OP Malhotra Class 12 Maths Solutions Chapter 16 Definite Integrals Ex 16(d)

Continuous practice using ISC Class 12 OP Malhotra Solutions Chapter 16 Definite Integrals Ex 16(d) can lead to a stronger grasp of mathematical concepts.

S Chand Class 12 ICSE Maths Solutions Chapter 16 Definite Integrals Ex 16(d)

Evaluate the following integrals as limit of sum:
Question 1.
(i) \(\int_0^3\)(x + 5) d x
(ii) \(\int_{-1}^1\)(x + 3) d x
(iii) \(\int_0^5\)(x – 1) d x
Answer:

Question 2.
(i) \(\int_1^2\)(3 x – 2) d x
(ii) \(\int_3^5\)(2 – x) d x
(iii) \(\int_1^3\)(1 – 2 x) d x
Answer:

Question 3.
(i) \(\int_2^3\) x² d x
(ii) \(\int_0^2\)(x² + 3) dx
(iii) \(\int_2^5\)(3 x² – 5) dx
(iv) \(\int_0^3\)(2 x² + 3) dx
Answer:

Question 4.
Prove that :
(i) \(\int_1^4\)(x² – x) dx
(ii) \(\int_0^3\)(x² + 2 x) dx
(iii) \(\int_1^3\)(2 x² + 5 x) dx
(iv) \(\int_0^2\)(x² + 2 x + 1) dx
Answer:

Question 5.
(i) \(\int_{-1}^1\) e^x dx
(ii) \(\int_0^2\) e^-x dx
(iii) \(\int_0^2\) e^3x+1 dx
(iv) \(\int_1^3\) a^x dx
(v) \(\int_0^4\)(x + e^2x) dx
Answer:

Question 6.
(i) \(\int_a^b\) cos x dx
(ii) \(\int_0^{\pi / 2}\) sin x dx
(iii) \(\int_0^{\pi / 2}\) cos x dx
Answer:

Examples:

Evaluate :
Question 1.
\(\int_2^3\) \(\frac{x^3+1}{x(x-1)}\) d x
Answer:

Question 2.
\(\int x(\log x)^2\) d x
Answer:
Let I = \(\int^{(2)}\) x(log x)² d x
= (log x)² .\(\frac{x^2}{2}\) – ∫2 log x.\(\frac{1}{x}\) . \(\frac{x^2}{2}\) d x
= \(\frac{x^2}{2}\)(log x)² – \(\int^{(2)}\) x(log x) d x
= \(\frac{x^2}{2}\)(log x)² – [log x .\(\frac{x^2}{2}\) – ∫\(\frac{1}{x}\).\(\frac{x^2}{2}\) d x]
= \(\frac{x^2}{2}\)(log x)² – \(\frac{x^2}{2}\) log x + \(\frac{x^2}{4}\) + C

Question 3.
\(\int_1^2\) \(\frac{2}{4 x^2-1}\) d x
Answer:

Question 4.
∫e^x (log x + \(\frac{1}{x}\)) d x
Answer:
Let I = ∫e^x(log x + \(\frac{1}{x}\)) d x
= \(\int^{(2)}\) e^x log x + d x e^x \(\frac{1}{x}\) d x
= log e^x – ∫\(\frac{1}{x}\)e^x d x + ∫e^x.\(\frac{1}{x}\) d x
= e^x log x + C

Question 5.
∫e^x cos x d x
Answer:
Let I = ∫e² cos x d x
= cos x e^x – ∫sin x e^x d x
= cos x e^x + [sin x e^x – ∫cos x e^x d x] + C
∴ I = cos x e^x + sin x e^x – I;
where C’ = \(\frac{C}{2}\)
I = \(\frac{e^x}{2}\)(cos x^x + sin x) + C

Question 6.
\(\int_0^1\) \(\frac{e^{-x}}{1+e^x}\) dx
Answer:

Question 7.
∫ \(\frac{\log (\log x)}{x}\) d x
Answer:
Let I = \(\int \frac{\log (\log x)}{x}\) d x
put log x = t
⇒ \(\frac{1}{x}\) d x = d t
= ∫log t .1 d t
= log t . t – ∫\(\frac{1}{t}\).t d t
= t log t – t + C
= log x[log (log x) – 1] + C

Question 8.
\(\int_a^b\) \(\frac{\log x}{x}\) d x
Answer:
put log x = t
⇒ \(\frac{1}{x}\) d x = d t
when x = a
⇒ t = log a;
when x = b
⇒ t = log b
Let I = \(\int_a^b\) \(\frac{\log x}{x}\) d x
= \(\int_{\log a}^{\log b}\) t . d t
= \(\frac{t^2}{2}]_{\log a}^{\log b}\)
= \(\frac{1}{2}\) [(log b)² – (log a)² ] d x
= \(\frac{1}{2}\)[log b – log a][log a + log b]
= \(\frac{1}{2}\) log a b log \(\frac{b}{a}\)

