Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(b)

I. Evaluate the following integrals.

Question 1.
∫e2x dx, x ∈ R.
Solution:
∫e2x dx = \(\frac{e^{2x}}{2}\) + C

Question 2.
∫sin 7x dx, x ∈ R.
Solution:
∫sin 7x dx = \(\frac{\cos 7x}{7}\) + C

Question 3.
∫\(\frac{x}{1+x^2}\) dx, x ∈ R.
Solution:
∫\(\frac{xdx}{1+x^2}\) dx = \(\frac{1}{2}\)∫\(\frac{2xdx}{1+x^2}\) = \(\frac{1}{2}\) log(1+x²) + C

Question 4.
∫2xsin(x²+1) dx, x ∈ R.
Solution:
∫2x.sin(x²+1) dx, x ∈ R.
t = x² + 1 ⇒ dt = 2x dx
∫2x. sin(x²+1) dx = ∫ sin t dt = -cos t + C
= -cos (x²+1) + C

Question 5.
∫\(\frac{(logx)^2}{x}\) dx on I ⊂ (0, ∞).
Solution:
∫\(\frac{(logx)^2}{x}\) dx
t = log x ⇒ dt = \(\frac{1}{x}\) dx
∫\(\frac{(logx)^2}{x}\) dx = ∫t² dt
= \(\frac{(t^3}{3}\) + C = \(\frac{(logx)^3}{3}\) + C

Question 6.
Inter 2nd Year Maths 2B Integration Solutions Ex 6(a) 21 dx on I ⊂ (1, ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(a) 20

Question 7.
∫\(\frac{\sin(Tan^{-1}x)}{1+x^2}\) dx, x ∈ R.
Solution:
∫\(\frac{\sin(tan^{-1}x)}{1+x^2}\) dx
t = tan-1 x ⇒ dt = \(\frac{dx}{1+x^2}\)
∫\(\frac{\sin(tan^{-1}x)}{1+x^2}\)dx = ∫ sin t
= -cos t + t
= -cos (tan-1 x) + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 8.
∫\(\frac{1}{8+2x^2}\) dx on R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 1

Question 9.
∫\(\frac{3x^2}{1+x^6}\)dx, on R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 2

Question 10.
∫\(\frac{2}{\sqrt{25+9x^2}}\)dx on R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 3

Question 11.
∫\(\frac{3}{\sqrt{9x^2-1}}\)dx on (\(\frac{1}{3}\), ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 4
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 5

Question 12.
∫sin mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx cos nx = \(\frac{1}{2}\)∫ 2 sin mx. cos nx dx
= \(\frac{1}{2}\)∫sin(m+n)x + sin(m-n)x) dx
= –\(\frac{1}{2}\)(\(\frac{\cos (m+n) x}{m+n}+ cos\frac{(m-n)x}{m-n}\)) + C

Question 13.
∫sin mx sin nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx. sin nx dx = \(\frac{1}{2}\) ∫2 sin mx. sin nx dx
= \(\frac{1}{2}\)∫cos (m – n)x – cos (m + n)x dx
= \(\frac{1}{2}\)(\(\frac{\sin (m-n) x}{m-n}+\frac{\sin (m+n)x}{m+n}\)) + C

Question 14.
∫cos mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫cos mx. cos nx dx = \(\frac{1}{2}\) ∫2 cos mx.cos nx dx
= \(\frac{1}{2}\)∫cos (m + n)x – cos (m – n)x dx
= \(\frac{1}{2}\)(\(\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n)x}{m-n}\)) + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 15.
∫sin x sin 2x. sin 3x dx on R.
Solution:
sin 2x . sin 3x = \(\frac{1}{2}\)(2 sin 3x. sin 2x)
= \(\frac{1}{2}\)(cos x – cos 5x)
sin x sin 2x sin 3x
= \(\frac{1}{2}\)(sin x . cos x – cos 5x . sin x)
= \(\frac{1}{2}\)(\(\frac{1}{2}\) sin 2x – \(\frac{1.2}{2}\)cos 5x . sin x)
= \(\frac{1}{2}\)(\(\frac{1}{2}\) sin 2x – \(\frac{1}{2}\) – (sin 6x – sin 4x)
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 6

