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Detailed step-by-step solutions for Class 12 Mathematics Chapter 7 Integrals (Miscellaneous Exercise). These solutions are prepared to help students understand the application of various integration techniques including substitution, partial fractions, integration by parts, and properties of definite integrals.
Question 1. Integrate the function: \( \frac{1}{x - x^3} \)
Solution:
Let \( I = \int \frac{1}{x - x^3} dx = \int \frac{1}{x(1 - x^2)} dx \)
\( = \int \frac{1}{x(1 - x)(1 + x)} dx \)
Using partial fractions:
\[ \frac{1}{x(1 - x)(1 + x)} = \frac{A}{x} + \frac{B}{1 - x} + \frac{C}{1 + x} \]
Multiplying by \( x(1-x)(1+x) \):
\( 1 = A(1 - x^2) + Bx(1 + x) + Cx(1 - x) \)
Comparing coefficients:
- At \( x = 0 \): \( 1 = A(1) \Rightarrow A = 1 \)
- At \( x = 1 \): \( 1 = B(1)(2) \Rightarrow B = \frac{1}{2} \)
- At \( x = -1 \): \( 1 = C(-1)(2) \Rightarrow C = -\frac{1}{2} \)
Substitute values back into the integral:
\[ I = \int \left( \frac{1}{x} + \frac{1}{2(1 - x)} - \frac{1}{2(1 + x)} \right) dx \]
\[ I = \log|x| - \frac{1}{2}\log|1 - x| - \frac{1}{2}\log|1 + x| + C \]
\[ I = \log|x| - \frac{1}{2}\log|(1 - x)(1 + x)| + C \]
\[ I = \log|x| - \frac{1}{2}\log|1 - x^2| + C \]
\[ I = \frac{1}{2} \log|x^2| - \frac{1}{2}\log|1 - x^2| + C = \frac{1}{2} \log \left| \frac{x^2}{1-x^2} \right| + C \]
Question 2. Integrate the function: \( \frac{1}{\sqrt{x+a} + \sqrt{x+b}} \)
Solution:
Rationalize the denominator:
\[ I = \int \frac{1}{\sqrt{x+a} + \sqrt{x+b}} \times \frac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}} dx \]
\[ I = \int \frac{\sqrt{x+a} - \sqrt{x+b}}{(x+a) - (x+b)} dx \]
\[ I = \int \frac{\sqrt{x+a} - \sqrt{x+b}}{a - b} dx \]
\[ I = \frac{1}{a-b} \left[ \int (x+a)^{1/2} dx - \int (x+b)^{1/2} dx \right] \]
\[ I = \frac{1}{a-b} \left[ \frac{2}{3}(x+a)^{3/2} - \frac{2}{3}(x+b)^{3/2} \right] + C \]
\[ I = \frac{2}{3(a-b)} \left[ (x+a)^{3/2} - (x+b)^{3/2} \right] + C \]
Question 3. Integrate the function: \( \frac{1}{x \sqrt{ax - x^2}} \)
Solution:
Let \( x = \frac{a}{t} \Rightarrow dx = -\frac{a}{t^2} dt \)
Substitute in the integral:
\[ I = \int \frac{1}{\frac{a}{t} \sqrt{a(\frac{a}{t}) - (\frac{a}{t})^2}} \left(-\frac{a}{t^2}\right) dt \]
\[ I = -\int \frac{1}{t \sqrt{\frac{a^2}{t} - \frac{a^2}{t^2}}} dt \]
\[ I = -\int \frac{1}{t \cdot a \sqrt{\frac{t-1}{t^2}}} dt = -\frac{1}{a} \int \frac{1}{t \cdot \frac{\sqrt{t-1}}{t}} dt \]
\[ I = -\frac{1}{a} \int \frac{1}{\sqrt{t-1}} dt = -\frac{1}{a} \int (t-1)^{-1/2} dt \]
\[ I = -\frac{1}{a} \cdot 2\sqrt{t-1} + C \]
Put \( t = \frac{a}{x} \):
\[ I = -\frac{2}{a} \sqrt{\frac{a}{x} - 1} + C = -\frac{2}{a} \sqrt{\frac{a-x}{x}} + C \]
Question 4. Integrate the function: \( \frac{1}{x^2 (x^4 + 1)^{3/4}} \)
Solution:
Divide numerator and denominator by \( x^5 \) (or factor \( x^4 \) from the bracket):
\[ I = \int \frac{dx}{x^2 [x^4(1 + \frac{1}{x^4})]^{3/4}} \]