Question 9.
∫ \(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}\) d x
Answer:
Let I = ∫\(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}\) d x
put sin^-1 x = t
⇒ x = sin t
⇒ d x = cos t d t
= ∫\(\frac{t \sin t \cdot \cos t d t}{\sqrt{1-\sin ^2 t}}\)
= \(\int^{(1)}\) t . sin t d t
= t(-cos t) – ∫(-cos t) d t + C
= -t cos t + sin t + C
= \(-\sqrt{1-x^2} \sin ^{-1}\) x + x + C

Question 10.
\(\int_0^{2 \pi}\) \(\sqrt{1+\sin \frac{x}{2}}\) d x
Answer:

Question 11.
∫\(\sin ^{-1} \frac{2 x}{1+x^2}\) d x
Answer:

Question 12.
∫\(\frac{x-1}{(x-3)(x-2)^2}\) d x
Answer:
Let \(\frac{x-1}{(x-3)(x-2)^2}\) = \(\frac{\mathrm{A}}{x-3}\) + \(\frac{\mathrm{B}}{x-2}\) + \(\frac{\mathrm{C}}{(x-2)^2}\) …………..(1)
Multiplying both sides of eqn (1) by (x – 3)(x – 2)² ; we get
x – 1 = A(x – 2)² + B(x – 3)(x – 2) + C(x – 3) …………(2)
putting x = 2 in eqn (2); we get
1 = C(2 – 3)
⇒ C = -1
putting x = 3 in eqn (2); we have
2 = A
∴ Coeff. of x²; 0 = A + B
⇒ B = -A = -2
Thus from eqn. (1); we have
\(\frac{x-1}{(x-3)(x-2)^2}\) = \(\frac{2}{x-3}\) – \(\frac{2}{x-2}\) – \(\frac{1}{(x-2)^2}\)
∴ ∫\(\frac{(x-1) d x}{(x-3)(x-2)^2}\) = 2 log |x – 3| – 2 log (x -2) + (x – 2)^-1 + C

Question 13.
∫e^x (\(\frac{1+\sin x}{1+\cos x}\)) d x
Answer:

Question 14.
Prove that
\(\int_0^{\pi / 2}\) \(\frac{\sin x}{\sqrt{\sin x}+\sqrt{\cos x}}\) d x = \(\frac{\pi}{4}\)
Answer:

Question 15.
Prove that
\(\int_0^{\pi / 2}\) sin 2 x log tan x d x = 0
Answer:

Question 16.
∫e^2x sin x d x
Answer:
Let I = ∫e^2x sin x d x
= sin \(\frac{e^{-2 x}}{-2}\) – ∫cos x \(\frac{e^{(2)}}{-2}\) d x
= \(-\frac{1}{2}\) sin x e^2x + \(\frac{1}{2}\)[cos x \(\frac{e^{-2 x}}{-2}\) – ∫-sin x \(\frac{e^{-2 x}}{-2}\) d x]
= \(-\frac{1}{2}\) sin x e^-2x – \(\frac{1}{4}\) cos x e^-2x – \(\frac{1}{4}\) I
I + \(\frac{\mathrm{I}}{4}\) = \(-\frac{1}{4}\)[2 sin x + cos x] e^-2x
⇒ I = \(-\frac{1}{5}\)[2 sin x + cos x] e^-2x + C

Question 17.
\(\int_0^{\pi / 4}\) \(\frac{2 \cos 2 x}{1+\sin 2 x}\) d x
Answer:
Let I = \(\int_0^{\pi / 4}\) \(\frac{2 \cos 2 x d x}{1+\sin 2 x}\)
put sin 2 x = t
⇒ 2 cos 2 x d x = d t
When x = 0
⇒ t = 0;
when x = \(\frac{\pi}{4}\)
⇒ t = sin \(\frac{\pi}{2}\) = 1
∴ I = \(\int_0^1 \frac{d t}{1+t}\)
= log (1 + t)]¹ ₀
= log 2 – log 1
= log 2

Question 18.
\(\int_0^1\) x tan ^-1 x d x
Answer:

Question 19.
\(\int_0^{\pi / 4}\)(tan x + cot x)^-1 d x
Answer:

Question 20.
∫ \(\frac{1+\tan ^2 x}{\sqrt{1-\tan ^2 x}}\) d x
Answer:
Let I = ∫\(\frac{(1+\tan ^2 x) d x}{\sqrt{1-\tan ^2 x}}\)
= ∫\(\frac{\sec ^2 x d x}{\sqrt{1-\tan ^2 x}}\)
put tan x = t
⇒ sec² x d x = d t
∴ I = ∫\(\frac{d t}{\sqrt{1^2-t^2}}\)
= sin^-1 (\(\frac{t}{1}\)) + C
= sin^-1 (tan x) + C