Question 16.
∫\(\frac{\sin x}{\sin(a+x)}\) dx on I ⊂ R\{nπ – a : n ∈ Z}.
Solution:
sin x = sin (a + x – a)
= sin (a + x) . cos a – cos (a + x) sin a
∫\(\frac{\sin x}{\sin(a+x)}\)dx
= cos a ∫ dx – sin a ∫\(\frac{\cos (a+x)}{\sin(a+x)}\) dx
= x . cos a – sin a . log |sin (a + x)| + c

II. Evaluate the following integrals.

Question 1.
∫(3x – 2)½ dx on (\(\frac{2}{3}\), ∞)
Solution:
t = 3x – 2 ⇒ dt = 3 dx
∫(3x – 2)½ dx = \(\frac{1}{3}\) ∫t½ dt = \(\frac{1}{3}\) \(\frac{t^{3/2}}{3/2}\) + C
= \(\frac{2}{9}\)(3x – 2)3/2 + C

Question 2.
∫\(\frac{1}{7x+3}\) dx on I ⊂ R\{-\(\frac{3}{7}\)}.
Solution:
∫\(\frac{1}{7x+3}\) dx
t = 7x + 3
⇒ dt = 7 dx
= ∫\(\frac{1}{7x+3}\) dx = \(\frac{1}{7}\) ∫\(\frac{dt}{t}\)
= \(\frac{1}{7}\) log |t| + C = \(\frac{1}{7}\) log|7x + 3| + C

Question 3.
∫\(\frac{log(1+x)}{1+x}\) dx on (-1, ∞).
Solution:
∫\(\frac{log(1+x)}{1+x}\) dx
t = 1 + x ⇒ dt = dx

Question 4.
∫(3x² – 4)x dx on R.
Solution:
∫(3x² – 4)x dx
t = 3x² – 4 ⇒ dt = 6x dx
∫(3x² – 4)x dx = \(\frac{1}{6}\)∫t dt = \(\frac{1}{6}\).\(\frac{t^2}{2}\) + C
= \(\frac{(3x^2-4)^2}{12}\) + C

Question 5.
∫\(\frac{dx}{\sqrt{1+5x}}\) dx on (-\(\frac{1}{5}\), ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 7

Question 6.
∫(1 – 2x³)x² dx on R.
Solution:
∫(1 – 2x³)x² dx
t = 1 – 2x³ ⇒ dt = -6x² dx
∫(1 – 2x³)x² dx = –\(\frac{1}{6}\)∫t dt
= –\(\frac{1}{6}\) . \(\frac{t^2}{2}\) + C
= \(\frac{-(1-2x^3)^2}{12}\) + C

Question 7.
∫\(\frac{\sec^2 x}{(1+tan x)^3}\) dx on I ⊂ R \ {nπ – \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
∫\(\frac{\sec^2 x}{(1+tan x)^3}\) dx
t = 1 + tan x ⇒ dt = sec² x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 8

Question 8.
∫x³ sinx4 dx on R.
Solution:
∫x³ . sinx4 dx
t = x4 ⇒ dt = 4x³ dx
∫x³ . sinx4 dx = \(\frac{1}{4}\)∫sin t. dt
= –\(\frac{1}{4}\)cos t + C
= –\(\frac{1}{4}\). cos x4 + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 9.
∫\(\frac{\cos x}{(1+sin x)^2}\) dx on I ⊂ R\{2nπ + \(\frac{3 \pi}{2}\) : n ∈ Z}.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 9

Question 10.
∫\(\sqrt[3]{sin x}\) cos x dx on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 10

Question 11.
∫ 2x e dx on R.
Solution:
∫ 2x. e dx
t = x² ⇒ dt = 2x dx
∫ 2xe dt = ∫ et dt = et + C
= e + C