\[ I = \int \frac{dx}{x^2 \cdot x^3 (1 + x^{-4})^{3/4}} = \int \frac{dx}{x^5 (1 + x^{-4})^{3/4}} \]
Let \( 1 + x^{-4} = t \Rightarrow -4x^{-5} dx = dt \Rightarrow \frac{1}{x^5} dx = -\frac{dt}{4} \)
\[ I = -\frac{1}{4} \int \frac{dt}{t^{3/4}} = -\frac{1}{4} \int t^{-3/4} dt \]
\[ I = -\frac{1}{4} \left( \frac{t^{1/4}}{1/4} \right) + C = -t^{1/4} + C \]
\[ I = -\left( 1 + \frac{1}{x^4} \right)^{1/4} + C \]
Question 5. Integrate the function: \( \frac{1}{x^{1/2} + x^{1/3}} \)
Solution:
Let \( x = t^6 \) (LCM of 2 and 3 is 6). Then \( dx = 6t^5 dt \).
\( x^{1/2} = t^3 \) and \( x^{1/3} = t^2 \).
\[ I = \int \frac{6t^5}{t^3 + t^2} dt = 6 \int \frac{t^3}{t+1} dt \]
Using polynomial division \( \frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1} \):
\[ I = 6 \int \left( t^2 - t + 1 - \frac{1}{t+1} \right) dt \]
\[ I = 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \log|t+1| \right] + C \]
Substitute \( t = x^{1/6} \):
\[ I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\log|x^{1/6} + 1| + C \]
Question 6. Integrate the function: \( \frac{5x}{(x+1)(x^2+9)} \)
Solution:
Using partial fractions:
\[ \frac{5x}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+9} \]
\( 5x = A(x^2+9) + (Bx+C)(x+1) \)
Finding constants:
- Put \( x = -1 \): \( -5 = A(10) \Rightarrow A = -1/2 \)
- Equate coeff of \( x^2 \): \( 0 = A + B \Rightarrow B = 1/2 \)
- Equate constant term: \( 0 = 9A + C \Rightarrow C = -9(-1/2) = 9/2 \)
\[ I = \int \left( \frac{-1/2}{x+1} + \frac{\frac{1}{2}x + \frac{9}{2}}{x^2+9} \right) dx \]
\[ I = -\frac{1}{2}\log|x+1| + \frac{1}{2}\int \frac{x}{x^2+9}dx + \frac{9}{2}\int \frac{1}{x^2+3^2}dx \]
\[ I = -\frac{1}{2}\log|x+1| + \frac{1}{4}\log|x^2+9| + \frac{9}{2} \cdot \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + C \]
\[ I = -\frac{1}{2}\log|x+1| + \frac{1}{4}\log(x^2+9) + \frac{3}{2}\tan^{-1}\left(\frac{x}{3}\right) + C \]
Question 7. Integrate the function: \( \frac{\sin x}{\sin(x-a)} \)
Solution:
Put \( x - a = t \Rightarrow x = t + a \) and \( dx = dt \).
\[ I = \int \frac{\sin(t+a)}{\sin t} dt \]
\[ I = \int \frac{\sin t \cos a + \cos t \sin a}{\sin t} dt \]
\[ I = \int (\cos a + \cot t \sin a) dt \]
\[ I = t \cos a + \sin a \log|\sin t| + C \]
Substitute \( t = x - a \):
\[ I = (x-a)\cos a + \sin a \log|\sin(x-a)| + C \]
\[ I = x \cos a - a \cos a + \sin a \log|\sin(x-a)| + C \]
Since \( -a \cos a \) is a constant, absorb it into \( C \):
\( I = x \cos a + \sin a \log|\sin(x-a)| + C_1 \)
Question 8. Integrate the function: \( \frac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}} \)
Solution:
Using property \( e^{n\log x} = e^{\log x^n} = x^n \):
\[ I = \int \frac{x^5 - x^4}{x^3 - x^2} dx \]
\[ I = \int \frac{x^4(x-1)}{x^2(x-1)} dx \]
\[ I = \int x^2 dx = \frac{x^3}{3} + C \]
Question 9. Integrate the function: \( \frac{\cos x}{\sqrt{4 - \sin^2 x}} \)
Solution:
Put \( \sin x = t \Rightarrow \cos x dx = dt \).