Question 21.
\(\int_0^{1 / 2}\) \(\frac{\sin ^{-1} x}{(1-x^2)^{3 / 2}}\) dx
Answer:
\(\int_0^{1 / 2}\) \(\frac{\sin ^{-1} x}{(1-x^2)^{3 / 2}}\) dx
put x = sinθ
⇒ θ = sin^-1x
⇒ d x = cosθ dθ
When x = 0
⇒ θ = 0 ;
when x = \(\frac{1}{2}\)
⇒ sinθ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{6}\)
= \(\int_0^{\pi / 6}\) \(\frac{\theta \cos \theta d \theta}{[1-\sin ^2 \theta)^{3 / 2}}\)
= \(\int_0^{\pi / 6}\) \(\frac{\theta \cos \theta d \theta}{\cos ^3 \theta}\)
= \(\int_0^{\pi / 6}\) θ sec² θ dθ
= θ tan θ]₀ ^π/6 – \(int_0^{\pi / 6}\) tanθ dθ
= (\(\frac{\pi}{6}\) tan \(\frac{\pi}{6}\) – 0) + log |cosθ|]₀ ^π/6
= \(\frac{\pi}{6}\) \(\frac{1}{\sqrt{3}}\) + log cos \(\frac{\pi}{6}\) – log cos 0
= \(\frac{\pi}{6 \sqrt{3}} \) + log \(\frac{\sqrt{3}}{2}\) – log 1
= \(\frac{\pi}{6 \sqrt{3}}\) + log \(\frac{\sqrt{3}}{2}\)

Question 22.
∫ \(\frac{\cos x}{\sin x+\sqrt{\sin x}}\) d x
Answer:
Let I = ∫ \(\frac{\cos x d x}{\sin x+\sqrt{\sin x}}\)
put \(\sqrt{\sin x}\) = t;
on squaring; we have
sin x = t² ; diff both sides;
we have cos x d x = 2 t d t
∴ I = ∫ \(\frac{2 t d t}{t^2+t}\)
= 2 ∫ \(\frac{d t}{t+1}\)
= 2 log |t + 1|+ C
= 2 log |1 + \(\sqrt{\sin x}\)| + C

Question 23.
Prove that \(\int_0^{\pi / 2}\) \(\frac{3 \sin \theta+4 \cos \theta}{\sin \theta+\cos \theta}\) dθ = \(\frac{7 \pi}{4}\)
Answer:

Question 24.
∫ x² (e^x3 ) cos (2 e^x3 ) d x
Answer:
Let I = ∫ x² (e^x3 ) cos (2 e^x3 ) d x
put e^x3 = t
⇒ ee^x3 3 x² d x = d t
= ∫cos 2 t .\(\frac{d t}{3}\)
= \(\frac{\sin 2 t}{6}\) + C
= \(\frac{1}{6}\) sin (2 e^x3 ) + C

Question 25.
Prove that \(\int_0^{2 \pi}\) \(\frac{x \cos x}{1+\cos x}\) d x = 2π²
Answer:

Question 26.
\(\int_0^{\pi / 4}\) log (1 + tan x) d x
Answer:

Question 27.
∫ \(\frac{1}{x \cos ^2(1+\log x)}\) d x
Answer:
Let I = ∫ \(\frac{1}{x \cos ^2(1+\log x)}\) d x
put 1 + log x = t
⇒ \(\frac{1}{x}\) d x = d t
= ∫\(\frac{d t}{\cos ^2 t}\) = ∫sec² d t
= tan t + C
= tan (1 + log x) + C

Question 28.
∫ \(\frac{x^2}{x^2-4}\) d x
Answer:

Question 29.
Evalute \(\int_0^9\) f(x) d x where f(x) is defined by
f(x) = {sinx, if 0 < x < \(\frac{\pi}{2}\)
1, if \(\frac{\pi}{2}\) < x < 5
e^x-5, if 5 < x < 9
Answer:

Question 30.
∫ \(\frac{e^{2 x}}{2+e^x}\) d x
Answer:
Let I = ∫ \(\frac{e^{2 x}}{2+e^x}\) d x
put
e^x = t
⇒ e^x d x = dt
= ∫\(\frac{t d t}{t+2}\)
= ∫[1 – \(\frac{2}{t+2}\)] dt
= t – 2log |t + 2| + C
= e^x – 2 log |e^x + 2| + C

Question 31.
∫\(\frac{x}{(x+1)^2}\) d x
Answer:
∫\(\frac{x d x}{(x+1)^2}\)
= ∫\(\frac{(x+1-1) d x}{(x+1)^2}\)
= ∫[\(\frac{1}{x+1}\) – \(\frac{1}{(x+1)^2}\)] d x
= log |x + 1| – \(\frac{(x+1)^{-2+1}}{-2+1}\) + C
= log |x + 1| + \(\frac{1}{x+1}\) + C

