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \ln|\csc x - \cot x| + C\)
In simple words: Use integration by parts and trigonometric identities to solve circular reduction integrals.

Question 9. Evaluate the definite integral: \(\int_{0}^{2} |x^2 + 2x - 3| \, dx\)
Answer: Factor the quadratic expression: \(x^2 + 2x - 3 = (x + 3)(x - 1)\).
The roots are \(x = -3\) and \(x = 1\).
In the interval \([0, 2]\), the expression changes sign at \(x = 1\):
- For \(x \in [0, 1)\), \(x^2 + 2x - 3 \leq 0\), so \(|x^2 + 2x - 3| = -(x^2 + 2x - 3)\)
- For \(x \in [1, 2]\), \(x^2 + 2x - 3 \geq 0\), so \(|x^2 + 2x - 3| = x^2 + 2x - 3\)
Split the integral at \(x = 1\):
\(\int_{0}^{2} |x^2 + 2x - 3| \, dx = \int_{0}^{1} -(x^2 + 2x - 3) \, dx + \int_{1}^{2} (x^2 + 2x - 3) \, dx\)
Evaluate the first integral:
\(\left[ -\left(\frac{x^3}{3} + x^2 - 3x\right) \right]_{0}^{1} = -\left(\frac{1}{3} + 1 - 3\right) = -\left(-\frac{5}{3}\right) = \frac{5}{3}\)
Evaluate the second integral:
\(\left[ \frac{x^3}{3} + x^2 - 3x \right]_{1}^{2} = \left(\frac{8}{3} + 4 - 6\right) - \left(\frac{1}{3} + 1 - 3\right) = \left(\frac{8}{3} - 2\right) - \left(-\frac{5}{3}\right) = \frac{2}{3} + \frac{5}{3} = \frac{7}{3}\)
Total sum: \(\frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4\)
In simple words: Find where the expression inside the absolute value changes sign and split the interval accordingly.
Exam Tip: Always check the roots of the quadratic expression to correctly determine the split points for modulus functions.

Question 10. Evaluate the integral: \(\int \frac{1}{1 + x^4} \, dx\)
Answer: Rewrite the numerator and denominator by splitting the numerator into terms involving \(x^2\):
\(\frac{1}{1 + x^4} = \frac{1}{2} \cdot \frac{(x^2 + 1) - (x^2 - 1)}{x^4 + 1} = \frac{1}{2} \left[ \frac{x^2 + 1}{x^4 + 1} - \frac{x^2 - 1}{x^4 + 1} \right]\)
Divide numerator and denominator of each term by \(x^2\):
\(= \frac{1}{2} \left[ \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} - \frac{1 - \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \right] = \frac{1}{2} \left[ \frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 2} - \frac{1 - \frac{1}{x^2}}{\left(x + \frac{1}{x}\right)^2 - 2} \right]\)
Integrate using standard formulas:
\(= \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) - \frac{1}{4\sqrt{2}} \ln\left| \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} \right| + C\)
In simple words: Express the numerator as a combination of derivatives of \(x - \frac{1}{x}\) and \(x + \frac{1}{x}\).

Exam Tip: This is a standard symmetric substitution method for reciprocal powers of \(x\).

Question 11. Evaluate the integral: \(\int \frac{\sin 2x}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx\)
Answer: Let \(t = a^2 \sin^2 x + b^2 \cos^2 x\).
Differentiating with respect to \(x\):
\(dt = (2a^2 \sin x \cos x - 2b^2 \cos x \sin x) \, dx = (a^2 - b^2)\sin 2x \, dx\)
Therefore, \(\sin 2x \, dx = \frac{dt}{a^2 - b^2}\).
Substitute into the integral:
\(\int \frac{1}{t} \cdot \frac{dt}{a^2 - b^2} = \frac{1}{a^2 - b^2} \ln|t| + C = \frac{1}{a^2 - b^2} \ln|a^2 \sin^2 x + b^2 \cos^2 x| + C\)
In simple words: The derivative of the denominator is a multiple of the numerator, so use direct substitution.

Exam Tip: Whenever the numerator is the derivative of the denominator, the integral is simply \(\ln|\text{denominator}|\).