Question 12.
∫\(\frac{e^{log x}}{x}\) dx on (0, ∞).
Solution:
∫\(\frac{e^{log x}}{x}\) dx
t = log x ⇒ dt = \(\frac{1}{x}\).dx
∫\(\frac{e^{log x}}{x}\) dx = ∫et. dt
= et + C
= elog x + C
= x + C

Question 13.
∫\(\frac{x^2}{\sqrt{1-x^6}}\) dx on I = (-1, 1).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 11

Question 14.
∫\(\frac{2x^3}{1+x^8}\) dx on R.
Solution:
t = x4 ⇒ dt = 4x³ dx
∫\(\frac{2x^3}{1+x^8}\) = \(\frac{1}{2}\)∫\(\frac{dt}{1+t^2}\)
= \(\frac{1}{2}\) tan-1 t + C
= \(\frac{1}{2}\) tan-1(x4) + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 15.
∫\(\frac{x^8}{1+x^18}\) dx on R.
Solution:
t = x9 ⇒ dt = 9x8 dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 12

Question 16.
∫\(\frac{e^x(1+x)}{\cos^2(xe^x)}\) dx on I ⊂ R/{x ∈ R : cos (xex) = 0}.
Solution:
t = x . ex
dt = (x . ex + ex)dx = ex(1 + x)dx
∫\(\frac{e^x(1+x)}{\cos^2(xe^x)}\) dx = ∫\(\frac{dt}{\cos^2 t}\) dx
= ∫sin² t dt
= tan t + C
= tan (x. ex) = C

Question 17.
∫\(\frac{cosec^2 x}{(a+b cot x)^5}\) dx on I ⊂ R\ {x ∈ R : a + b cot x = 0}, where a, b ∈ R, b ≠ 0.
Solution:
Let t = a + b cot x
dt = -b cosec² x dx
∫\(\frac{cosec^2 x}{(a+b cot x)^5}\) dx = –\(\frac{1}{b}\) ∫\(\frac{bt}{t^5}\)
= –\(\frac{1}{b}\) ∫t-5 dt
= –\(\frac{1}{b}\) \(\frac{t^{-4}}{-4}\) + C
= \(\frac{1}{4b t^{4}}\) + C
= \(\frac{1}{4b(a+b cotx)^{4}}\) + C

Question 18.
∫ex sin ex dx on R.
Solution:
t = ex ⇒ dt = ex dx
∫ex . sin ex dx = ∫ sin t dt
= -cos t + C
= -cos (ex) + C

Question 19.
∫\(\frac{\sin(log x)}{x}\) dx on (-1, ∞).
Solution:
t = log x ⇒ dt = \(\frac{1}{x}\) dx
∫\(\frac{\sin(log x)}{x}\)dx = ∫sin t dt
= -cos t + C
= -cos (log x) + C

Question 20.
∫\(\frac{1}{x log x}\) dx on (-1, ∞).
Solution:
t = log x
dt = \(\frac{1}{x}\) . dx
∫\(\frac{1}{x log x}\) dx = ∫\(\frac{1}{t}\)dt = log t + C = log(log x + C)

Question 21.
∫\(\frac{(1+log x)^n}{x}\) dx on (e-1, ∞). n ≠ -1.
Solution:
t = 1 + log x
dt = \(\frac{1}{x}\) dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 13

Question 22.
∫\(\frac{\cos(log x)}{x}\) dx on (0, ∞).
Solution:
t = log x
dt = \(\frac{1}{x}\) dx
∫\(\frac{\cos(log x)dx}{x}\) = ∫cos t dt
= sin t + C
= sin (log x) + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 23.
∫\(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx on (0, ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 14

Question 24.
∫\(\frac{2x+1}{x^2+x+1}\) dx on R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 15