Question 32.
\(\int_{-3}^3\)|x + 2| d x
Answer:
Let I = \(\int_{-3}^3\)|x + 2| d x
= \(\int_{-3}^{-2}\)|x + 2| d x + \(\int_{-2}^3\)|x + 2| d x
when -3 < x < -2
⇒ x + 2 < 0
∴ |x + 2| = -(x + 2)
when -2 < x < 3 ⇒ x + 2 > 0
⇒ |x + 2| = x + 2
∴ I = \(\int_{-3}^{-2}\)(x + 2) d x + \(\int_{-2}^3\)(x + 2) d x
= \(-\frac{(x+2)^2}{2}\)^-2 ₃ + \(\frac{(x+2)^2}{2}\)³ _-2
= \(-\frac{1}{2}\)[0 – 1] + \(\frac{1}{2}\)[25 – 0]
= \(\frac{1}{2}\) + \(\frac{25}{2}\) = 13

Question 33.
∫ \(\frac{2 \sin 2 \theta-\cos \theta}{6-\cos ^2 \theta-4 \sin \theta}\) dθ
Answer:

Question 34.
∫ \(\frac{cosecx}{\log\tan(\frac{x}{2})}\) d x
Answer:

Question 35.
\(\int_0^1\) \(\frac{x e^x}{(1+x)^2}\) d x
Answer:

Question 36.
∫ \(\frac{x^2-5 x-1}{x^4+x^2+1}\) d x
Answer:

Question 37.
\(\int_1^2\) \(\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\) d x
Answer:

Question 38.
∫ \(\frac{1}{x[6(\log x)^2+7 \log x+2]}\) d x
Answer:

Question 39.
∫ \(\frac{1}{x+\sqrt{x}}\) d x
Answer:
Let I = ∫ \(\frac{1}{x+\sqrt{x}}\) d x
put \(\sqrt{x}\) = t
⇒ x = t²
⇒ dx = 2 t dt
= ∫\(\frac{2 t d t}{t^2+t}\) = ∫\(\frac{2 d t}{t+1}\)
= 2 log |t + 1| + C

Question 40.
\(\int_0^1\) log (\(\frac{1}{x}\) – 1) dx
Answer:

Question 41.
∫\(\frac{\cos ^{-1} x}{x^2}\) d x
Answer:

Question 42.
∫e^x \(\frac{(2+\sin 2 x)}{\cos ^2 x}\) d x
Answer:
Let I = ∫e^x \(\frac{(2+\sin 2 x)}{\cos ^2 x}\) d x
= ∫e^x [2 sec^2x + 2 tan x] d x
= ∫2 e^x sec^2x d x + ∫2 e^xtan x d x
= 2[e^x tan x – ∫e^xtan x d x] + ∫2 e^xtan x d x
= 2 e^xtan x + C

Question 43.
Using properties of definite integrate,
evalute \(\int_0^{\pi / 2}\) \(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) d x .
Answer:

Evaluate the following integrals
Question 44.
∫\(\frac{x+\sin x}{1+\cos x}\) d x
Answer:

Question 45.
∫\(\frac{2 y^2}{y^2+4}\) d y
Answer:

Question 46.
\(\int_0^3\) f(x) d x
where
f(x) = {cos 2x, 0 < x < \(\frac{\pi}{2}\)
3, \(\frac{\pi}{2}\) < x < 3
Answer:

Question 47.
∫\(\frac{\sec x}{1+cosecx}\)
Answer:

Question 48.
∫tan^3x d x
Answer:
Let I = ∫tan^3xd x
= ∫tan x(sec² x – 1) d x
= ∫tan x sec² x d x – ∫tan x d x
= ∫t d t + log |cos x| + C
[put tan x = t ⇒ sec²x d x = dt]
= \(\frac{t^2}{2}\) + log |cos x| + C
= \(\frac{\tan ^2 x}{2}\) + log |cos x| + C

Question 49.
∫\(\frac{\sin x+\cos x}{\sqrt{9+16 \sin 2 x}}\)
Answer:

Question 50.
Using properties of definite integrals, evaluate \(\int_0^{\pi / 2}\) \(\frac{\sin x-\cos x}{1+\sin x \cos x}\) dx.
Answer:

Question 51.
∫(1 – x) \(\sqrt{x}\) d x=
Answer:
∫(1 – x) \(\sqrt{x}\) d x
= \(\int x^{\frac{1}{2}}\) d x – \(\int x^{\frac{3}{2}}\) d x
= \(\frac{2}{3}\) x^{\(\frac{3}{2}\)} – \(\frac{2}{5}\) x^{\(\frac{5}{2}\)} + C

Question 52.
∫cos^-1 (sin x) d x=
Answer:
∫cos^-1 (sin x) d x
= ∫cos^-1 {cos(\(\frac{\pi}{2}\) – x)} d x
= ∫(\(\frac{\pi}{2}\) – x) d x
= \(\frac{\pi}{2}\)x – \(\frac{x^2}{2}\) + C

