Inter 2nd Year Maths 2B Integration Important Questions

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Integration Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Integration Important Questions

Question 1.
\(\int\left(\frac{1}{1-x^{2}}+\frac{2}{1+x^{2}}\right)\) [May 11]
Solution:

Question 2.
\(\int\) sec²x cosec²x dx on I ⊂ R \ ({nπ : n ∈ Z} ∪ {(2n + 1) \(\frac{\pi}{2}\) : n ∈ Z}) [T.S. Mar. 16; Mar, May 07]
Solution:

Question 3.
\(\int \frac{1+\cos ^{2} x}{1-\cos 2 x}\) dx on I ⊂ R \ {nπ : n ∈ Z} [Mar. 13]
Solution:

Question 4.
\(\int \sqrt{1-\cos 2 x}\) dx on I ⊂ [2nπ, (n + 1) π], n ∈ Z [May 06]
Solution:
\(\int \sqrt{1-\cos 2 x}\) dx = \(\int \sqrt{2}\) sin x dx
= –\(\sqrt{2}\) cos x + C

Question 5.
\(\int \frac{1}{\cosh x+\sinh x}\) dx on R. [A.P. Mar. 16]
Solution:
\(\int \frac{1}{\cosh x+\sinh x}\) dx
= \(\int \frac{\cosh x-\sinh x}{\cosh ^{2} x-\sinh ^{2} x}\) dx
= \(\int\) (cosh x – sinh x) dx
= sinh x – cosh x + C

Question 6.
\(\int \frac{1}{1+\cos x}\) dx on I ⊂ R \ {(2n + 1)π : n ∈ Z} [T.S. Mar. 15]
Solution:

Question 7.
\(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}\)dx, x ∈ R. [A.P. Mar. 15]
Solution:
\(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}\)dx
t = tan^-1 x ⇒ dt = \(\frac{d x}{1+x^{2}}\)
\(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}\)dx = \(\int\) sin t dt
= – cos t + t
= -cos (tan^-1 x) + C

Question 8.
\(\int \frac{\log (1+x)}{1+x}\)dx on (-1, ∞) [T.S. Mar. 15]
Solution:
\(\int \frac{\log (1+x)}{1+x}\)dx
t = 1 + x ⇒ dt = dx

Question 9.
\(\int \frac{x^{2}}{\sqrt{1-x^{6}}}\) dx on I = (-1, 1). [May 05]
Solution:
\(\int \frac{x^{2}}{\sqrt{1-x^{6}}}\)
t = x³ ⇒ dt = mx² dx
\(\int \frac{x^{2}}{\sqrt{1-x^{6}}}\) = \(\frac{1}{3} \int \frac{d \mathrm{t}}{\sqrt{1-\mathrm{t}^{2}}}\)
= \(\frac{1}{3}\) sin^-1 t + C
= \(\frac{1}{3}\) sin^-1 (x³) + C

Question 10.
\(\int \frac{x^{8}}{1+x^{18}}\) dx on R. [A.P. Mar. 16]
Solution:
t = x⁹ ⇒ dt = 9x⁸ dx
\(\int \frac{x^{8} d x}{1+x^{18}}=\int \frac{x^{8}}{1+\left(x^{9}\right)^{2}} d x\)
= \(\frac{1}{9} \int \frac{d t}{1+t^{2}}\) = = \(\frac{1}{9}\) tan^-1 t + C
= \(\frac{1}{9}\) tan^-1 (x⁹) + C

Question 11.
\(\int \frac{1}{x \log x[\log (\log x)]}\) dx on (1, ∞) [Mar. 11]
Solution:
t = log (log x)
dt = \(\frac{1}{\log x} \cdot \frac{1}{x}\) dx
\(\int \frac{1}{x \log x[\log (\log x)]}\) dx = \(\int \frac{d t}{t}\)
= log |t| + C
= log |log(log x)| + C

Question 12.
\(\int \frac{1}{(x+3) \sqrt{x+2}}\)dx on I ⊂ (-2, ∞) [Mar. 14]
Solution:
x + 2 = t²
dx = 2t dt
\(\int \frac{d x}{(x+3) \sqrt{x+2}}=\int \frac{2 t d t}{t\left(t^{2}+1\right)}\)
= \(2 \int \frac{d t}{t^{2}+1}\)
= 2 tan^-1 (t) + C
= 2 tan^-1 (\(\sqrt{x+2}\)) + C

Question 13.
\(\int \frac{\cot (\log x)}{x}\) dx, x ∈ I ⊂ (0, ∞) \ {e^nπ : n ∈ Z). [Mar. 05]
Solution:
t = log x ⇒ dt = \(\frac{\mathrm{dx}}{\mathrm{x}}\) \(\int\)
\(\int \frac{\cot (\log x)}{x}\) dx = \(\int\) cot t dt = log (sin t) + C
= log |sin (log x)| + C