Question 25.
∫\(\frac{ax^{n-1}}{bx^n+c}\) dx, where n ∈ N, a, b, c are real nembers, b ≠ 0 and x ∈ I ⊂ {x ∈ R : xn ≠ –\(\frac{c}{b}\)}.
Solution:
t = bxn + C
dt = nbxn-1dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 16

Question 26.
∫\(\frac{1}{x log x[log(log x)]}\) dx on (1, ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 17

Question 27.
∫cot hx dx on R.
Solution:
t = sinh x ⇒ dt = cosh x dx
∫cot hx dx = ∫\(\frac{dt}{t}\)
= log |t| +C
= log |log(log x)| + C

Question 28.
∫\(\frac{1}{\sqrt{1-4x^2}}\) dx on (-\(\frac{1}{2}\), \(\frac{1}{2}\)).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 18

Question 29.
∫\(\frac{dx}{\sqrt{25+x^2}}\) dx on R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 19

Question 30.
∫\(\frac{1}{(x+3\sqrt{x+2}}\) dx on I ⊂ (-2, ∞).
Solution:
x + 2 = t²
dx = 2t dt
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 20

Question 31.
∫\(\frac{1}{1+\sin 2x}\) dx on I ⊂ R\{\(\frac{n \pi}{2}\) + (-1)n \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 21

Question 32.
∫\(\frac{x^2+1}{x^4+1}\) dx on R.
Solution:
∫\(\frac{x^2+1}{x^4+1}\) dx
Dividing Nr and Dr by x²
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 22
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 32

Question 33.
∫\(\frac{dx}{\cos^2+\sin 2x}\) on I ⊂ R\({(2n+1)\(\frac{\pi}{2}\) : n ∈ Z}∪{2nπ + tan-1 \(\frac{1}{2}\) : n ∈ Z})
Solution:
∫\(\frac{dx}{\cos^2+\sin 2x}\)
Dividing Nr and Dr by cos² x
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 33

Question 34.
∫\(\sqrt{1-sin 2x}\) dx on I ⊂ {2nπ – \(\frac{3 \pi}{4}\), 2nπ + \(\frac{\pi}{4}\)}, n ∈ Z.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 34

Question 35.
∫\(\sqrt{1+cos 2x}\) dx I ⊂ {2nπ – \(\frac{\pi}{2}\), 2nπ + \(\frac{\pi}{2}\)}, n ∈ Z.
Solution:
∫\(\sqrt{1+cos 2x}\) dx = ∫\(\sqrt{2cos^2 x}\) dx
= √2 ∫ cos x dx
= √2 sin x + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 36.
∫\(\frac{\cos x + \sin x}{\sqrt{1+\sin 2x}}\) dx on I ⊂ {2nπ – \(\frac{\pi}{4}\), 2nπ + \(\frac{3\pi}{4}\)}, n ∈ Z.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 35

Question 37.
∫\(\frac{\sin 2x}{(a+b\cos x)^2}\) dx on {R, if |a| > |b| I ⊂ {x ∈ R : a + b cos x ≠ x}, if |a| < |b|.
Solution:
Put a + b cos x = t ⇒ cos x = \(\frac{t-a}{b}\)
Then b(-sin x)dx = dt
⇒ sin dx = \(\frac{-1}{b}\) dt
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 36

Question 38.
∫\(\frac{\sec x}{(\sec x+\tan x)^2}\) dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
Put sec x + tan x = t
Then (sec x tan x + sec² x) dx = dt
sec x (sec x + tan x) dx = dt
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 37

Question 39.
∫\(\frac{dx}{a^2\sec^2 x+b^2\cos^2 x}\) on R, a ≠ 0, b ≠ 0.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 38

Question 40.
∫\(\frac{dx}{\sin(x-a)\sin(x-b)}\) on I ⊂ R\({a + nπ : n ∈ Z} ∪ {b + nπ n ∈ Z}).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 39
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 40

Question 41.
∫\(\frac{dx}{\sin(x-a)\sin(x-b)}\) on I ⊂ R\ ({a + \(\frac{(2n+1)\pi}{2}\) : n ∈ Z} ∪ {b + \(\frac{(2n+1)}{2}\)π : n ∈ Z}).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 41

