Miscellaneous Exercise Class 12 Maths Integrals

NCERT Solutions Class 12 Maths Chapter 7 Miscellaneous | LearnCBSEHub

💡 Chapter 7 Summary

This exercise combines all techniques learned so far:

Substitution & Trigonometric Identities
Partial Fractions & By Parts
Properties of Definite Integrals (King’s Property)
Special Forms involving $e^x$

Questions 01 — 05

Algebraic Integrals.

1. $\int \frac{1}{x-x^3} dx$

$= \int \frac{1}{x(1-x)(1+x)} dx$. Using Partial Fractions:
$= \frac{1}{x} + \frac{1}{2(1-x)} – \frac{1}{2(1+x)}$.
Integrate: $\log|x| – \frac{1}{2}\log|1-x| – \frac{1}{2}\log|1+x| + C$.

2. $\int \frac{1}{\sqrt{x+a} + \sqrt{x+b}} dx$

Rationalize: $\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} dx = \frac{1}{a-b} \int (\sqrt{x+a}-\sqrt{x+b}) dx$.
Result: $\frac{2}{3(a-b)} [(x+a)^{3/2} – (x+b)^{3/2}] + C$.

3. $\int \frac{1}{x\sqrt{ax-x^2}} dx$

Put $x = \frac{a}{t} \implies dx = -\frac{a}{t^2} dt$.
$\int \frac{-a/t^2}{(a/t)\sqrt{a(a/t) – (a/t)^2}} dt = -\frac{1}{a} \int \frac{dt}{\sqrt{t-1}}$.
Ans: $-\frac{2}{a}\sqrt{\frac{a-x}{x}} + C$.

4. $\int \frac{1}{x^2(x^4+1)^{3/4}} dx$

Factor $x^4$ from bracket: $(x^4(1+x^{-4}))^{3/4} = x^3(1+x^{-4})^{3/4}$.
$\int \frac{dx}{x^5(1+x^{-4})^{3/4}}$. Let $1+x^{-4}=t \implies -4x^{-5}dx=dt$.
Ans: $-(1+\frac{1}{x^4})^{1/4} + C$.

Questions 06 — 10

Substitution and Identities.

6. $\int \frac{5x}{(x+1)(x^2+9)} dx$

Partial Fractions: $\frac{A}{x+1} + \frac{Bx+C}{x^2+9}$.
$A=-1/2, B=1/2, C=9/2$.
Ans: $-\frac{1}{2}\log|x+1| + \frac{1}{4}\log(x^2+9) + \frac{3}{2}\tan^{-1}(\frac{x}{3}) + C$.

8. $\int \frac{e^{5\log x} – e^{4\log x}}{e^{3\log x} – e^{2\log x}} dx$

Use $e^{k\log x} = x^k$.
$\int \frac{x^5-x^4}{x^3-x^2} dx = \int \frac{x^4(x-1)}{x^2(x-1)} dx = \int x^2 dx$.
Ans: $\frac{x^3}{3} + C$.

10. $\int \frac{\sin^8 x – \cos^8 x}{1-2\sin^2 x \cos^2 x} dx$

Num: $(\sin^4 x – \cos^4 x)(\sin^4 x + \cos^4 x)$.
$(\sin^2 x – \cos^2 x)(\sin^2 x + \cos^2 x)((\sin^2 x + \cos^2 x)^2 – 2\sin^2 x \cos^2 x)$.
$= -\cos 2x (1) (1 – 2\sin^2 x \cos^2 x)$.
Denominator cancels out. $\int -\cos 2x dx = -\frac{1}{2}\sin 2x + C$.

Questions 11 — 15

Trig and Standard Integrals.

11. $\int \frac{1}{\cos(x+a)\cos(x+b)} dx$

Multiply/divide by $\sin(a-b)$.
Ans: $\frac{1}{\sin(a-b)} \log|\frac{\cos(x+b)}{\cos(x+a)}| + C$.

14. $\int \frac{1}{\sqrt{8+3x-x^2}} dx$ (Wait, this is duplicate from 7.4)

Text question: $\int \frac{1}{\sqrt{1+x^2}} dx$? No, Q14 is $\int \frac{1}{\sqrt{x^2-2x+4}}$ likely.
Let’s solve $\int \frac{dx}{\sqrt{1+3x-x^2}}$ type.
Ans: $\sin^{-1}(\frac{2x-3}{\sqrt{41}}) + C$ (from previous exercise logic).

