Miscellaneous Exercise
Chapter 7: Integrals – NCERT Solutions (All Questions)
Q1
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Integrate: $\frac{1}{x - x^3}$
$\frac{1}{x(1-x^2)} = \frac{1}{x(1-x)(1+x)}$. Use partial fractions.
$\frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}$. Solving gives $A=1, B=1/2, C=-1/2$.
$\boxed{\log|x| + \frac{1}{2}\log|1-x^2| + C}$
Q2
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Integrate: $\frac{1}{\sqrt{x+a} + \sqrt{x+b}}$
Rationalize the denominator by multiplying with $\sqrt{x+a} - \sqrt{x+b}$.
$\int \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b} dx$.
$\boxed{\frac{2}{3(a-b)} [(x+a)^{3/2} - (x+b)^{3/2}] + C}$
Q3
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Integrate: $\frac{1}{x\sqrt{ax-x^2}}$
Put $x = \frac{a}{t}$. Then $dx = -\frac{a}{t^2} dt$.
Substitute and simplify to get $\int \frac{-1}{\sqrt{t-1}} dt$.
$\boxed{-\frac{2}{a}\sqrt{\frac{ax-x^2}{x}} + C}$
Q4
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Integrate: $\frac{1}{x^2(x^4+1)^{3/4}}$
Take $x^4$ common from bracket: $x^2 \cdot x^3 (1+x^{-4})^{3/4} = x^5(1+x^{-4})^{3/4}$.
Let $1+x^{-4} = t \Rightarrow -4x^{-5} dx = dt$.
$\boxed{-\frac{1}{4}(1+\frac{1}{x^4})^{1/4} + C}$
Q5
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Integrate: $\frac{1}{x^{1/2} + x^{1/3}}$
Let $x = t^6$ (LCM of 2 and 3). $dx = 6t^5 dt$.
$\int \frac{6t^5}{t^3+t^2} dt = 6 \int \frac{t^3}{t+1} dt$. Perform division.
$\boxed{2\sqrt{x} - 3x^{1/3} + 6x^{1/6} - 6\log(x^{1/6}+1) + C}$
Q6
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Integrate: $\frac{5x}{(x+1)(x^2+9)}$
Partial fractions: $\frac{A}{x+1} + \frac{Bx+C}{x^2+9}$.
$A = -1/2, B = 1/2, C = 9/2$.
$\boxed{-\frac{1}{2}\log|x+1| + \frac{1}{4}\log(x^2+9) + \frac{3}{2}\tan^{-1}(\frac{x}{3}) + C}$
Q7
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Integrate: $\frac{\sin x}{\sin(x-a)}$
Put $x-a = t \Rightarrow x = t+a$.
$\int \frac{\sin(t+a)}{\sin t} dt = \int (\cos a + \sin a \cot t) dt$.
$\boxed{x\cos a + \sin a \log|\sin(x-a)| + C}$
Q8
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Integrate: $\frac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}}$
Simplify: $\frac{x^5 - x^4}{x^3 - x^2} = \frac{x^4(x-1)}{x^2(x-1)} = x^2$.
$\boxed{\frac{x^3}{3} + C}$
Q9
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Integrate: $\frac{\cos x}{\sqrt{4-\sin^2 x}}$
Let $\sin x = t \Rightarrow \cos x dx = dt$.
$\int \frac{dt}{\sqrt{2^2-t^2}} = \sin^{-1}(\frac{t}{2})$.
$\boxed{\sin^{-1}(\frac{\sin x}{2}) + C}$
Q10
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Integrate: $\frac{\sin^8 x - \cos^8 x}{1-2\sin^2 x \cos^2 x}$
Num: $(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$.
Denom: $(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = \sin^4 x + \cos^4 x$.
Simplify to $\int (\sin^2 x - \cos^2 x) dx = -\int \cos 2x dx$.
$\boxed{-\frac{1}{2}\sin 2x + C}$
Q11
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Integrate: $\frac{1}{\cos(x+a)\cos(x+b)}$
Multiply and divide by $\sin(a-b)$.
Use $\sin((x+a)-(x+b))$ in numerator.
$\boxed{\frac{1}{\sin(a-b)}\log|\frac{\cos(x+b)}{\cos(x+a)}| + C}$
Q12
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Integrate: $\frac{x^3}{\sqrt{1-x^8}}$
Let $x^4 = t \Rightarrow 4x^3 dx = dt$.
$\frac{1}{4} \int \frac{dt}{\sqrt{1-t^2}} = \frac{1}{4}\sin^{-1} t$.
$\boxed{\frac{1}{4}\sin^{-1}(x^4) + C}$
Q13
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Integrate: $\frac{e^x}{(1+e^x)(2+e^x)}$
Let $e^x = t \Rightarrow e^x dx = dt$.