Question 53.
∫\(\frac{1-\sin x}{\cos ^2 x}\) d x
Answer:
∫\(\frac{1-\sin x}{\cos ^2 x}\) d x
= ∫[sec² x – tan xsec x] d x
= tan x – sec x + C

Question 54.
∫ \(\frac{(\log x)^2}{x}\) d x =
Answer:
Let I = ∫ \(\frac{(\log x)^2}{x}\) d x;
put log x = t
⇒ \(\frac{1}{x}\) d x = d t
= ∫t² d t
= \(\frac{t^3}{3}\) + C
= \(\frac{(\log x)^3}{3}\) + C

Question 55.
∫ \(\frac{\sin ^6 x}{\cos ^8 x}\) d x
Answer:
= ∫\(\frac{\sin ^6 x}{\cos ^8 x}\) d x
= ∫\(\frac{\sin ^6 x}{\cos ^6 x}\) \(\frac{1}{\cos ^2 x}\) d x
= ∫tan⁶ x sec² x d x
put tan x = t
⇒ sec²x d x = d t
= ∫t⁶d t = \(\frac{t^7}{7}\) + c
= \(\frac{\tan ^7 x}{7}\) + c

Question 56.
∫x sec² x d x
Answer:
∫x sec² x d x
= x tan x – ∫tan x d x
= x tan x + log |cos x| + C
{[∵ ∫tan x d x = -log |cos x| + C]}

Question 57.
∫sin^-1 x d x=
Answer:

Question 58.
\(\int_{-\pi}^\pi\) sin³x cos²x d x
Answer:
Let I = \(\int_{-\pi}^\pi\) sin³x cos²x d x
Here f(x) = sin³x cos²x
∴ f f(-x) = sin³(-x) cos²(-x)
= [sin (-x)]³ [cos (-x)]²
= -sin³x cos²x = -f(x)
Thus, f(x) be an odd function.
∴ \(\int_{-\pi}^\pi\) f(x) d x = 0
⇒ \(\int_{-\pi}^\pi\) sin³x cos² x d x = 0

Question 59.
\(\frac{x^4+1}{x^2+1}\) d x
Answer:
∫\(\frac{x^4+1}{x^2+1}\) d x
= ∫\(\frac{x^4-1+2}{x^2+1}\) d x
= ∫[\(\frac{x^4-1}{x^2+1}\) + \(\frac{2}{x^2+1}\)]
= ∫(x² – 1) d x + 2∫\(\frac{d x}{x^2+1}\)
= \(\frac{x^3}{3}\) – x + 2 tan^-1 x + C

Question 60.
The value of the integral
\(\int_0^{\pi / 2}\) \(\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\) d x is
Answer:

Question 61.
∫(\(\sqrt{x}\) + \(\frac{1}{\sqrt{x}}\))² d x is equal to
(a) \(\frac{x^2}{2}\) + 2 x + log |x|+ c
(b) \(\frac{x^2}{2}\) + 2 + log |x| + c
(c) \(\frac{x^2}{2}\) + x + log |x| + c
(d) \(\frac{x^2}{2}\) + 2 x + 2|log | + c
Answer:
∫(\(\sqrt{x}\) + \(\frac{1}{\sqrt{x}}\))² d x
= ∫(x + \(\frac{1}{x}\) + 2) d x
= \(\frac{x^2}{2}\) + log |x| + 2 x + C
∴ (a) \(\frac{x^2}{2}\) + 2 x + log |x|+ c

Question 62.
∫\(\frac{\sec x}{\sec x+\tan x}\) d x
(a) sec x – tan x + c
(b) log sin x + log cos x + c
(c) tan x – sec x + c
(d) log (1 + sin x) + c
Answer:
∫\(\frac{sec x}{sec x+tan x}\) d x
= ∫\(\frac{\sec x(\sec x-\tan x)}{\sec ^2 x-\tan ^2 x}\) d x
= ∫\(sec² x d x – ∫sec x tan x d x + C
= tan x – sec x + C
∴ (c) tan x – sec x + c

Question 63.
∫[latex]\frac{d x}{e^x+e^{-x}+2}\) d x is equal to
(a) \(\frac{1}{e^x+1}\) + c
(b) \(\frac{-1}{e^x+1}\) + c
(c) \(\frac{1}{e^x+1}\) + c
(d) \(\frac{1}{e^{-x}-1}\) + c
Answer:

Question 64.
∫\(\frac{\sec ^8 x}{cosecx}\) dx
(a) \(\frac{\sec ^8 x}{8}\) + c
(b) \(\frac{\sec ^7 x}{7}\) + c
(c) \(\frac{\sec ^6 x}{6}\) + c
(d) \(\frac{\sec ^9 x}{9}\) + c
Answer:
put cos x = t
⇒ -sin x d x = d t
= -∫\(\frac{d t}{t^8}\)
= –\(\frac{t^{-8+1}}{-8+1}\) + C
= \(\frac{1}{7 t^7}\) + C
= \(\frac{1}{7 \cos ^7 x}\) + C
= \(\frac{\sec ^7 x}{7}\) + C
∴ (b) \(\frac{\sec ^7 x}{7}\) + c