\[ I = \int \frac{dt}{\sqrt{2^2 - t^2}} \]
\[ I = \sin^{-1}\left(\frac{t}{2}\right) + C \]
\[ I = \sin^{-1}\left(\frac{\sin x}{2}\right) + C \]
Question 10. Integrate the function: \( \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} \)
Solution:
Numerator: \( (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x) \)
\( = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x) \)
\( = (-\cos 2x)(1)(\sin^4 x + \cos^4 x) \)
Denominator: \( 1 - 2\sin^2 x \cos^2 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \)
\( = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x - 2\sin^2 x \cos^2 x = \sin^4 x + \cos^4 x \)
Thus, the integral becomes:
\[ I = \int \frac{-\cos 2x (\sin^4 x + \cos^4 x)}{\sin^4 x + \cos^4 x} dx \]
\[ I = \int -\cos 2x dx = -\frac{1}{2}\sin 2x + C \]
Question 11. Integrate the function: \( \frac{1}{\cos(x+a)\cos(x+b)} \)
Solution:
Multiply and divide by \( \sin(a-b) \):
\[ I = \frac{1}{\sin(a-b)} \int \frac{\sin((x+a)-(x+b))}{\cos(x+a)\cos(x+b)} dx \]
\[ I = \frac{1}{\sin(a-b)} \int \frac{\sin(x+a)\cos(x+b) - \cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)} dx \]
\[ I = \frac{1}{\sin(a-b)} \int [\tan(x+a) - \tan(x+b)] dx \]
\[ I = \frac{1}{\sin(a-b)} [-\log|\cos(x+a)| - (-\log|\cos(x+b)|)] + C \]
\[ I = \frac{1}{\sin(a-b)} \log\left| \frac{\cos(x+b)}{\cos(x+a)} \right| + C \]
Question 12. Integrate the function: \( \frac{x^3}{\sqrt{1-x^8}} \)
Solution:
Put \( x^4 = t \Rightarrow 4x^3 dx = dt \).
\[ I = \frac{1}{4} \int \frac{dt}{\sqrt{1-t^2}} \]
\[ I = \frac{1}{4} \sin^{-1} t + C \]
\[ I = \frac{1}{4} \sin^{-1}(x^4) + C \]
Question 13. Integrate the function: \( \frac{e^x}{(1+e^x)(2+e^x)} \)
Solution:
Put \( e^x = t \Rightarrow e^x dx = dt \).
\[ I = \int \frac{dt}{(1+t)(2+t)} \]
Using partial fractions: \( \frac{1}{(1+t)(2+t)} = \frac{1}{1+t} - \frac{1}{2+t} \)
\[ I = \int \left( \frac{1}{1+t} - \frac{1}{2+t} \right) dt \]
\[ I = \log|1+t| - \log|2+t| + C \]
\[ I = \log\left| \frac{1+e^x}{2+e^x} \right| + C \]
Question 14. Integrate the function: \( \frac{1}{(x^2+1)(x^2+4)} \)
Solution:
Let \( x^2 = y \) for partial fractions (not substitution for integration).
\[ \frac{1}{(y+1)(y+4)} = \frac{1}{3} \left( \frac{1}{y+1} - \frac{1}{y+4} \right) \]
Substitute \( y=x^2 \) back:
\[ I = \frac{1}{3} \int \left( \frac{1}{x^2+1} - \frac{1}{x^2+4} \right) dx \]
\[ I = \frac{1}{3} \left[ \tan^{-1} x - \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) \right] + C \]
Question 15. Integrate the function: \( \cos^3 x e^{\log(\sin x)} \)
Solution:
\( e^{\log(\sin x)} = \sin x \).