III. Evaluate the following integers.

Question 1.
∫\(\frac{\sin 2x}{a \cos^2 x + b \sin^2 x}\) dx on I ⊂ R\ {x ∈ R |a cos² x + b sin² x = 0}.
Solution:
t = a cos² x + b sin² x
dt = (a(2cos x)(-sin x) + b(2sin x cos x))dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 42

Question 2.
∫\(\frac{1-\tan x}{1+\tan x}\) dx for x ∈ I ⊂ R\ {nπ – \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 43

Question 3.
∫\(\frac{\cot(log x)}{x}\) dx, x ∈ I ⊂ (0, ∞)\{e : n ∈ Z}.
Solution:
t = log x ⇒ dt = \(\frac{dx}{x}\)
∫\(\frac{1-\tan x}{1+\tan x}\) dx = ∫cot t dt
= log(sin t) + C
= log |sin (log x)| + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 4.
∫ex . cot ex dx, x ∈ I ⊂ R\ {log n π : n ∈ Z}.
Solution:
t = ex ⇒ dt = ex dx
∫ex . cot ex dx = ∫cot t dt
= log (sin t) + C
= log |sin(log x)| + C

Question 5.
∫sec (tan x) sec² x dx, on I ⊂ {x ∈ E : tan x ≠ \(\frac{(2k+1)\pi}{2}\) for any k ∈ Z}. where E = R / {\(\frac{(2n+1)\pi}{2}\) : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫sec(tan x) sec² x dx = ∫sec t. dt
= log tan (\(\frac{\pi}{4}+\frac{t}{2}\) + C
= log (tan(\(\frac{\pi}{4}+\frac{\tan x}{2}\))) + C

Question 6.
∫\(\sqrt{sin x}\)cos dx on {2nπ, (2n + 1)π}, (n ∈ Z).
Solution:
t = sin x ⇒ dt = cos x dx
∫\(\sqrt{sin x}\)cos dx = ∫√t dt
= \(\frac{2}{3}\) = t3/2 + C
= \(\frac{2}{3}\) = (sin x)3/2 + C

Question 7.
∫tan4 x sec² x dx, x ∈ I ⊂ R\{\(\frac{(2n+1)\pi}{2}\) : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫tan4 x sec² x dx = ∫t4dt
= \(\frac{t^5}{5}\) + C = \(\frac{(\tan x)^2}{5}\) + C

Question 8.
∫\(\frac{2x+3}{\sqrt{x^2+3x-4}}\) dx, x ∈ T ⊂ \{-4, 1}
Solution:
t = x² + 3x – 4
dt = (2x + 3)dx
∫\(\frac{2x+3}{\sqrt{x^2+3x-4}}\) = ∫\(\frac{dt}{\sqrt{t}}\)
= 2√t + C
= \(\sqrt{x^2+3x-4}\) + C

Question 9.
∫cosec² x\(\sqrt{\cot x}\) dx on (0, \(\frac{\pi}{2}\)).
Solution:
t = cot x ⇒ dt = -cosec² x dx
∫cosec² x\(\sqrt{\cot x}\) dx = -∫√t dt
= –\(\frac{2}{3}\) t√t + C
= –\(\frac{2}{3}\) cot(x)3/2 + C

Question 10.
∫sec x log(sec x + tan x)dx on (0, \(\frac{\pi}{2}\)).
Solution:
t = log(sec x + tanx)
dt = \(\frac{(\sec x.\tan x+\sec^2 x)}{(\sec x +\tan x}\) = sec x dx
∫sec x .log(sec x + tan x)dx = ∫t dt
= \(\frac{t^2}{2}\) + C
= \(\frac{(log(\sec x+\tan x)^2}{2}\) + C