Question 65.
∫\(\frac{(1+x) e^x}{\sin ^2(2 e^x)}\) d x is equal to
(a) -cot (e^x ) + c
(b) tan (x e^x ) + c
(c) tan (e^x ) + c
(d) -cot(x e^x ) + c
Answer:
put x e^x = t
⇒ (x e^x + e^x ) d x = d t
⇒ (x + 1) e^x d x = d t
∴ I = ∫\(\frac{(1+x) e^x d x}{\sin ^2(x e^x)}\)
= ∫\(\frac{d t}{\sin ^2 t}\)
= ∫cosec² t d t
= -cot t + C
= -cot(x e^x) + C
∴ Ans. (d) }

Question 66.
∫\(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}\) d x is equal to
(a) x – sin^-1 x+c
(b) x – \(\sqrt{1-x^2}\) sin^-1 x + c
(c) x + sin^-1 x+c
(d) x + \(\sqrt{1-x^2}\) sin^-1 x + c
Answer:

Question 67.
∫e^x (\(\frac{1-x}{1+x^2}\))² d x is equal to
(a) \(\frac{e^x(1-x)}{1+x^2}\) + C
(b) \(\frac{e^x}{(1+x^2)^2}\) + C
(c) \(\frac{e^x}{1+x}\) + C
(d) \(\frac{e^x}{(1+x^2)}\) + C
Answer:

Question 68.
\(\int_{-5}^5\)|x + 2| d x is equal to
(a) 28
(b) 29
(c) 27
(d) 30
Answer:

Question 69.
\(\int_0^3\)[x] d x = ………………….., where |x| is greatest integer function
(a) 3
(b) 0
(c) 2
(d) 1
Answer:

Question 70.
The value of \(\int_0^{\pi / 2}\) \(\frac{d x}{1+\tan x}\) is
(a) \(\frac{\pi}{2}\)
(b) 0
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{8}\)
Answer:

Question 71.
∫x² e^x3 d x equals
(a) \(\frac{1}{3} e^{x^3}\) + C
(b) \(\frac{1}{3} e^{x^4}\) + C
(c) \(\frac{1}{2} e^{x^3}\) + C
(d) \(\frac{1}{2} e^{x^2}\) + C
Answer:
I = \(\int x^2 e^{x^3}\) d x;
put x³ = t
⇒ 3 x² d x = d t
⇒ I = ∫e^t \(\frac{d t}{3}\)
= \(\frac{1}{3}\) e^t + C
= \(\frac{1}{3} e^{x^3}\) + C
∴ Ans. (a)

Question 72.
\(\int_0^{\pi / 8}\) tan²(2 x) d x is equal to
(a) \(\frac{4-\pi}{8}\)
(b) \(\frac{4+\pi}{8}\)
(c) \(\frac{4-\pi}{4}\)
(d) \(\frac{4-\pi}{2}\)
Answer:
Let I = \(\int_0^{\pi / 8}\) tan² 2 x d x
= \(\int_0^{\pi / 8}\)(sec²2 x – 1) d x
= \(\frac{\tan 2 x}{2}-x]_0^{\pi / 8}\)
= \(\frac{1}{2}\) – \(\frac{\pi}{8}\)
= \(\frac{4-\pi}{8}\)
∴ Ans. (a)

Question 73.
∫4^x 3^x d x equals
(a) \(\frac{12^x}{\log 12}\) + c
(b) \(\frac{4^x}{\log 4}\) + c
(c) (\(\frac{4^x 3^x}{\log 4 \log 3}\) + c)
(d) (\(\frac{3^x}{\log 3}\) + c)
Answer:
Let I = ∫4^x 3^x d x
= ∫(4 × 3)^x d x
= ∫12^x d x = \(\frac{12^x}{\log 12}\) + C
{[∵∫a^x d x = \(\frac{a^x}{\log a}\)]
∴ Ans. (a)

Question 74.
∫\(\frac{d x}{\sqrt{9-25 x^2}}\) equals
(a) sin^-1(\(\frac{5 x}{3}\)) + c
(b) \(\frac{1}{5}\) sin^-1(\(\frac{5 x}{3}\)) + c
(c) \(\frac{1}{6}\)log (\(\frac{3+5 x}{3-5 x}\)) + c
(d) \(\frac{1}{30}\)log (\(\frac{3+5 x}{3-5 x}\)) + c
Answer:

Question 75.
∫\(\frac{\sec ^2(\sin ^{-1} xt)}{\sqrt{1-x^2}}\) d x is equal to
(a) sin (tan^-1 xt) + c
(b) tan (sec^-1xt) + c
(c) tan (sin^-1xt) + c
(d) -tan (cos^-1xt) + c
Answer:
put sin^-1x = t
⇒ \(\frac{1}{\sqrt{1-x^2}}\) d x = d t
∴ I = ∫sec² t d t
= tan t + C
= tan (sin^-1xt) + C
∴ Ans. (c)