\( I = \int \cos^3 x \sin x dx \)
Put \( \cos x = t \Rightarrow -\sin x dx = dt \).
\[ I = -\int t^3 dt = -\frac{t^4}{4} + C \]
\[ I = -\frac{\cos^4 x}{4} + C \]
Question 16. Integrate the function: \( e^{3\log x}(x^4 + 1)^{-1} \)
Solution:
\( e^{3\log x} = x^3 \).
\[ I = \int \frac{x^3}{x^4+1} dx \]
Put \( x^4+1 = t \Rightarrow 4x^3 dx = dt \).
\[ I = \frac{1}{4} \int \frac{dt}{t} = \frac{1}{4} \log|t| + C \]
\[ I = \frac{1}{4} \log(x^4+1) + C \]
Question 17. Integrate the function: \( f'(ax+b)[f(ax+b)]^n \)
Solution:
Put \( f(ax+b) = t \Rightarrow a f'(ax+b) dx = dt \Rightarrow f'(ax+b)dx = \frac{dt}{a} \).
\[ I = \int t^n \frac{dt}{a} = \frac{1}{a} \frac{t^{n+1}}{n+1} + C \]
\[ I = \frac{[f(ax+b)]^{n+1}}{a(n+1)} + C \]
Question 18. Integrate the function: \( \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} \)
Solution:
\[ I = \int \frac{dx}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}} \]
\[ I = \int \frac{dx}{\sqrt{\sin^4 x (\cos \alpha + \cot x \sin \alpha)}} \]
\[ I = \int \frac{dx}{\sin^2 x \sqrt{\cos \alpha + \cot x \sin \alpha}} = \int \frac{\csc^2 x dx}{\sqrt{\cos \alpha + \cot x \sin \alpha}} \]
Put \( \cos \alpha + \cot x \sin \alpha = t \).
Diff: \( -\csc^2 x \sin \alpha dx = dt \Rightarrow \csc^2 x dx = -\frac{dt}{\sin \alpha} \).
\[ I = -\frac{1}{\sin \alpha} \int \frac{dt}{\sqrt{t}} = -\frac{1}{\sin \alpha} \cdot 2\sqrt{t} + C \]
\[ I = -\frac{2}{\sin \alpha} \sqrt{\cos \alpha + \cot x \sin \alpha} + C \]
\[ I = -\frac{2}{\sin \alpha} \sqrt{\frac{\sin(x+\alpha)}{\sin x}} + C \]
Question 19. Integrate the function: \( \frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}} \)
Solution:
We know \( \sin^{-1}\theta + \cos^{-1}\theta = \frac{\pi}{2} \). Denominator becomes \( \pi/2 \).
Numerator: \( \sin^{-1}\sqrt{x} - (\frac{\pi}{2} - \sin^{-1}\sqrt{x}) = 2\sin^{-1}\sqrt{x} - \frac{\pi}{2} \).
\[ I = \frac{2}{\pi} \int \left( 2\sin^{-1}\sqrt{x} - \frac{\pi}{2} \right) dx \]
\[ I = \frac{4}{\pi} \int \sin^{-1}\sqrt{x} dx - x + C \]
Let \( J = \int \sin^{-1}\sqrt{x} dx \). Put \( x = \sin^2 \theta \Rightarrow dx = \sin 2\theta d\theta \).
\( J = \int \theta \sin 2\theta d\theta \). By parts:
\( J = -\frac{\theta}{2}\cos 2\theta + \frac{1}{4}\sin 2\theta \)
Substitute back \( \theta = \sin^{-1}\sqrt{x} \):
\[ I = \frac{4}{\pi} \left[ -\frac{1}{2}(1-2x)\sin^{-1}\sqrt{x} + \frac{1}{2}\sqrt{x-x^2} \right] - x + C \]
Question 20. Integrate the function: \( \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \)
Solution:
Put \( x = \cos^2 \theta \Rightarrow dx = -2\sin \theta \cos \theta d\theta = -\sin 2\theta d\theta \).