Question 14.
\(\int\) (tan x + log sec x)e^x dx on ((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π) n ∈ Z. [May 07, Mar. 08]
Solution:
t = log |sec x| ⇒ dt = \(\frac{1}{\sec x}\) . sec x . tan x dx
= tan x dx
\(\int\) (tan x + log sec x)e^x dx = e^x . log|sec x| + C

Question 15.
\(\int \sqrt{x}\) log x dx on (0, ∞) [T.S. Mar. 16]
Solution:
\(\int \sqrt{x}\) log x dx
= (log x) \(\frac{2}{3}\) x^3/2 – \(\frac{2}{3}\) \(\int\) x^3/2 \(\frac{1}{x}\) dx
= \(\frac{2}{3}\) x^3/2 (log x) – \(\frac{2}{3}\) \(\int\) x^1/2 dx
= \(\frac{2}{3}\) x^3/2 (log x) – \(\frac{2}{3}\) \(\frac{x^{3 / 2}}{3 / 2}\) + C
= \(\frac{2}{3}\) x^3/2 (log x) – \(\frac{4}{9}\) x^3/2 + C

Question 16.
\(\int\) e^x (tan x + sec² x)dx on I ⊂ R \ {(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z} [Mar 06]
Solution:
f(x) = tanx = f'(x) ⇒ sec² x dx
I = \(\int\) e^x [f(x) + f'(x)] dx = e^x. f(x) + C
= e^x . tan x + C

Question 17.
\(\int e^{x}\left(\frac{1+x \log x}{x}\right)\) dx on (0, ∞) [A.P. Mar. 15, Mar. 13]
Solution:
\(\int e^{x}\left(\frac{1+x \log x}{x}\right)\) dx = \(\int\) e^x (log x + \(\frac{1}{x}\))dx
= e^x . log x + C

Question 18.
\(\int \frac{\cos x}{\sin ^{2} x+4 \sin x+5}\)dx [Mar. 07]
Solution:
t = sin x ⇒ dt = cos x dx
I = \(\int \frac{d t}{t^{2}+4 t+5}=\int \frac{d t}{(t+2)^{2}+1}\)
= tan^-1(t + 2) + C
= tan^-1(sin x + 2) + C

Question 19.
\(\int \frac{d x}{(x+1)(x+2)}\) [Mar. 14, May 11]
Solution:

Question 20.
\(\int \frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)}\) dx on I ⊂ R \ {x ∈ R : cos (xe^x) = 0} [T.S. Mar. 17]
Solution:
t = x . e^x
dt = (x . e^x + e^x) dx = e^x (1 + x) dx
\(\int \frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)}\) dx = \(\int \frac{d t}{\cos ^{2} t}=\int \sec ^{2} t d t\)
= tan t + C
= tan (x . e^x) + C

Question 21.
\(\int\) x tan^-1 x dx, x∈ R [Mar. 05]
Solution:

Question 22.
\(\int \sqrt{1+3 x-x^{2}} d x\) [May 11]
Solution:
\(\int \sqrt{1+3 x-x^{2}} d x=\int \sqrt{1-\left(x^{2}-3 x\right)} d x\)
= \(\int \sqrt{1-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)} d x\)

Question 23.
\(\int \frac{9 \cos x-\sin x}{4 \sin x+5 \cos x} d x\) [T.S. Mar. 17; Mar. 08]
Solution:

Question 24.
\(\int \frac{d x}{5+4 \cos 2 x}\) [Mar. 11]
Solution:

Question 25.
Obtain reduction formula for I_n = \(\int\) cotⁿ x dx, n being a positive integer. n ≥ 2 and deduce the value of \(\int\) cot⁴ x dx. [T.S. Mar. 19] [A.P. Mar. 16; May 11]
Solution:

Question 26.
Obtain reduction formula for I_n = \(\int\) cosecⁿ x dx, n being a positive integer. n ≥ 2 and deduce the value of \(\int\) cosec⁵ x dx. [T.S. Mar. 16]
Solution:
I_n = \(\int\) cosecⁿ x dx = \(\int\) cosec^{n – 2}x . cosec² x dx
= cosec^{n – 2}x (-cot x) + \(\int\) cot x . (n – 2) cosec^{n – 3} . (cot x) dx
= – cosec^{n – 2}x . cot x + (n – 2) \(\int\) cosec^{n – 2} x . (cosec² x -1) dx
= – cosec^{n – 2}x . cot x + (n – 2) I_{n – 2} – (n – 2)I_n
I_n (1 + n – 2) = – cosec^{n – 2} x . cot x + (n – 2) I_{n – 2}