Question 76.
Evaluate:
∫\(\frac{e^{\tan ^{-1} x}}{1+x^2}\) d x.
Answer:
Let I = ∫\(\frac{e^{\tan ^{-1} x}}{1+x^2}\) d x put
tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) d x = d t
∴ I = ∫e^t d t = e^t + c
= e^tan-1x + c

Question 77.
Evaluate: ∫sec x(sec x + tan x) d x.
Answer:
Let I = ∫sec x(sec x + tan x) d x.
= ∫sec² x d x + ∫sec x tan x d x
= tan x + sec x + c

Question 78.
∫\(\sqrt{1+\cos x}\) d x
Answer:
∫\(\sqrt{1+\cos x}\) d x
= ∫\(\sqrt{2 \cos ^2 \frac{x}{2}}\) d x
= \(\sqrt{2}\) ∫cos \(\frac{x}{2}\) d x
= \(\sqrt{2}\) \(\frac{\sin \frac{x}{2}}{\frac{1}{2}}\) + C
= 2 \(\sqrt{2}\) sin \(\frac{x}{2}\) + C

Question 79.
∫(cos²2 x – sin²2 x) d x
Answer:
∫(cos²2 x – sin²2 x) d x = ∫cos (4 x) d x
= \(\frac{\sin 4 x}{4}\) + C
[∵ cos²θ – sin²θ cos 2θ]

Question 80.
∫\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) d x
Answer:
Let I = ∫\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx ;
put e^2x + e^-2x = t
⇒ 2(e^2x – e^2x ) d x
= ∫\(\frac{d t}{2 t}\) = \(\frac{1}{2}\) log |t| + C
= \(\frac{1}{2}\) log |e^2x + e^-2x| + C

Question 81.
∫\(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) d x
Answer:
Let I = ∫\(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx;
put tan x = t
⇒ sec²x d x = d t
= ∫\(\frac{d t}{\sqrt{t^2+2^2}}\)
= log |t + \(\sqrt{t^2+4}\)| + C
[∵ ∫\(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + c]
= log |tan x + \(\sqrt{\tan ^2 x+4}\)| + C

Question 82.
If 0 < x < \(\frac{\pi}{4}\), then ∫\(\sqrt{1-\sin 2 x}\) dx is equal to?
Answer:
Let I = ∫\(\sqrt{1-\sin 2 x}\) d x
= ∫\(\sqrt{\cos ^2 x+\sin ^2 x-2 \sin x \cos x}\) d x
= ∫\(\sqrt{(\cos x-\sin x)^2}\) d x
= ∫|cos x – sin x| d x
0 < x < \(\frac{\pi}{4}\) ∴ cos x > sin x
⇒ cos x – sin x > 0
⇒ |cos x – sin x| = cos x – sin x
= ∫(cos x – sin x) d x
= sin x + cos x + C

Question 83.
∫\(\frac{3+3 \cos x}{x+\sin x}\) d x
Answer:
Let I = ∫\(\frac{3(1+\cos x) d x}{x+\sin x}\)
put x + sin x = t
⇒ (1 + cos x) d x = d t
= ∫\(\frac{3 d t}{t}\) = 3 log |t| + C
= 3 log |x + sin x| + C

Question 84.
∫x e^{(1 + x2)} d x
Answer:
put (1 + x²) = t
⇒ 2 x d x = d t
∴ I = ∫e^t \(\frac{d t}{2}\)
= \(\frac{1}{2}\) e^t + C
= \(\frac{1}{2}\) e^{(1 + x2)} + C

Question 85.
∫\(\frac{d x}{\sqrt{x}+x}\)
Answer:
Let I = ∫\(\frac{d x}{\sqrt{x}+x}\)
put
x = t²
⇒ d x = 2 t d t
= ∫\(\frac{2 t d t}{t+t^2}\)
= ∫\(\frac{2 d t}{1+t}\)
= 2 log |1 + t| + C
= 2 log |1 + \(\sqrt{x}\)| + C

Question 86.
∫sin \(\frac{5 x}{2}\) cos(\(\frac{x}{2}\)) d x
Answer:
Let I = ∫sin \(\frac{5 x}{2}\) cos \(\frac{x}{2}\) d x
= \(\frac{1}{2}\) ∫(sin 3 x +sin 2 x) d x
= \(\frac{1}{2}\)[-\(\frac{\cos 3 x}{3}\) – \(\frac{\cos 2 x}{2}\)] + C

Question 87.
∫\(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) d x
Sol.
Let
I = ∫\(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) d x
= ∫\(\frac{(1-2 \sin ^2 x+2 \sin ^2 x)}{\cos ^2 x}\) d x
= ∫sec² x d x
= tan x + C