Question 11.
∫sin³ x dx on R.
Solution:
sin 3x = 3 sin x – 4 sin³ x
sin³ x = \(\frac{1}{4}\)(3 sin x – sin 3x)
∫sin³ x dx = \(\frac{3}{4}\)∫sin x – \(\frac{1}{4}\)∫sin 3x dx
= –\(\frac{3}{4}\) cos x + \(\frac{1}{12}\) cos 3x + C
= \(\frac{1}{12}\)(cos 3x – 9 cos x) + C

Question 12.
∫cos³ x dx on R.
Solution:
cos 3x = 4 cos³ x – 3 cos x
∫cos³ x dx = \(\frac{3}{4}\)∫cos x + \(\frac{1}{4}\)∫cos 3x dx
= –\(\frac{3}{4}\) sin x + \(\frac{1}{12}\) sin 3x + C
= \(\frac{1}{12}\)(9 sin x + 9 sin 3x) + C

Question 13.
∫cos x cos 2x dx on R.
Solution:
cos 2x cos x = \(\frac{1}{2}\)(2cos 2x cos x)
∫cos x cos 2x dx = \(\frac{1}{2}\)∫(cos 3x + cos x) dx
= \(\frac{1}{2}\)∫cos 3x + \(\frac{1}{2}\)∫cos x
= \(\frac{1}{2}\)(\(\frac{\sin 3x}{3}\) + sin x) + C
= \(\frac{\sin 3x+3\sin x}{6}\) + C

Question 14.
∫cos x cos 3x dx on R.
Solution:
cos 3x cos x = \(\frac{1}{2}\)(2cos 3x . cos x)
= \(\frac{1}{2}\)(cos 4x + cos 2x)
∫cos x cos 3x dx = \(\frac{1}{2}\)∫cos 4x dx + \(\frac{1}{2}\)∫cos 2x dx
= \(\frac{1}{2}\)(\(\frac{\sin 4x}{4}+\frac{\sin 2x}{2}\) + C
= \(\frac{1}{8}\)(sin 4x + 2sin 2x) + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 15.
∫cos4 x dx on R.
Solution:
cos4 x = (cos² x)² = (\(\frac{1+\cos 2x}{2}\))²
= \(\frac{1}{4}\) (1 + 2 cos 2x+ cos² 2x)
= \(\frac{1}{4}\) (1 + 2 cos 2x + = \(\frac{1+\cos 4x}{2}\))
= \(\frac{1}{8}\) (2 + 4 cos 2x + 1 + cos 4x)
= \(\frac{1}{8}\) (3 + 4 cos 2x + cos 4x)
= \(\frac{1}{8}\)(3∫dx + 4∫cos 2x dx + ∫cos 4x dx)
= \(\frac{1}{8}\)(3x + 4\(\frac{\sin 2x}{2}+\frac{\sin 4x}{4}\)) + C
= \(\frac{1}{32}\)(12x + 8 sin 2x + sin 4x) + C

Question 16.
∫x \(\sqrt{4x+3}\) dx on (-\(\frac{3}{4}\), ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 44

Question 17.
∫\(\frac{dx}{\sqrt{a^2-(b+cx)^2}}\) on {x ∈ R : |b + cx| < a}. where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 45
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 46

Question 18.
∫\(\frac{dx}{a^2+(b+cx)^2}\) on R, a, b, c are real numbers, c ≠ 0 and a > 0.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 47

Question 19.
∫\(\frac{dx}{1+e^x}\), x ∈ R
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 48

Inter 2nd Year Maths 2B Integration Solutions Ex 6(b)

Question 20.
∫\(\frac{x^2}{(a+bx)^x}\) dx, x ∈ I ⊂ R\{-\(\frac{a}{b}\)}, where a, b are real nembers, b ≠ 0.
Solution:
Put t = a + bx
dt = b dx ⇒ dx = \(\frac{1}{b}\). dt
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 49
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 50

Question 21.
∫\(\frac{x^2}{\sqrt{1-x}}\) dx, x ∈ (-∞, 1).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(b) 51