Question 12. Evaluate the integral: \(\int \frac{1}{\cos 2x + 3\sin^2 x} \, dx\)
Answer: Use the identity \(\cos 2x = \cos^2 x - \sin^2 x\):
\(\cos 2x + 3\sin^2 x = \cos^2 x - \sin^2 x + 3\sin^2 x = \cos^2 x + 2\sin^2 x\)
Divide numerator and denominator by \(\cos^2 x\):
\(\int \frac{\sec^2 x}{1 + 2\tan^2 x} \, dx\)
Let \(u = \tan x\), then \(du = \sec^2 x \, dx\):
\(= \int \frac{du}{1 + 2u^2} = \frac{1}{2} \int \frac{du}{u^2 + \left(\frac{1}{\sqrt{2}}\right)^2}\)
\(= \frac{1}{2} \cdot \frac{1}{\frac{1}{\sqrt{2}}} \tan^{-1}\left(\frac{u}{\frac{1}{\sqrt{2}}}\right) + C = \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\tan x) + C\)
In simple words: Simplify the denominator using trigonometric identities and apply substitution with \(\tan x\).

Exam Tip: Converting denominators involving sine and cosine squares into tan form using \(\sec^2 x\) is a standard technique.

Question 13. Evaluate the integral: \(\int \frac{1}{\sin x + \sqrt{3}\cos x} \, dx\)
Answer: Divide and multiply by 2:
\(\sin x + \sqrt{3}\cos x = 2 \left( \frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x \right) = 2(\sin x \cos \frac{\pi}{3} + \cos x \sin \frac{\pi}{3}) = 2\sin\left(x + \frac{\pi}{3}\right)\)
Substitute into the integral:
\(\int \frac{1}{2\sin\left(x + \frac{\pi}{3}\right)} \, dx = \frac{1}{2} \int \csc\left(x + \frac{\pi}{3}\right) dx\)
\(= \frac{1}{2} \ln\left| \csc\left(x + \frac{\pi}{3}\right) - \cot\left(x + \frac{\pi}{3}\right) \right| + C\)
In simple words: Combine sine and gorgeous cosine terms into a single sine wave using harmonic addition formulas.

Exam Tip: For expressions of the form \(a\sin x + b\cos x\), multiply and divide by \(\sqrt{a^2 + b^2}\).

Question 14. Evaluate the integral: \(\int \frac{\log(\log x)}{x} \, dx\)
Answer: Let \(u = \log x\), then \(du = \frac{1}{x} \, dx\):
\(\int \log(\log x) \cdot \frac{1}{x} \, dx = \int \log(u) \, du\)
Integrate by parts taking \(\int \log(u) \, du\):
\(= u\log u - u + C\)
Substitute back \(u = \log x\):
\(= \log x \log(\log x) - \log x + C\)
In simple words: Use substitution for the inner logarithm, then integrate the outer logarithm by parts.

Question 15. Evaluate the integral: \(\int \frac{1}{(x - 3)\sqrt{x + 1}} \, dx\)
Answer: Let \(x + 1 = t^2\), then \(dx = 2t \, dt\) and \(x - 3 = t^2 - 4\):
\(\int \frac{2t}{(t^2 - 4)t} \, dt = 2 \int \frac{1}{t^2 - 2^2} \, dt\)
Using partial fractions or standard formulas:
\(= 2 \cdot \frac{1}{2(2)} \ln\left| \frac{t - 2}{t + 2} \right| + C = \frac{1}{2} \ln\left| \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} \right| + C\)
In simple words: Substitute \(x + 1 = t^2\) to eliminate the square root and integrate using standard partial fraction formulas.
Exam Tip: Rationalizing substitutions are extremely effective for integrands with linear terms under square roots.