Question 27.
Evaluate \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\) dx. [T.S. Mar. 15]
Solution:
We write
2x + 5 = A \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² – 2x + 10) + B
= A(2x – 2) + B
On comparing the coefficients of the like powers of x on both sides of the above equation, we get A = 1 and B = 7.
Thus 2x + 5 = (2x – 2) + 7
Hence \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\) dx
= \(\int \frac{2 x-2}{\sqrt{x^{2}-2 x+10}} d x+7 \int \frac{d x}{\sqrt{x^{2}-2 x+10}}+C\)
= \(2 \sqrt{x^{2}-2 x+10}+7 \int \frac{d x}{\sqrt{(x-1)^{2}+3^{2}}}\) + C
= \(2 \sqrt{x^{2}-2 x+10}+7 \sinh ^{-1}\left(\frac{x-1}{3}\right)\) + C

Question 28.
Evaluate \(\int\) sin⁴ x dx. [Mar. 14]
Solution:
I_n = sinⁿ x dx = \(-\frac{\sin ^{n-1} x \cdot \cos x}{n}+\frac{n-1}{n} \cdot I_{n-2}\)

Question 29.
\(\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x\) [A.P. Mar. 15]
Solution:
Let 2 cos x + 3 sin x = A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x)
Equating the coefficient of sin x and cos x we get
4A + 5B = 2
5A – 4B = 3

Question 30
\(\int \frac{1}{1+\sin x+\cos x}\) dx [T.S. Mar. 15]
Solution:

Question 31.
\(\int\) (6x + 5) \(\sqrt{6-2 x^{2}+x}\) dx [May 06]
Solution:
Let 6x + 5 = A(1 – 4x) + B
Equating the co-efficients of x
6 = -4 A ⇒ A = \(\frac{-3}{2}\)
Equating the constants
A + B = 5
B = 5 – A = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)

Question 32.
\(\int \frac{d x}{4+5 \sin x}\) [Mar. 05]
Solution:
t = tan \(\frac{x}{2}\) ⇒ dt = sec² \(\frac{x}{2}\) . \(\frac{1}{2}\) dx

Question 33.
\(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^{2}}}\) [May 05]
Solution:

Question 34.
\(\int \frac{d x}{4 \cos x+3 \sin x}\) [Mar. 06]
Solution:

Question 35.
\(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5} d x\) [Mar. 14, 11] [A.P. & T.S. Mar. 16]
Solution:
Let 2 sin x + 3 cos x + 4
= A(3 sin x+4 cos x + 5) + 3(3 cos x – 4sin x) + C
Equating the co-efficients of
sin x. we get 3A – 4B = 2
cos x, we get 4A + 3B = 3


Substituting in (1)
I = \(\frac{18}{25}\) . x + \(\frac{1}{25}\) log |3 sin x + 4 cos x + 5| – \(\frac{4}{5\left(3+\tan \frac{x}{2}\right)}\) + C

Question 36.
\(\int \frac{x+3}{(x-1)\left(x^{2}+1\right)} d x\) dx [May 07]
Solution:
Let \(\frac{x+3}{(x-1)\left(x^{2}+1\right)}\) = \(\frac{A}{x-1}\) + \(\frac{B x+C}{x^{2}+1}\)
⇒ (x + 3) = A(x² + 1) + (Bx + C)(x – 1) ………………… (1)
Put x = 1 in (1)
Then 4 = A(1 + 1) + 0 ⇒ A = 2
Put x = 0 in (1)
3 = A(1) + C(-1)
⇒ A – C = 3 ⇒ C = A – 3 = 2 – 3 = -1
Equating coefficient of x² in (1)
0 = A + B
⇒ B = -A = -2

Question 37.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\) [Mar. 13]
Solution:

Question 38.
Find \(\int\) 2x⁷ dx on R.
Solution:
\(\int\) 2x⁷ dx = 2 \(\int\) x⁷ dx
= 2 . \(\frac{x^{8}}{8}\) + C
= \(\frac{x^{8}}{4}\) + C

Question 39.
Evaluate \(\int\) cot²x dx on I ⊂ R \ {nπ : n ∈ Z}.
Solution:
\(\int\) cot²x dx = \(\int\) (cosec²x – 1) dx
= \(\int\) cosec² x dx – \(\int\) dx
= -cot x – x + C

Question 40.
Evaluate \(\int\left(\frac{x^{6}-1}{1+x^{2}}\right)\) dx for x ∈ R.
Solution:
\(\int\left(\frac{x^{6}-1}{1+x^{2}}\right)\) dx = \(\int\)[(x⁴ – x² + 1) – \(\frac{2}{1+x^{2}}\)] dx
= \(\int\) x⁴ dx – \(\int\) x² dx + \(\int\) dx – 2 \(\int \frac{d x}{1+x^{2}}\)
= \(\frac{x^{5}}{5}\) – \(\frac{x^{3}}{3}\) + x – 2 tan^-1 x + C.