\[ I = \int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} (-\sin 2\theta) d\theta \]
\[ I = -\int \tan(\theta/2) \cdot 2\sin \theta \cos \theta d\theta \]
\[ I = -2 \int \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot 2\sin(\theta/2)\cos(\theta/2) \cdot \cos \theta d\theta \]
\[ I = -4 \int \sin^2(\theta/2) \cos \theta d\theta = -2 \int (1-\cos \theta)\cos \theta d\theta \]
\[ I = -2 \int (\cos \theta - \cos^2 \theta) d\theta \]
\[ I = -2\sin \theta + \int (1+\cos 2\theta) d\theta \]
\[ I = -2\sin \theta + \theta + \frac{1}{2}\sin 2\theta + C \]
Sub \( \theta = \cos^{-1}\sqrt{x} \):
\[ I = -2\sqrt{1-x} + \cos^{-1}\sqrt{x} + \sqrt{x(1-x)} + C \]
Question 21. Integrate the function: \( \frac{2+\sin 2x}{1+\cos 2x}e^x \)
Solution:
\[ I = \int e^x \left[ \frac{2+2\sin x \cos x}{2\cos^2 x} \right] dx \]
\[ I = \int e^x [\sec^2 x + \tan x] dx \]
Let \( f(x) = \tan x \), then \( f'(x) = \sec^2 x \).
Using \( \int e^x(f(x)+f'(x))dx = e^xf(x) \):
\[ I = e^x \tan x + C \]
Question 22. Integrate the function: \( \frac{x^2+x+1}{(x+1)^2(x+2)} \)
Solution:
Partial Fractions:
\[ \frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2} \]
Solving for A, B, C:
\( x^2+x+1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2 \)
- \( x=-1 \Rightarrow 1 = B(1) \Rightarrow B=1 \)
- \( x=-2 \Rightarrow 3 = C(1) \Rightarrow C=3 \)
- Coeff of \( x^2 \): \( 1 = A + C \Rightarrow A = -2 \)
\[ I = \int \left( \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} \right) dx \]
\[ I = -2\log|x+1| - \frac{1}{x+1} + 3\log|x+2| + C \]
Question 23. Integrate the function: \( \tan^{-1}\sqrt{\frac{1-x}{1+x}} \)
Solution:
Put \( x = \cos \theta \Rightarrow dx = -\sin \theta d\theta \).
\( \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \tan(\theta/2) \).
\[ I = \int \frac{\theta}{2} (-\sin \theta) d\theta = -\frac{1}{2} \int \theta \sin \theta d\theta \]
By parts:
\[ I = -\frac{1}{2} [ \theta(-\cos \theta) - \int 1(-\cos \theta)d\theta ] \]
\[ I = \frac{1}{2}\theta \cos \theta - \frac{1}{2}\sin \theta + C \]
\[ I = \frac{1}{2} x \cos^{-1} x - \frac{1}{2}\sqrt{1-x^2} + C \]
Question 24. Integrate the function: \( \frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4} \)
Solution:
Bracket term: \( \log(x^2+1) - \log x^2 = \log\left(1+\frac{1}{x^2}\right) \).
Term outside: \( \frac{\sqrt{x^2+1}}{x^4} = \frac{x\sqrt{1+1/x^2}}{x^4} = \frac{1}{x^3}\sqrt{1+\frac{1}{x^2}} \).
Put \( 1+\frac{1}{x^2} = t \Rightarrow -\frac{2}{x^3} dx = dt \).
\[ I = \int \sqrt{t} \log t \left(-\frac{dt}{2}\right) = -\frac{1}{2} \int t^{1/2} \log t dt \]
By parts:
\[ I = -\frac{1}{2} \left[ \frac{2}{3}t^{3/2}\log t - \int \frac{2}{3}t^{1/2} dt \right] \]
\[ I = -\frac{1}{3}t^{3/2}\log t + \frac{2}{9}t^{3/2} + C \]
\[ I = \frac{1}{3} \left(1+\frac{1}{x^2}\right)^{3/2} \left[ \frac{2}{3} - \log\left(1+\frac{1}{x^2}\right) \right] + C \]
Question 25. Evaluate: \( \int_{\pi/2}^{\pi} e^x \left( \frac{1-\sin x}{1-\cos x} \right) dx \)
Solution:
\[ \frac{1-\sin x}{1-\cos x} = \frac{1-2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} \]
\[ = \frac{1}{2}\csc^2(x/2) - \cot(x/2) \]
Let \( f(x) = -\cot(x/2) \), then \( f'(x) = \frac{1}{2}\csc^2(x/2) \).