Question 16. Evaluate the integral: \(\int \frac{x - 3}{(x - 1)^3} e^x \, dx\)
Answer: Rewrite the numerator in terms of \(x - 1\):
\(x - 3 = (x - 1) - 2\)
Substitute into the integral:
\(\int \frac{(x - 1) - 2}{(x - 1)^3} e^x \, dx = \int \left( \frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3} \right) e^x \, dx\)
Let \(f(x) = \frac{-1}{(x - 1)^2}\), then \(f'(x) = \frac{2}{(x - 1)^3}\). Using the formula \(\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C\):
\(= e^x \left( -\frac{1}{(x - 1)^2} \right) + C = -\frac{e^x}{(x - 1)^2} + C\)
In simple words: Split the fraction into two terms matching the derivative form of \(\int e^x [f(x) + f'(x)] \, dx\).

Exam Tip: Look for exponential integrals containing rational functions that match the derivative pattern of negative powers.

Question 17. Evaluate the integral: \(\int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) dx\)
Answer: Let \(\log x = t\), so \(x = e^t\) and \(dx = e^t \, dt\):
\(\int \left( \log t + \frac{1}{t^2} \right) e^t \, dt\)
Let \(f(t) = \log t\), then \(f'(t) = \frac{1}{t}\). We can rewrite \(\frac{1}{t^2}\) using integration by parts or note that:
\(\int e^t \log t \, dt + \int \frac{e^t}{t^2} \, dt\)
Integrate the first part by parts with \(\log t\):
\(\int e^t \log t \, dt = e^t \log t - \int \frac{e^t}{t} \, dt\)
Adding the second term \(\int \frac{e^t}{t^2} \, dt\):
Integrating \(\frac{e^t}{t^2}\) by parts:
\(\int \frac{e^t}{t^2} \, dt = -\frac{e^t}{t} + \int \frac{e^t}{t} \, dt\)
Combining both parts:
\(= e^t \log t - \frac{e^t}{t} + C\)
Substitute back \(t = \log x\):
\(= x \log(\log x) - \frac{x}{\log x} + C\)
In simple words: Substitute \(x = e^t\) to convert into an exponential integral, then apply integration by parts.

Exam Tip: Substitution simplifies complex logarithmic integrands into standard exponential forms.

Question 18. Evaluate the integral: \(\int (x + 1)\sqrt{1 - x - x^2} \, dx\)
Answer: Derivative of the quadratic expression \(1 - x - x^2\) is \(-1 - 2x = -2(x + \frac{1}{2})\).
Express \(x + 1\) in terms of the derivative:
\(x + 1 = -\frac{1}{2}(-2x - 1) + \frac{1}{2} = -\frac{1}{2}(-1 - 2x) + \frac{1}{2}\)
Substitute into the integral:
\(\int \left( -\frac{1}{2}(-1 - 2x) + \frac{1}{2} \right) \sqrt{1 - x - x^2} \, dx\)
\(= -\frac{1}{2} \int (-1 - 2x)\sqrt{1 - x - x^2} \, dx + \frac{1}{2} \int \sqrt{1 - x - x^2} \, dx\)
For the first integral, substitute \(u = 1 - x - x^2\), \(du = (-1 - 2x)dx\):
\(-\frac{1}{2} \int u^{1/2} \, du = -\frac{1}{2} \cdot \frac{2}{3} u^{3/2} = -\frac{1}{3}(1 - x - x^2)^{3/2}\)
For the second integral, complete the square: \(1 - x - x^2 = \frac{5}{4} - \left(x + \frac{1}{2}\right)^2\):
\(\frac{1}{2} \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - \left(x + \frac{1}{2}\right)^2} dx\)
\(= \frac{1}{2} \left[ \frac{x + \frac{1}{2}}{2}\sqrt{1 - x - x^2} + \frac{5}{8}\sin^{-1}\left(\frac{x + \frac{1}{2}}{\frac{\sqrt{5}}{2}}\right) \right]\)
Combine both results for the final answer.
In simple words: Express the linear factor as a derivative of the quadratic expression plus a constant.

Exam Tip: For integrands of the form \((px + q)\sqrt{ax^2 + bx + c}\), always adjust the linear term to match the derivative of the quadratic.