Question 41.
Find \(\int\) (1 – x) (4 – 3x) (3 + 2x) dx, x ∈ R.
Solution:
(1 – x) (4 – 3x) (3 + 2x) = 6x³ – 5x² – 13x + 12
\(\int\)(1 – x) (4 – 3x) (3 + 2x) dx
= \(\int\) (6x³ – 5x² – 13x + 12) dx
= 6\(\int\) x³dx – 5 \(\int\) x² dx – 13 \(\int\) x dx + 12 \(\int\) dx
= \(\frac{6 x^{4}}{4}\) – 5 \(\frac{x^{3}}{3}\) – \(\frac{13 x^{2}}{2}\) + 12x + C
= \(\frac{3}{2}\)x⁴ – \(\frac{5}{3}\)x³ – \(\frac{13}{2}\)x² + 12x + C.

Question 42.
Evaluate \(\int\left(x+\frac{1}{x}\right)^{3}\) dx, x > 0.
Solution:
(x + \(\frac{1}{x}\))³ = x³ + 3x + \(\frac{3}{x}\) + \(\frac{1}{x^{3}}\)
\(\int\) (x + + \(\frac{1}{x}\))³ dx = \(\int\) (x³ + 3x + \(\frac{3}{x}\) + \(\frac{1}{x^{3}}\)) dx
= \(\int\) x³ dx + 3 \(\int\) x dx + 3 \(\int\) \(\frac{\mathrm{dx}}{\mathrm{x}}\) + \(\int\) \(\frac{d x}{x^{3}}\)
= \(\frac{x^{4}}{4}\) + \(\frac{3 x^{2}}{2}\) + 3 log x – \(\frac{1}{2 x^{2}}\) + C

Question 43.
Find \(\int \sqrt{1+\sin 2 x}\) dx on R.
Solution:
1 + sin 2x = sin² x + cos² x + 2 sin x . cos x
= (sin x + cos x)²
\(\sqrt{1+\sin 2 x}\) = sin x + cos x
If 2nπ – \(\frac{\pi}{4}\) ≤ x ≤ 2nπ + \(\frac{3\pi}{4}\)
= -(sin x + cos x). otherwise
If 2nπ – \(\frac{\pi}{4}\) ≤ x ≤ 2nπ + \(\frac{3\pi}{4}\), then
\(\int \sqrt{1+\sin 2 x}\) dx = \(\int\) (sin x + cos x)dx
= -cos x + sin x + C
If 2nπ + \(\frac{3\pi}{4}\) ≤ x ≤ 2nπ + \(\frac{7\pi}{4}\)
\(\int \sqrt{1+\sin 2 x}\)
= \(\int\) -(sin x + cos x)dx
= –\(\int\) sin x dx – \(\int\) cos x dx
= cos x – sin x + c

Question 44.
Evaluate \(\int \frac{2 x^{3}-3 x+5}{2 x^{2}}\) dx for x > 0 and verify the result by differentiation.
Solution:

and it is the given expression and hence the result is correct.

Question 45.
Evaluate \(\int \frac{x^{5}}{1+x^{12}}\) dx on R.
Solution:
We define f : R → R by f(t) = \(\frac{1}{1+t^{2}}\)
g : R → R by g(x) = x⁶
Then g'(x) = 6x⁵
Define F : R → R by F(t) = tan^-1 t
F is the primitive of f
\(\int \frac{x^{5}}{1+x^{12}}\) dx = \(\frac{1}{6}\) \(\int\) f(g(x)) g'(x) dx
= \(\frac{1}{6}\) (F(t) + C)_t=g(x)
= \(\frac{1}{6}\) [tan^-1 t + C]_t=x⁶
= \(\frac{1}{6}\) tan^-1 x⁶ + C

Question 46.
Evaluate \(\int\) cos³ x sin x dx on R.
Solution:
We define : f = R → R by f(x) = cosx
∴ f'(x) = – sin x
\(\int\) cos³ x sin x dx = \(\int\) (f(x))³ [-f'(x)] dx
= \(\frac{-[f(x)]^{4}}{4}\) + C
= \(\frac{-\cos ^{4} x}{4}\) + C