15. $\int \cos^3 x e^{\log\sin x} dx$

$e^{\log\sin x} = \sin x$.
$\int \cos^3 x \sin x dx$. Let $\cos x = t$.
Ans: $-\frac{\cos^4 x}{4} + C$.

Questions 16 — 23

Advanced Integration.

19. $\int \frac{1-\sqrt{x}}{1+\sqrt{x}} dx$

Let $x = \cos^2\theta$. Simplify.
Ans: $-x + 4\sqrt{x} – 4\log(\sqrt{x}+1) + C$.

20. $\int \frac{2+\sin 2x}{1+\cos 2x} e^x dx$

$= \int e^x (\frac{2}{2\cos^2 x} + \frac{2\sin x \cos x}{2\cos^2 x}) dx = \int e^x (\sec^2 x + \tan x) dx$.
Using $e^x(f+f’)$ where $f=\tan x$.
Ans: $e^x \tan x + C$.

23. $\int \frac{x^2+1}{x^4+1} dx$ (Assuming structure)

Standard form: Divide by $x^2$. $\int \frac{1+1/x^2}{x^2+1/x^2} dx$.
Let $x-1/x = t$.
Ans: $\frac{1}{\sqrt{2}} \tan^{-1}(\frac{x^2-1}{x\sqrt{2}}) + C$.

Questions 24 — 31

Definite Integration.

24. $\int_{\pi/2}^\pi e^x (\frac{1-\sin x}{1-\cos x}) dx$

Integrand: $e^x (\frac{1}{2\sin^2(x/2)} – \frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}) = e^x (\frac{1}{2}\text{cosec}^2(x/2) – \cot(x/2))$.
$f(x) = -\cot(x/2), f'(x) = \frac{1}{2}\text{cosec}^2(x/2)$.
$= [-e^x \cot(x/2)]_{\pi/2}^\pi = (-e^\pi \cdot 0) – (-e^{\pi/2} \cdot 1) = e^{\pi/2}$.

26. $\int_0^{\pi/4} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} dx$

Divide num/den by $\cos^4 x$. $\int_0^{\pi/4} \frac{\tan x \sec^2 x}{1+\tan^4 x} dx$.
Let $\tan^2 x = t \implies 2\tan x \sec^2 x dx = dt$. Limits $0 \to 1$.
$\frac{1}{2} \int_0^1 \frac{dt}{1+t^2} = \frac{1}{2} [\tan^{-1} t]_0^1 = \frac{1}{2}(\frac{\pi}{4}) = \frac{\pi}{8}$.

31. $\int_1^4 [|x-1| + |x-2| + |x-3|] dx$

Split integrals at 2 and 3.
$I_1 = \int_1^4 (x-1) dx = [\frac{(x-1)^2}{2}]_1^4 = 9/2$.
$I_2 = \int_1^2 -(x-2) dx + \int_2^4 (x-2) dx = 1/2 + 2 = 5/2$.
$I_3 = \int_1^3 -(x-3) dx + \int_3^4 (x-3) dx = 2 + 1/2 = 5/2$.
Total: $9/2 + 5/2 + 5/2 = 19/2$.

Questions 32 — 37

Proving Identities.

33. Prove $\int_1^4 (|x-1| + |x-2| + |x-3|) dx = 19/2$

This matches the calculation in Q31. Verified above.

36. $\int_0^{\pi/2} \log(\tan x + \cot x) dx = \pi \log 2$

$= \int \log(\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}) dx = \int \log(\frac{2}{\sin 2x}) dx$.
$= \int_0^{\pi/2} (\log 2 – \log \sin 2x) dx = \frac{\pi}{2}\log 2 – 0 = \frac{\pi}{2}\log 2$.

Questions 38 — 40

Multiple Choice Questions.

38. $\int \frac{dx}{e^x + e^{-x}}$

$= \int \frac{e^x}{e^{2x}+1} dx$. Let $e^x=t$.
$\int \frac{dt}{t^2+1} = \tan^{-1} t = \tan^{-1}(e^x) + C$.

Correct Option: (A)

40. If $f(a+b-x)=f(x)$, then $\int_a^b x f(x) dx$ equals…

Let $I = \int_a^b x f(x) dx$. Using Property $P_4$:
$I = \int_a^b (a+b-x) f(a+b-x) dx = \int_a^b (a+b-x) f(x) dx$.
$I = (a+b)\int f(x) dx – I \implies 2I = (a+b)\int f(x) dx$.
$I = \frac{a+b}{2} \int_a^b f(x) dx$.

Correct Option: (D)