Question 19. Evaluate the integral: \(\int \frac{x^3}{(x - 1)(x - 2)} \, dx\)
Answer: Since the degree of the numerator is greater than the denominator, perform polynomial division:
\(x^3 = (x^2 - 3x + 2)(x + 3) + (7x - 6)\)
Thus:
\(\frac{x^3}{(x-1)(x-2)} = x + 3 + \frac{7x - 6}{(x-1)(x-2)}\)
Resolve \(\frac{7x - 6}{(x-1)(x-2)}\) into partial fractions:
\(\frac{7x - 6}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\)
\(7x - 6 = A(x-2) + B(x-1)\)
- Put \(x = 1 \implies 7(1) - 6 = A(-1) \implies A = -1\)
- Put \(x = 2 \implies 7(2) - 6 = B(1) \implies B = 8\)
So the integral becomes:
\(\int \left( x + 3 - \frac{1}{x-1} + \frac{8}{x-2} \right) dx = \frac{x^2}{2} + 3x - \ln|x-1| + 8\ln|x-2| + C\)
In simple words: Divide first, then use partial fractions for the remaining proper rational function.

Exam Tip: Always perform long division before partial fraction decomposition when the numerator degree exceeds the denominator.

Question 20. Evaluate the integral: \(\int 5^{5^{5^x}} \cdot 5^{5^x} \cdot 5^x \, dx\)
Answer: Let \(u = 5^{5^{5^x}}\).
Applying the chain rule for exponential differentiation \(\frac{d}{dx}(a^{f(x)}) = a^{f(x)} \ln a \cdot f'(x)\):
Let \(t = 5^{5^x}\), then \(dt = 5^{5^x} \ln 5 \cdot 5^x \ln 5 \, dx\).
Using successive substitutions:
1. Let \(w = 5^x\), \(dw = 5^x \ln 5 \, dx\)
2. Let \(v = 5^w\), \(dv = 5^w \ln 5 \, dw\)
The integral evaluates directly to:
\(= \frac{5^{5^{5^x}}}{(\ln 5)^3} + C\)
In simple words: Use repeated substitution for nested exponential functions.

Exam Tip: Keep track of the \(\ln 5\) factors generated during exponential differentiation and integration.

Question 21. Evaluate the integral: \(\int \frac{\sin x}{\sin 4x} \, dx\)
Answer: Express \(\sin 4x\) in terms of multiple angles or write \(\sin 4x = 2\sin 2x \cos 2x = 4\sin x \cos x \cos 2x\):
\(\int \frac{\sin x}{4\sin x \cos x \cos 2x} \, dx = \frac{1}{4} \int \frac{1}{\cos x \cos 2x} \, dx\)
Multiply numerator and denominator by \(\sin x\):
\(= \frac{1}{4} \int \frac{\sin x}{\sin x \cos x \cos 2x} \, dx = \frac{1}{8} \int \frac{2\sin x}{\sin 2x \cos 2x} \, dx = \frac{1}{8} \int \frac{2\sin x}{\frac{1}{2}\sin 4x} \, dx\)
Alternatively, use partial fractions or factor \(\cos 2x\):
\(\frac{1}{\cos x \cos 2x} = \frac{\cos x}{\cos^2 x \cos 2x} = \frac{\cos x}{(1 - \sin^2 x)(1 - 2\sin^2 x)}\)
Let \(u = \sin x\), \(du = \cos x \, dx\):
\(\frac{1}{4} \int \frac{du}{(1 - u^2)(1 - 2u^2)} = \frac{1}{4} \int \left( \frac{2}{1 - 2u^2} - \frac{1}{1 - u^2} \right) du\)
Integrating using standard formulas:
\(= \frac{1}{4} \left[ \frac{2}{2\sqrt{2}} \ln\left| \frac{1 + \sqrt{2}u}{1 - \sqrt{2}u} \right| - \frac{1}{2} \ln\left| \frac{1 + u}{1 - u} \right| \right] + C\)
Substitute \(u = \sin x\) for the final solution.
In simple words: Expand the denominator using multiple angle formulas and apply partial fraction decomposition.

Exam Tip: Transforming trigonometric products into rational algebraic functions via substitution simplifies integration.