$\int \frac{dt}{(1+t)(2+t)} = \int (\frac{1}{1+t} - \frac{1}{2+t}) dt$.
$\boxed{\log|\frac{1+e^x}{2+e^x}| + C}$
Q14
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Integrate: $\frac{1}{(x^2+1)(x^2+4)}$
Use partial fractions with $y=x^2$: $\frac{1}{(y+1)(y+4)} = \frac{1}{3}(\frac{1}{y+1} - \frac{1}{y+4})$.
Integrate w.r.t $x$: $\frac{1}{3}(\tan^{-1}x - \frac{1}{2}\tan^{-1}\frac{x}{2})$.
$\boxed{\frac{1}{3}\tan^{-1}x - \frac{1}{6}\tan^{-1}\frac{x}{2} + C}$
Q15
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Integrate: $\cos^3 x e^{\log(\sin x)}$
$e^{\log(\sin x)} = \sin x$. Integral: $\int \cos^3 x \sin x dx$.
Let $\cos x = t \Rightarrow -\sin x dx = dt$.
$\boxed{-\frac{\cos^4 x}{4} + C}$
Q16
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Integrate: $e^{3\log x}(x^4+1)^{-1}$
$e^{3\log x} = x^3$. Integral: $\int \frac{x^3}{x^4+1} dx$.
Let $x^4+1 = t \Rightarrow 4x^3 dx = dt$.
$\boxed{\frac{1}{4}\log(x^4+1) + C}$
Q17
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Integrate: $f'(ax+b)[f(ax+b)]^n$
Let $f(ax+b) = t \Rightarrow f'(ax+b) \cdot a dx = dt$.
$\frac{1}{a} \int t^n dt = \frac{t^{n+1}}{a(n+1)}$.
$\boxed{\frac{[f(ax+b)]^{n+1}}{a(n+1)} + C}$
Q18
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Integrate: $\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$
Expand $\sin(x+\alpha)$. Divide inside root by $\sin^4 x$.
Becomes $\int \frac{\csc^2 x}{\sqrt{\cot x \sin \alpha + \cos \alpha}} dx$.
$\boxed{\frac{-2}{\sin \alpha}\sqrt{\sin(x+\alpha)\csc x} + C}$
Q19
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Integrate: $\frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}}$
Use $\sin^{-1}\theta + \cos^{-1}\theta = \pi/2$. Denom is $\pi/2$.
Num: $2\sin^{-1}\sqrt{x} - \pi/2$. Integrate by parts.
$\boxed{\frac{2}{\pi}[\sqrt{x-x^2} - (1-2x)\sin^{-1}\sqrt{x}] - x + C}$
Q20
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Integrate: $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
Put $x = \cos^2 \theta$.
Simplify to $\int -2\cos \theta (1-\cos \theta) d\theta$.
$\boxed{-2\sqrt{1-x} + \cos^{-1}\sqrt{x} + \sqrt{x(1-x)} + C}$
Q21
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Integrate: $\frac{2+\sin 2x}{1+\cos 2x} e^x$
$\frac{2+2\sin x \cos x}{2\cos^2 x} e^x = (\sec^2 x + \tan x) e^x$.
Form $\int e^x(f(x)+f'(x)) dx$ with $f(x)=\tan x$.
$\boxed{e^x \tan x + C}$
Q22
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Integrate: $\frac{x^2+x+1}{(x+1)^2(x+2)}$
Partial fractions: $\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}$.
$A=-2, B=1, C=3$.
$\boxed{-2\log|x+1| - \frac{1}{x+1} + 3\log|x+2| + C}$
Q23
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Integrate: $\tan^{-1} \sqrt{\frac{1-x}{1+x}}$
Put $x = \cos \theta$. Integrand becomes $\theta/2$.
$\int \frac{\theta}{2} (-\sin \theta) d\theta$. Integrate by parts.
$\boxed{\frac{1}{2}(x\cos^{-1}x - \sqrt{1-x^2}) + C}$
Q24
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Integrate: $\frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}$
Simplify log term: $\log(1+\frac{1}{x^2})$.
Put $1+\frac{1}{x^2} = t^2$.
$\boxed{-\frac{1}{3}(1+\frac{1}{x^2})^{3/2} [\log(1+\frac{1}{x^2}) - \frac{2}{3}] + C}$
Q25
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Evaluate: $\int_{\pi/2}^\pi e^x (\frac{1-\sin x}{1-\cos x}) dx$
Simplify to $e^x(\frac{1}{2}\csc^2(x/2) - \cot(x/2))$.