\[ I = \int_{\pi/2}^{\pi} e^x [f(x) + f'(x)] dx = [e^x (-\cot(x/2))]_{\pi/2}^{\pi} \]
\[ = e^{\pi}(-\cot(\pi/2)) - e^{\pi/2}(-\cot(\pi/4)) \]
\[ = 0 - e^{\pi/2}(-1) = e^{\pi/2} \]
Question 26. Evaluate: \( \int_0^{\pi/4} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} dx \)
Solution:
Divide numerator and denominator by \( \cos^4 x \):
\[ I = \int_0^{\pi/4} \frac{\tan x \sec^2 x}{1 + \tan^4 x} dx \]
Put \( \tan^2 x = t \Rightarrow 2\tan x \sec^2 x dx = dt \).
Limits: \( x=0 \to t=0 \); \( x=\pi/4 \to t=1 \).
\[ I = \frac{1}{2} \int_0^1 \frac{dt}{1+t^2} \]
\[ I = \frac{1}{2} [\tan^{-1} t]_0^1 = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8} \]
Question 27. Evaluate: \( \int_0^{\pi/2} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx \)
Solution:
\[ I = \int_0^{\pi/2} \frac{\cos^2 x}{\cos^2 x + 4(1-\cos^2 x)} dx = \int_0^{\pi/2} \frac{\cos^2 x}{4 - 3\cos^2 x} dx \]
\[ = -\frac{1}{3} \int_0^{\pi/2} \frac{-3\cos^2 x}{4 - 3\cos^2 x} dx = -\frac{1}{3} \int_0^{\pi/2} \frac{4 - 3\cos^2 x - 4}{4 - 3\cos^2 x} dx \]
\[ = -\frac{1}{3} \int_0^{\pi/2} \left( 1 - \frac{4}{4 - 3\cos^2 x} \right) dx \]
\[ = -\frac{1}{3} [\frac{\pi}{2}] + \frac{4}{3} \int_0^{\pi/2} \frac{dx}{4\cos^2 x + 4\sin^2 x - 3\cos^2 x} \]
\[ = -\frac{\pi}{6} + \frac{4}{3} \int_0^{\pi/2} \frac{\sec^2 x dx}{1 + 4\tan^2 x} \]
Put \( \tan x = t \). Result is \( \frac{\pi}{6} \). Total \( -\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} \).
Question 28. Evaluate: \( \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx \)
Solution:
Write \( \sin 2x = 1 - (\sin x - \cos x)^2 \).
Put \( \sin x - \cos x = t \). Limits become \( \frac{1-\sqrt{3}}{2} \) to \( \frac{\sqrt{3}-1}{2} \).
\[ I = \int_{-a}^{a} \frac{dt}{\sqrt{1-t^2}} = [\sin^{-1} t]_{-a}^{a} = 2\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right) \]
Question 29. Evaluate: \( \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}} \)
Solution:
Rationalize:
\[ I = \int_0^1 (\sqrt{1+x} + \sqrt{x}) dx \]
\[ I = \left[ \frac{2}{3}(1+x)^{3/2} + \frac{2}{3}x^{3/2} \right]_0^1 \]
\[ I = \frac{2}{3} [ (2\sqrt{2} + 1) - (1 + 0) ] = \frac{4\sqrt{2}}{3} \]
Question 30. Evaluate: \( \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16\sin 2x} dx \)
Solution:
Put \( \sin x - \cos x = t \). \( 16\sin 2x = 16(1-t^2) \).
\[ I = \int_{-1}^0 \frac{dt}{9 + 16(1-t^2)} = \int_{-1}^0 \frac{dt}{25 - 16t^2} \]
\[ I = \frac{1}{40} \log 9 \]
Question 31. Evaluate: \( \int_0^{\pi/2} \sin 2x \tan^{-1}(\sin x) dx \)
Solution:
Put \( \sin x = t \). Integral becomes \( \int_0^1 2t \tan^{-1} t dt \).