Question 22. Evaluate the integral: \(\int \frac{x^2 + 4}{x^4 + 16} \, dx\)
Answer: Divide numerator and denominator by \(x^2\):
\(\int \frac{1 + \frac{4}{x^2}}{x^2 + \frac{16}{x^2}} \, dx = \int \frac{1 + \frac{4}{x^2}}{\left(x - \frac{4}{x}\right)^2 + 8} \, dx\)
Let \(u = x - \frac{4}{x}\), then \(du = \left(1 + \frac{4}{x^2}\right) dx\):
\(= \int \frac{du}{u^2 + (\sqrt{8})^2} = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{u}{2\sqrt{2}}\right) + C\)
Substitute back \(u = x - \frac{4}{x}\):
\(= \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x - \frac{4}{x}}{2\sqrt{2}}\right) + C = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x^2 - 4}{2\sqrt{2}x}\right) + C\)
In simple words: Divide by \(x^2\) and use the substitution \(u = x - \frac{4}{x}\).

Exam Tip: This is a standard reciprocal substitution pattern for biquadratic denominators.

Question 23. Evaluate the integral: \(\int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx\)
Answer: Let \(\log x = t\), so \(x = e^t\) and \(dx = e^t \, dt\):
\(\int \left( \frac{1}{t} - \frac{1}{t^2} \right) e^t \, dt\)
Let \(f(t) = \frac{1}{t}\), then \(f'(t) = -\frac{1}{t^2}\). Using the formula \(\int e^t [f(t) + f'(t)] \, dt = e^t f(t) + C\):
\(= e^t \left( \frac{1}{t} \right) + C\)
Substitute back \(t = \log x\):
\(= \frac{e^{\log x}}{\log x} + C = \frac{x}{\log x} + C\)
In simple words: Substitute \(x = e^t\) to transform the integral into the standard form \(\int e^t [f(t) + f'(t)] \, dt\).

Exam Tip: Recognizing the \(e^t[f(t) + f'(t)]\) pattern saves time compared to integration by parts.

Question 24. Evaluate the integral: \(\int \sqrt{\tan x} \, dx\)
Answer: Let \(t^2 = \tan x\), then \(2t \, dt = \sec^2 x \, dx = (1 + \tan^2 x) \, dx = (1 + t^4) \, dx\), so \(dx = \frac{2t}{1 + t^4} \, dt\):
\(\int t \cdot \frac{2t}{1 + t^4} \, dt = \int \frac{2t^2}{1 + t^4} \, dt\)
Rewrite \(2t^2 = (t^2 + 1) + (t^2 - 1)\):
\(= \int \frac{t^2 + 1}{t^4 + 1} \, dt + \int \frac{t^2 - 1}{t^4 + 1} \, dt\)
Divide numerator and denominator by \(t^2\) for both integrals:
\(= \int \frac{1 + \frac{1}{t^2}}{\left(t - \frac{1}{t}\right)^2 + 2} \, dt + \int \frac{1 - \frac{1}{t^2}}{\left(t + \frac{1}{t}\right)^2 - 2} \, dt\)
Evaluate using standard inverse trigonometric and logarithmic formulas:
\(= \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t - \frac{1}{t}}{\sqrt{2}}\right) + \frac{1}{2\sqrt{2}} \ln\left| \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} \right| + C\)
Substitute back \(t = \sqrt{\tan x}\) to obtain the final answer.
In simple words: Substitute \(t = \sqrt{\tan x}\), then use algebraic splitting and standard rational integration techniques.

Exam Tip: Integrals of square roots of trigonometric functions almost always require the substitution \(t^2 = \tan x\).

Question 25. Evaluate the integral: \(\int \tan x \tan 2x \tan 3x \, dx\)
Answer: Use the relation \(\tan 3x = \tan(2x + x)\):
\(\tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}\)
Cross-multiplying gives:
\(\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x\)
\(\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x\)
Rearranging terms:
\(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\)
Now integrate both sides:
\(\int \tan x \tan 2x \tan 3x \, dx = \int (\tan 3x - \tan 2x - \tan x) \, dx\)
\(= \frac{\ln|\sec 3x|}{3} - \frac{\ln|\sec 2x|}{2} - \ln|\sec x| + C\)
In simple words: Use the addition formula for tangents to express the product as a sum of individual tangent functions.

Exam Tip: For products of three tangents where angles satisfy \(A + B = C\), always use the compound angle tangent identity.