Question 47.
Find \(\int\) (1 – \(\frac{1}{x^{2}}\)) e^{(x + \(\frac{1}{x}\))} dx on I where I = (0, ∞)
Solution:
Let J = I = (0, ∞)
Define f : I → R by f(t) = e^t and g : J → R by g(x) = x + \(\frac{1}{x}\)
Then g(J) ⊂ I, g'(x) = 1 – \(\frac{1}{x^{2}}\)
\(\int\) (1 – \(\frac{1}{x^{2}}\)) e^{(x + \(\frac{1}{x}\))} dx = \(\int\) f (g(x)) g'(x) dx
= \(\int\) [f(t)dt]_{t = g(x)}
= [\(\int\)e^t dt]_{t = g(x)}
= [e^t + c]_{t=x+\(\frac{1}{x}\)}
= e^{(x+\(\frac{1}{x}\))} + C

Question 48.
Evaluate \(\int \frac{1}{\sqrt{\sin ^{-1} x} \sqrt{1-x^{2}}}\) dx on I = (0, 1).
Solution:
We define f : I → R by f(x) = sin^-1x
f'(x) = \(\frac{1}{\sqrt{1-x^{2}}}\)
\(\int \frac{1}{\sqrt{\sin ^{-1} x} \sqrt{1-x^{2}}}\) dx = \(\int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x\)
= 2\(\sqrt{f(x)}\) + C
= 2 \(\sqrt{\sin ^{-1} x}\) + C

Question 49.
Evaluate \(\int \frac{\sin ^{4} x}{\cos ^{6} x}\) dx, x ∈ I ⊂ R \ {\(\frac{(2 n+1) \pi}{2}\) : n ∈ z}
Solution:
\(\int \frac{\sin ^{4} \dot{x}}{\cos ^{6} x}\) dx = \(\int\) tan⁴ x . sec² x dx
We define f : I → R by f(x)
f(x) = tan x, then f'(x) = sec²x
\(\int \frac{\sin ^{4} \dot{x}}{\cos ^{6} x}\) dx = \(\int\) [f(x)]⁴ . f'(x) dx = \(\frac{[f(x)]^{5}}{5}\) + C.
= \(\frac{1}{5}\) tan⁵x + C.

Question 50.
Evaluate \(\int\) sin² x dx on R.
Solution:
\(\int\) sin²x dx = \(\int \frac{(1-\cos 2 x)}{2}\) dx
= \(\frac{1}{2}\) \(\int\) dx – \(\frac{1}{2}\) \(\int\) cos 2x dx
= \(\frac{1}{2}\) x – \(\frac{1}{4}\) sin 2x + C.
(since \(\int\) cos 2x dx = \(\frac{1}{2}\) sin 2x + C)

Question 51.
Evaluate \(\int \frac{1}{a \sin x+b \cos x}\) dx where a, b ∈ R and a² + b² ≠ 0 on R.
Solution:
We can find real numbers r and θ such that
a = r cos θ, b = r sin θ
Then r = \(\sqrt{a^{2}+b^{2}}\), cos θ = \(\frac{a}{r}\) and sin θ = \(\frac{b}{r}\)
a sin x + b cos x = r . cos θ sin x + r sin θ cos x
= r[cos θ sin x + sin θ cos x]
= r sin (x + θ)
\(
= [latex]\frac{1}{r}\) (cosec (x + θ) dx
= \(\frac{1}{r}\) log |tan \(\frac{1}{2}\)(x + θ)| + C
= \(\frac{1}{\sqrt{a^{2}+b^{2}}} \log \left|\tan \frac{1}{2}(\tilde{x}+\theta)\right|+c\)
For all x ∈ I where I is an interval disjoint with {nπ – θ : n ∈ z}.

Question 52.
Find \(\int \frac{x^{2}}{\sqrt{x+5}}\) dx on (-5, ∞)
Solution:
Put t = x + 5 so that t > 0 on (-5, ∞)
dx = dt and x = t – 5

Question 53.
Find \(\int \frac{x}{\sqrt{1-x}}\) dx, x ∈ I = (0, 1)
Solution:
We define f : I → R by f(x) = \(\frac{x}{\sqrt{1-x}}\)
Let J = (0, \(\frac{\pi}{2}\))
Define Φ : J → I by Φ(θ) = sin² θ
Then Φ is a bijective mapping from J to I Further Φ and Φ^-1 are differentiable on their respective domains.
put x = Φ(θ) = sin²θ
dx = 2 sin θ . cos θ dθ

Question 54.
Evaluate \(\int \frac{d x}{(x+5) \sqrt{x+4}}\) on (-4, ∞)
Solution:
Let I = (-4, ∞)
Define f on I as f(x) = \(\frac{d x}{(x+5) \sqrt{x+4}}\)
Let J = (0, ∞)
We define Φ : J → I by Φ(t) = t² – 4
Then Φ is differentiable and is a bijection
Put x = Φ(t) = t² – 4
then t = \(\sqrt{x+4}\) ⇒ dx = 2t dt
Thus \(\int \frac{d x}{(x+5) \sqrt{x+4}}\) = \(\int \frac{2 t d t}{\left(t^{2}+1\right) t}\)
= \(\int \frac{2 d t}{t^{2}+1}\)
= 2 tan^-1 t + C
= 2 tan^-1 (\(\sqrt{x+4}\)) + C