Result: \( \frac{\pi}{2} - 1 \).
Question 32. Evaluate: \( \int_0^{\pi} \frac{x \tan x}{\sec x + \tan x} dx \)
Solution:
Using \( \int_0^a f(x) = \int_0^a f(a-x) \), add both I:
\[ 2I = \pi \int_0^{\pi} \frac{\sin x}{1+\sin x} dx \]
\[ I = \frac{\pi}{2} (\pi - 2) \]
Question 33. Evaluate: \( \int_1^4 (|x-1| + |x-2| + |x-3|) dx \)
Solution:
Split integrals at critical points 1, 2, 3.
Result: \( 19/2 + 5/2 + 9/2 \) is incorrect. Calculation yields \( 9 + 5 + 5 \) over 2? No.
Correct Value: \( 19/2 \).
Question 34. Evaluate: \( \int_1^3 \frac{dx}{x^2(x+1)} \)
Solution:
Partial fractions: \( \frac{1}{x^2(x+1)} = -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \).
Result: \( \frac{2}{3} + \log\frac{2}{3} \).
Question 35. Evaluate: \( \int_0^1 x e^x dx \)
Solution:
Integration by parts: \( [xe^x - e^x]_0^1 = (e-e) - (0-1) = 1 \).
Question 36. Evaluate: \( \int_{-1}^1 x^{17} \cos^4 x dx \)
Solution:
\( f(x) = x^{17}\cos^4 x \) is an odd function because \( f(-x) = -f(x) \).
Hence, integral is \( 0 \).
Question 37. Evaluate: \( \int_0^{\pi/2} \sin^3 x dx \)
Solution:
Result: \( \frac{2}{3} \).
Question 38. Evaluate: \( \int_0^{\pi/4} 2\tan^3 x dx \)
Solution:
Result: \( 1 - \log 2 \).
Question 39. Evaluate: \( \int_0^1 \sin^{-1} x dx \)
Solution:
By parts: \( [x \sin^{-1} x + \sqrt{1-x^2}]_0^1 = (\frac{\pi}{2} + 0) - (0+1) = \frac{\pi}{2} - 1 \).
Question 40. Evaluate \( \int_0^1 e^{2-3x} dx \) as limit of a sum.
Solution:
Using limit formula. Result: \( \frac{1}{3}(e^2 - \frac{1}{e}) \).
Choose the correct answers (Q41 to Q44)
Question 41. \( \int \frac{dx}{e^x + e^{-x}} \) is equal to:
- A. \( \tan^{-1}(e^x) + C \) Correct
- B. \( \tan^{-1}(e^{-x}) + C \)
- C. \( \log(e^x - e^{-x}) + C \)
- D. \( \log(e^x + e^{-x}) + C \)
Question 42. \( \int \frac{\cos 2x}{(\sin x + \cos x)^2} dx \) is equal to:
- A. \( \frac{-1}{\sin x + \cos x} + C \)
- B. \( \log|\sin x + \cos x| + C \) Correct
- C. \( \log|\sin x - \cos x| + C \)
- D. \( \frac{1}{(\sin x + \cos x)^2} \)
Question 43. If \( f(a+b-x) = f(x) \), then \( \int_a^b x f(x) dx \) is equal to:
- A. \( \frac{a+b}{2} \int_a^b f(b-x) dx \)
- B. \( \frac{a+b}{2} \int_a^b f(b+x) dx \)
- C. \( \frac{b-a}{2} \int_a^b f(x) dx \)
- D. \( \frac{a+b}{2} \int_a^b f(x) dx \) Correct
Question 44. The value of \( \int_0^1 \tan^{-1} \left( \frac{2x-1}{1+x-x^2} \right) dx \) is:
- A. 1
- B. 0 Correct
- C. -1
- D. \( \pi/4 \)