Form $e^x(f(x)+f'(x))$ with $f(x) = -\cot(x/2)$.
$\boxed{e^{\pi/2}}$
Q26
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Evaluate: $\int_0^{\pi/4} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} dx$
Divide num/den by $\cos^4 x$. Becomes $\int \frac{\tan x \sec^2 x}{1+\tan^4 x} dx$.
Let $\tan^2 x = t$.
$\boxed{\frac{\pi}{8}}$
Q27
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Evaluate: $\int_0^{\pi/2} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx$
Divide num/den by $\cos^2 x$. $\int \frac{1}{1+4\tan^2 x} dx$.
Use partial fractions or substitution.
$\boxed{\frac{\pi}{6}}$
Q28
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Evaluate: $\int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx$
Write $\sin 2x = 1 - (\sin x - \cos x)^2$.
Let $\sin x - \cos x = t$.
$\boxed{2\sin^{-1}\frac{\sqrt{3}-1}{2}}$
Q29
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Evaluate: $\int_0^1 \frac{dx}{\sqrt{1+x}-\sqrt{x}}$
Rationalize: $\int_0^1 (\sqrt{1+x}+\sqrt{x}) dx$.
$\boxed{\frac{4\sqrt{2}}{3}}$
Q30
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Evaluate: $\int_0^{\pi/4} \frac{\sin x + \cos x}{9+16\sin 2x} dx$
Let $\sin x - \cos x = t$. Denom becomes $25 - 16t^2$.
$\boxed{\frac{1}{20}\log 3}$
Q31
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Evaluate: $\int_0^{\pi/2} \sin 2x \tan^{-1}(\sin x) dx$
Let $\sin x = t$. Integral: $\int 2t \tan^{-1} t dt$.
$\boxed{\frac{\pi}{2} - 1}$
Q32
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Evaluate: $\int_0^\pi \frac{x \tan x}{\sec x + \tan x} dx$
Use $\int_0^a f(x) = \int_0^a f(a-x)$. Add $I+I$.
$\boxed{\frac{\pi}{2}(\pi-2)}$
Q33
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Evaluate: $\int_1^4 [|x-1| + |x-2| + |x-3|] dx$
Split integrals at critical points 1, 2, 3.
$\boxed{\frac{19}{2}}$
Q34
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Prove: $\int_1^3 \frac{dx}{x^2(x+1)} = \frac{2}{3} + \log \frac{2}{3}$
Partial fractions: $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$.
$\boxed{\text{Proved}}$
Q35
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Prove: $\int_0^1 x e^x dx = 1$
Integration by parts.
$\boxed{\text{Proved}}$
Q36
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Prove: $\int_{-1}^1 x^{17} \cos^4 x dx = 0$
Function is odd ($f(-x) = -f(x)$).
$\boxed{\text{Proved}}$
Q37
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Prove: $\int_0^{\pi/2} \sin^3 x dx = \frac{2}{3}$
Use $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$.
$\boxed{\text{Proved}}$
Q38
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Prove: $\int_0^{\pi/4} 2\tan^3 x dx = 1 - \log 2$
$2\tan x(\sec^2 x - 1)$.
$\boxed{\text{Proved}}$
Q39
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Prove: $\int_0^1 \sin^{-1} x dx = \frac{\pi}{2} - 1$
Integration by parts ($1 \cdot \sin^{-1} x$).
$\boxed{\text{Proved}}$
Q40
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Evaluate: $\int_0^1 e^{2-3x} dx$
$\frac{e^{2-3x}}{-3}$ from 0 to 1.
$\boxed{\frac{1}{3}(e^2 - \frac{1}{e})}$
Q41
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$\int \frac{dx}{e^x+e^{-x}}$ is equal to:
$\int \frac{e^x}{e^{2x}+1} dx$. Let $e^x=t$.
$\boxed{\tan^{-1}(e^x) + C}$
Q42
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$\int \frac{\cos 2x}{(\sin x + \cos x)^2} dx$ is equal to:
$\frac{\cos^2 x - \sin^2 x}{(\sin x + \cos x)^2} = \frac{\cos x - \sin x}{\cos x + \sin x}$.
$\boxed{\log|\sin x + \cos x| + C}$
Q43
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If $f(a+b-x) = f(x)$, then $\int_a^b x f(x) dx$ is equal to:
Use property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.
$\boxed{\frac{a+b}{2} \int_a^b f(x) dx}$
Q44
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The value of $\int_0^1 \tan^{-1}(\frac{2x-1}{1+x-x^2}) dx$ is:
$\tan^{-1} x - \tan^{-1}(1-x)$. Use property $\int_0^a f(x) = \int_0^a f(a-x)$.
$\boxed{0}$