Question 55.
Evaluate \(\int \frac{d x}{\sqrt{4-9 x^{2}}}\) on I = (-\(\frac{2}{3}\), \(\frac{2}{3}\))
Solution:

Question 56.
Evaluate \(\int \frac{1}{a^{2}-x^{2}}\) dx for x ∈ I = (-a, a)
Solution:

Question 57.
Evaluate \(\int \frac{1}{1+4 x^{2}}\) dx on R.
Solution:
\(\int \frac{1}{1+4 x^{2}}\) dx = \(\int \frac{d x}{4\left(\left(\frac{1}{2}\right)^{2}+x^{2}\right)}\)
= \(\frac{1}{4}\) (2 tan^-1 2x) + C
= \(\frac{1}{2}\) tan^-1 2x + C

Question 58.
Find \(\int \frac{1}{\sqrt{4-x^{2}}}\) dx on (-2, 2).
Solution:
\(\int \frac{1}{\sqrt{4-x^{2}}}\) dx = \(\int \frac{1}{\sqrt{2^{2}-x^{2}}}\) dx = sin^-1(\(\frac{x}{2}\) dx) + C

Question 59.
Evaluate \(\int \sqrt{4 x^{2}+9}\) dx on R.
Solution:

Question 60.
Evaluate \(\int \sqrt{9 x^{2}-25} d x\) on [\(\frac{5}{3}\), ∞)
Solution:

Question 61.
Evaluate \(\int \sqrt{16-25 x^{2}}\) dx on (\(\frac{-4}{5}\), \(\frac{4}{5}\))
Solution:

Question 62.
Evaluate \(\int\) x sin^-1x dx on (-1, 1).
Solution:
Let u(x) = sin^-1 x and v(x) = \(\frac{x^{2}}{2}\) so that
v'(x) = x
∴ u(x) v'(x) = x sin^-1x
Even though the domain of u is (-1, 1) the function u ¡s differentiable only on (-1, 1).
From the same formula, we have

Question 63.
Evaluate \(\int\) x² cos x dx.
Solution:
Let us take u(x) = x², v(x) = sin x
so that v'(x) = cos x
u(x) v'(x) = x² cosx
By using the formula for integration by parts, we have
\(\int\) x² cos x dx = x² sin x – \(\int\) sin x (x²)’ dx
= x² sin x – 2 \(\int\) x sin x dx + C.
Again, by applying the formula for integration by parts to
\(\int\) x sin x dx, we get
\(\int\) x. sin x dx = -x cos x – \(\int\) (-cos x) dx
= -x cos x + sin x + C₂
\(\int\) x² cos x dx = x² sin x – 2(sin x – x cos x) + C
= x² sin x – 2 sin x + 2x cosx + C
= (x² – 2) sin x + 2x cos x + C
In evaluating certain integrals by using the formula for integration by parts,, twice or more than twice, we come across the given integral with change of sign. This enables us to evaluate the given integral.

Question 64.
Evaluate \(\int\) e^x sin x dx on R.
Solution:
Let A = e^x sin x dx on R
A = \(\int\) e^x . sin x dx = \(\int\) e^x (-cos x)’ dx.
= e^x (-cos x) – \(\int\) (-cos x) (e^x)’ dx
= – e^x cos x + \(\int\) e^x cos x dx + C₁ ………(1)
\(\int\) e^x cos x dx = e^x. sin x – \(\int\) e^x . sin x dx
= e^x sinx – A …………….. (2)
From (1) and (2)
A = – e^x cos x + e^x sin x – A + C₁
2A = e^x (sin x – cos x) + C₁
A = \(\frac{1}{2}\) e^x (sin x – cos x) + C where
C = \(\frac{C_{1}}{2}\)
i.e., \(\int\) e^x sin x dx = \(\frac{1}{2}\) e^x (sin x – cos x) + C.