Question 88.
∫\(\frac{3 x}{3 x-1}\) d x
Answer:
Let I = ∫\(\frac{3 x}{3 x-1}\) d x
= ∫\(\frac{3 x-1+1}{3 x-1}\) d x
= ∫[1 + \(\frac{1}{3 x-1}\)] d x
= x + \(\frac{1}{3}\) log |3 x – 1| + C

Question 89.
∫\(\frac{x}{\sqrt{32-x^2}}\) d x
Answer:
put x² = t
⇒ 2 x d x = d t
∴ I = ∫\(\frac{d t}{2 \sqrt{32-t}}\)
= \( \frac{1}{2}\) \(\frac{(32-t)^{-\frac{1}{2}+1}}{(-\frac{1}{2}+1)(-1)}\)
= \(-\sqrt{32-t}\) + C
= \(-\sqrt{32-x^2}\) + C

Question 90.
∫\(\frac{1}{x(1+\log x)}\) d x
Answer:
Let I =∫\(\frac{d x}{x(1+\log x)}\)
put 1 + log x = t
⇒ \(\frac{1}{x}\) d x = d t
= ∫\(\frac{d t}{t}\) + C
= log |t| + C
= log |1 + log x| + C

Question 91.
∫\(\frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x}\) d x
Answer:
∫\(\frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x}\) d x
= ∫[\(\frac{1}{\cos ^2 x}\) – \(\frac{1}{\sin ^2 x}\)] d x
= ∫[sec²x – cosec}^2 x] d x
= tan x + cot x + C

Question 92.
∫\(\frac{\log (\sin x)}{\tan x}\) d x
Answer:
Let I = ∫ \(\frac{\log (\sin x)}{\tan x}\) dx
put log (sin x) = t
⇒ \(\frac{1}{\sin x}\) cos x d x = d t
⇒ \(\frac{d x}{\tan x}\) = d t
∴ I = ∫t d t = \(\frac{t^2}{2}\) + C
= \(\frac{1}{2}\)[log (sin x)]² + C

Question 93.
∫(cosec²x – cot x) e^x d x
Answer:
Let I = ∫(cosec²x – cot x) e^x d x
= ∫e^x cosec²x d x – ∫e^xcot x d x
= e^x(-cot x) – ∫e^x(-cot x) d x
= -e^xcot x + C

Question 94.
∫\(\frac{2^{x+1}-5^{x-1}}{10^x}\) d x
Answer:

Question 95.
\(\int_{-2}^2\)(x³ – 1) d x
Answer:
\(\int_{-2}^2\)(x³ – 1) d x
= \(\frac{x^4}{4}-x]_{-2}^2\)
= ( \(\frac{2^4}{4}\) – 2) – (\(\frac{(-2)^4}{4}\) + 2) = -4

Question 96.
\(\int_2^3\) 3x d x
Ans.
\(\int_2^3\) 3x d x
= \(\frac{3^x}{\log 3}\)]³₂
= \(\frac{1}{\log 3}\)[3³ – 3²]
= \(\frac{18}{\log _e 3}\)
= 18 log₃e[∵\(\frac{1}{\log _b a}\) = log_a b]

Question 97.
\(\int_e^{e^2}\) \(\frac{1}{x \log x}\) d x
Answer:
Let I = \(\int_e^{e^2}\) \(\frac{1}{x \log x}\) d x
put log x = t
⇒ \(\frac{1}{x}\) d x = d t
When x = e
⇒ t = log e = 1
When x = e²
⇒ t = log e² = 2 log e = 2
∴ I = \(\int_1^2\) \(\frac{d t}{t}\)
= log |t|
= log 2 – log 1 = log 2

Question 98.
\(\int_0^1\) \(\frac{1}{\sqrt{1-x^2}}\) d x
Answer:
\(\int_0^1\) \(\frac{1}{\sqrt{1-x^2}}\)
= sin^-1 x]¹ ₀
= sin^-1 1 – sin^-1 0
= \(\frac{\pi}{2}\) – 0
= \(\frac{\pi}{2}\)

Question 99.
\(\int_1^4\)|x – 5| d x
Answer:
Let I = \(\int_1^4\)|x – 5| d x
When 1 < x < 4 < 5
⇒ x – 5 < 0
∴ |x – 5| = -(x – 5)
= \(-\frac{1}{2}\)[(-1)² – 4²]
= \(\frac{15}{2}\)

Question 100.
\(\int_0^{3 / 2}\)[x^2] d x
Answer:

Question 101.
\(\int_0^{\pi / 2}\)e^x(sin x – cos x) d x
Answer:

Question 102.
\(\int_0^{2 \pi}\)|sin x| d x
Answer:

Question 103.
If \(\int_0^a \frac{1}{1+4 x^2}\) d x = \(\frac{\pi}{8}\), Find a.
Answer: