RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 7 Integrals Ex 7.2

Question 1.
$\frac{2 x}{1+x^{2}}$
Answer:
Let I = ∫ $\frac{2 x}{1+x^{2}}$ dx
Putting 1 + x² = t
⇒ 2x dx = dt
∴ ∫ $\frac{2 x}{1+x^{2}}$ dx = ∫ $\frac{d t}{t}$
= log |t| + C
= log |1 + x²| + C
= log (1 + x²) + C

Question 2.
$\frac{(\log x)^{2}}{x}$
Answer:

Question 3.
$\frac{1}{(x+x \log x)}$
Answer:

Question 4.
sin x sin (cos x)
Answer:
Let I = ∫sin x sin (cos x) dx
Putting cos x = t
⇒ - sin x dx = dt
⇒ sin x dx = - dt
∴ ∫ sin x sin (cos x) dx = - ∫ sin t dt
= + cos t + C
= cos (cos x) + C

Question 5.
sin (ax + b) cos (ax + b)
Answer:
Let I = ∫ sin (ax + b) cos (ax + b) dx
Putting sin (ax + b) = t
⇒ a cos (ax + b) dx = dt
⇒ cos (ax + b) dx = $\frac{d t}{a}$

Remark : Here, we can also evaluate integral by assuming cos (ax + b) = t.
Putting cos(ax + b) = t
⇒ - a sin(ax + b) dx = dt
⇒ sin (ax + b) dx = - $\frac{d t}{a}$
∴ ∫sin (ax + b) cos(ax + b)dx
= - $\int \frac{t d t}{a}=-\frac{1}{a} \int t$ dt
= - $\frac{1}{a} \cdot \frac{t^{2}}{2}$ + C
= - $\frac{\cos ^{2}(a x+b)}{2 a}$ + C

Second Method:
∫ sin (ax + b) cos (ax + b) dx
= $\frac{1}{2}$ ∫ 2 sin (ax + b) cos (ax + b) dx
[Multiplying numerator and denominator by 2]
= $\frac{1}{2}$ ∫ sin 2(ax + b) dx
= $\frac{1}{2}$ ∫ sin (2ax + 2b) dx
Putting 2ax + 2b = t
⇒ 2a dx = dt

Question 6.
x$\sqrt{a x+b}$
Answer:
Let I = ∫ $\sqrt{a x+b}$ dx = ∫ (ax + b)^1/2 dx
Putting ax + b = t
⇒ a dx = dt
⇒ dx = $\frac{d t}{a}$

Question 7.
x$\sqrt{x+2}$
Answer:

Question 8.
x$\sqrt{1+2 x^{2}}$
Answer:
Let I = ∫ x$\sqrt{1+2 x^{2}}$ dx
Putting 1 + 2x² = t
⇒ 4 dx = dt
⇒ x dx = $\frac{d t}{4}$

Question 9.
(4x + 2) $\sqrt{x^{2}+x+1}$
Answer:

Question 10.
$\frac{1}{x-\sqrt{x}}$
Answer:

Question 11.
$\frac{x}{\sqrt{x+4}}$, x > 0
Answer:
Let I = ∫ $\frac{x}{\sqrt{x+4}}$ dx
Putting x + 4 = t
⇒ dx = dt
and x = t - 4

Question 12.
(x³ - 1)^1/3 x⁵
Answer:
Let I = ∫ (x³ - 1)^1/3 x⁵
Putting x³ - 1 = t
⇒ 3x² dx = dt
⇒ x² dx = $\frac{d t}{3}$
and x³ = t + 1
∴ ∫ (x³ - 1)^1/3 x⁵ dx = ∫ (x³ - )^1/3 . x³ .x² dx

Question 13.
$\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}$
Answer:
Let I = ∫ $\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}$ dx
Putting 2 + 3x³ = t
⇒ 9x² dx = dt
⇒ x² dx = $\frac{1}{9}$ dt

Question 14.
$\frac{1}{x(\log x)^{m}}$, x > 0 , m ≠ 1
Answer:

Question 15.
$\frac{x}{9-4 x^{2}}$
Answer:
Let I = ∫ $\frac{x}{9-4 x^{2}}$ dx
Putting 9 - 4x² = t
⇒ - 8x dx = dt
⇒ x dx = - $\frac{1}{8}$dt

Question 16.
e^{2x + 3}
Answer:
Let I = ∫ e^{2x + 3} dx
= ∫ e^2x . e³ dx
= e³ ∫ e^2x dx
Putting 2x = t
⇒ 2 dx = dt
⇒ dx = $\frac{d t}{2}$
∴ ∫ e^{2x + 3} dx = e³ ∫ e^t

Alter: ∫ e^{2x + 3} dx
Let 2x + 3 = t
2 dx = dt
or dx = $\frac{d t}{2}$
∴ ∫ e^{2x + 3} dx = $\frac{1}{2}$ ∫ e^t dt = $\frac{1}{2}$ e^t + C
= $\frac{1}{2}$ e^{2x + 3} + C