Question 65.
Find \(\int\) e^ax cos (bx + c) dx on R, where a. b, c are real numbers and b ≠ 0.
Solution:
Let A = \(\int\) e^ax cos (bx + c)dx
Then from the formula for integration by parts
A = e^ax [latex]\frac{\sin (b x+c)}{b}[/latex] – \(\int\) a e^ax [latex]\frac{\sin (b x+c)}{b}[/latex] dx
= \(\frac{1}{b}\) e^ax sin(bx + c) – \(\frac{a}{b}\) \(\int\) e^ax . sin(bx + c) dx

By taking c = 0, we get
\(\int\) e^ax . cos bx dx = \(\frac{e^{a x}}{a^{2}+b^{2}}\) [a cos bx + b sin bx] + K

Question 66.
Evaluate \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\) dx, on (-1, 1)
Solution:
Put x = cos θ, θ ∈ (0, π) dx = -sin θ . dθ

Question 67.
Evaluate \(\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x\) on I ⊂ R \ {2nπ : n ∈ z}.
Solution:

= -e^x . cot \(\frac{x}{2}\) + C

Question 68.
Evaluate \(\int \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) d x\) on I ⊂ R \ {-1, 1}
Solution:
Let x = tan θ ⇒ dx = sec² θ dθ
\(\frac{2 x}{1-x^{2}}=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\) = tan 2θ
tan^-1 \(\left(\frac{2 x}{1-x^{2}}\right)\) = tan^-1 (tan 2θ ) = 2θ + nπ
Where n = 0 if |x| < 1 = -1 if x > 1
= 1 if x < -1
We have dθ = \(\frac{1}{1+x^{2}}\) dx and
1 + x² = 1 + tan² θ = sec²θ
∴ \(\int \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) d x\)
= \(\int\left(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)\left(1+x^{2}\right) \frac{1}{1+x^{2}} d x\)
= \(\int\) (2θ + nπ) \(\int\) sec²θ dθ
= 2 \(\int\) θ sec² θ dθ + nπ ) \(\int\) sec² θ dθ + c
= 2 (θ tan θ – \(\int\) tan θ dθ) nπ tan θ + c
= 2 (θ tan θ + log |cos θ| + nπ tan θ + c
= (2θ + nπ) tan θ + 2 log cos θ + c
= (2θ + nπ) tan θ + log cos² θ + c
= (2θ + nπ) tan θ + log sec² θ + c
= x tan^-1 \(\left(\frac{2 x}{1-x^{2}}\right)\) – log (1 + x²) + c

Question 69.
Find \(\int x^{2} \cdot \frac{\exp \left(m \sin ^{-1} x\right)}{\sqrt{1-x^{2}}}\)dx on (-1, 1) where m is a real number. (Here for y ∈ R, exp. {y} stands for e^y).
Solution:
Let t = sin^-1x, then

Question 70.
Evaluate \(\int \frac{d x}{4 x^{2}-4 x-7}\)
Solution:

Question 71.
Find \(\int \frac{d x}{5-2 x^{2}+4 x}\)
Solution:
5 – 2x² + 4x = -2 (x² – 2x – \(\frac{5}{2}\))
= -2 ((x – 1)² – \(\frac{5}{2}\) – 1)
= -2 ((x – 1)² – \(\left(\sqrt{\frac{7}{2}}\right)^{2}\))
\(\int \frac{d x}{5-2 x^{2}+4 x}\)
= \(-\frac{1}{2} \int \frac{1}{\left((x-1)^{2}-\sqrt{\frac{7}{2}}\right)^{2}} d x+C\)

Question 72.
Evaluate \(\int \frac{d x}{x^{2}+x+1}\)
Solution:

Question 73.
Evaluate \(\int \frac{d x}{\sqrt{x^{2}+2 x+10}}\)
Solution:

Question 74.
Evaluate \(\int \frac{d x}{\sqrt{1+x-x^{2}}}\)
Solution:

Question 75.
Evaluate \(\int \sqrt{3+8 x-3 x^{2}} d x\)
Solution:

Question 76.
Evaluate \(\int \frac{x+1}{x^{2}+3 x+12}\) dx.
Solution:
We write x + 1 = A(2x + 3) + B
Equating the co-efficients of x; we get 1 = 2A.
A = \(\frac{1}{2}\)
Equating the constants 3A + B = 1
B = 1 – 3A = 1 – \(\frac{3}{2}\) = –\(\frac{1}{2}\)
x + 1 = \(\frac{1}{2}\) (2x + 3) – \(\frac{1}{2}\)

Question 77.
Evaluate \(\int(3 x-2) \sqrt{2 x^{2}-x+1} d x\)
Solution:
Let (3x – 2) = A(4x – 1) + B
Equating the co-efficients of x, we get 3 = 4A
A = \(\frac{3}{4}\)
Equating the constants -2 = -A + B
B = -2 + A = -2 + \(\frac{3}{4}\)
= \(\frac{-5}{4}\)

Question 78.
Evaluate \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\)dx. [T.S. Mar. 15]
Solution:
We write
2x + 5 = A \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² – 2x + 10) + B
= A (2x – 2) + B
On comparing the coefficients of the like powers of x on both sides of the above equation, we get A = 1 and B = 7.
Thus 2x + 5 = (2x – 2) + 7
Hence \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\)dx
= \(\int \frac{2 x-2}{\sqrt{x^{2}-2 x+10}} d x+7 \int \frac{d x}{\sqrt{x^{2}-2 x+10}}+C\)

Question 79.
Evaluate \(\int \frac{d x}{(x+5) \sqrt{x+4}}\)
Solution:
Put t = \(\sqrt{x+4}\)
dt = \(\frac{1}{2 \sqrt{x+4}}\) dx
We have t² = x + 4
x + 5 = t² + 1
\(\int \frac{d x}{(x+5)(\sqrt{x+4})}=\int \frac{2}{t^{2}+1} d t\)
= 2 tan^-1 t + C
= 2 tan^-1 (\(\sqrt{x+4}\)) + C.

Question 80.
Evaluate \(\int \frac{d x}{5+4 \cos x}\)
Solution:

Question 81.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\) [Mar. 13]
Solution:
t = tan \(\frac{x}{2}\) ⇒ dx = \(\frac{2 d t}{1+t^{2}}\)

Question 82.
Find \(\int \frac{d x}{d+e \tan x}\)
Solution:

Question 83.
Evaluate \(\int \frac{\sin x}{d \cos x+e \sin x} d x\) and \(\int \frac{\cos x}{d \cos x+e \sin x} d x\).
Solution:

Question 84.
Evaluate \(\int \frac{\cos x+3 \sin x+7}{\cos x+\sin x+1} d x\)
Solution:
Let cos x + 3 sin x + 7 = A(cos x + sin x + 1)’ + B(cos x + sin x + 1) + C
Comparing the coefficients
A + B = 1, A – B = 3, B + C = 7
A = -1, B = 2, C = 5

Question 85.
Find \(\int \frac{x^{3}-2 x+3}{x^{2}+x-2}\) dx.
Solution:

Question 86.
Find \(\int \frac{d x}{x^{2}-81}\)
Solution:

Question 87.
Find \(\int \frac{2 x^{2}-5 x+1}{x^{2}\left(x^{2}-1\right)}\) dx.
Solution:
Let \(\frac{2 x^{2}-5 x+1}{x^{2}\left(x^{2}-1\right)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}+\frac{D}{x+1}\)
2x² – 5x + 1 = Ax(x² – 1) + B(x² – 1) + Cx² (x + 1) + Dx² (x – 1)
x = 1 ⇒ 2 – 5 + 1 = C (1 + 1) ⇒ 2C = -2
⇒ C = -1
x = -1 ⇒ 2 + 5 + 1 = D (-1 -1)
⇒ 8 = -2B ⇒ D = -4
x = 0 ⇒ 1 = B(-1) ⇒ B = -1
Equating the coefficients of x³
0 = A + C + D ⇒ A = -C – D = 1 + 4 = 5

Question 88.
Find \(\int \frac{3 x-5}{x\left(x^{2}+2 x+4\right)}\) dx.
Solution:
\(\frac{3 x-5}{x\left(x^{2}+2 x+4\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+2 x+4}\)
3x – 5 = A(x² + 2x + 4) + (Bx + C) . x
x = 0 ⇒ -5 = 4 A ⇒ A = –\(\frac{5}{4}\)
Equating the coefficients of x²
A + B = 0 ⇒ B = -A = \(\frac{5}{4}\)
Equating the coefficient of x
3 = 2 A + C
C = 3 – 2 A = 3 + 2 . \(\frac{5}{4}\) = \(\frac{11}{2}\)

Question 89.
Find \(\int \frac{2 x+1}{x\left(x^{2}+4\right)^{2}} d x\)
Solution:
Let \(\frac{2 x+1}{x\left(x^{2}+4\right)^{2}}=\frac{A}{x}+\frac{B x+C}{x^{2}+4}+\frac{D x+E}{\left(x^{2}+4\right)^{2}}\)
2x + 1 = A (x² + 4)² + (Bx + C) + x (x² + 4) + (Dx + E)x
Equating the coefficients of like power of x, we obtain
A + B = 0, C = 0, 8A + 4B + D = 0,
4C + E = 2, A = \(\frac{1}{16}\)
Solving these equation, we obtain

Question 90.
Evaluate \(\int\) x³ . e^5x dx.
Solution:
We take a = 5, use the reduction formula

Question 91.
Evaluate \(\int\) sin⁴ x dx. [Mar. 14]
Solution:

Question 92.
Evaluate \(\int\) tan⁶ x dx.
Solution:

Question 93.
\(\int\) sec⁵ x dx.
Solution:
Reduction formula is