Question 17.
$\frac{x}{e^{x^{2}}}$
Answer:

Question 18.
$\frac{e^{\tan ^{-1} x}}{1+x^{2}}$
Answer:

Question 19.
$\frac{e^{2 x}-1}{e^{2 x}+1}$
Answer:

Question 20.
$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$
Answer:
Let I = $\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$
Putting e^2x + e^-2x = t
⇒ (2e^2x - 2e^-2x) dx = dt
⇒ (e^2x - e^-2x) = $\frac{1}{2}$ dt
∴ ∫$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$ dx = $\frac{1}{2} \int \frac{1}{t}$ dt
= log |t| + C
= log |e^2x + e^-2x| + C

Question 21.
tan² (2x - 3)
Answer:
Let I = ∫ tan² (2x - 3) dx
= ∫ {sec² (2x - 3) - 1} dx
= ∫ sec² (2x - 3) dx - ∫ dx
Putting 2x - 3 = t
⇒ 2dx = dt
⇒ dx = $\frac{1}{2}$dt
∴ ∫ tan² (2x - 3) dx = ∫ sec² (2x - 3) dx = ∫ dx
= $\frac{1}{2}$ ∫ sec² t.dt - ∫ dx
= $\frac{1}{2}$ tan t - x + C
= $\frac{1}{2}$ tan (2x - 3) - x + C

Question 22.
sec² (7 - 4x)
Answer:
Let I = ∫ sec² (7 - 4x) dx
Putting 7 - 4x = t
⇒ - 4 dx = dt
⇒ dx = - $\frac{1}{4}$ dt
∴ ∫ sec² (7 - 4x) dx = - $\frac{1}{4}$ ∫sec² t dt
= - $\frac{1}{4}$ tan t + C
= - $\frac{1}{4}$ tan (7 - 4x) + C

Question 23.
$\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$
Answer:
Let I = ∫ $\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$ dx
Putting sin^-1 x = t

Question 24.
$\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$
Answer:

Putting 3 cos x + 2 sin x = t
⇒ (- 3 sin x + 2 cos x) dx = dt
⇒ (2 cos x - 3 sin x) dx = dt

Question 25.
Answer:

Question 26.
$\frac{\cos \sqrt{x}}{\sqrt{x}}$
Answer:

Question 27.
$\sqrt{\sin 2 x}$ cos 2x
Answer:
Let I = ∫ $\sqrt{\sin 2 x}$ cos 2x dx
Putting sin 2x = t
⇒ 2 cos 2x dx = dt
⇒ cos 2x dx = $\frac{1}{2}$ dt

Question 28.
$\frac{\cos x}{\sqrt{1+\sin x}}$
Answer:

Question 29.
cot x log sin x
Answer:
Let I = ∫ cot x log sin x dx
Putting log sin x = t
⇒ $\frac{1}{\sin x}$ × cos x dx = dt
⇒ cot x dx = dt
∴ ∫ cot x (log sin x) dx = ∫t dt = $\frac{t^{2}}{2}$ + C
= $\frac{(\log \sin x)^{2}}{2}$ + C

Question 30.
$\frac{\sin x}{1+\cos x}$
Answer:
Let I = ∫ $\frac{\sin x}{1+\cos x}$ dx
Putting 1 + cos x = t
⇒ - sin x dx = dt
⇒ sin x dx = - dt
∴ ∫ $\frac{\sin x}{1+\cos x}$ dx = - ∫ $\frac{1}{t}$ dt
= - log |t| + C
= - log |(1 + cos x)| + C

Question 31.
$\frac{\sin x}{(1+\cos x)^{2}}$
Answer:
Let I = ∫ $\frac{\sin x}{(1+\cos x)^{2}}$ dx
Putting 1 + cos x = t
⇒ - sin x dx = dt
⇒ sin x dx = - dt

Question 32.
$\frac{1}{(1+\cot x)}$
Answer:

Question 33.
$\frac{1}{1-\tan x}$
Answer:

Question 34.
$\frac{\sqrt{\tan x}}{\sin x \cos x}$
Answer:

Question 35.
$\frac{(1+\log x)^{2}}{x}$
Answer:

Question 36.
$\frac{(x+1)(x+\log x)^{2}}{x}$
Answer:

Question 37.
$\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$
Answer:

Question 38.
∫ $\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}$ dx equals:
(A) 10^x - x¹⁰ + C
(B) 10^x + x¹⁰ + C
(C) (10^x - x¹⁰)^-1
(D) log (10^x + x¹⁰) + C
Answer:
Let x¹⁰ + 10^x = t
⇒ (10x⁹ + 10^x log_e 10) dx = dt
∴ ∫ $\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}$ dx = ∫$\frac{d t}{t}$
= log |t| + C
= log (x¹⁰ + 10^x) + C
Hence, (D) is the correct answer.

Question 39.
∫ $\frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals:
(A) tan x + cot x + C
(B) tan x - cot x + C
(C) tan x cot x + C
(D) tan x - cot 2x + C
Answer:

Hence, (B) is the correct answer.