AP Inter 2nd Year Maths Exercise 7c Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7c

I.

Question 1.
Find the integral of sin²(2x + 5)
Solution:
sin²(2x + 5) = \(\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2}\) [∵ sin² A = \(\frac{1-\cos 2 \mathrm{~A}}{2}\)]
∴ ∫sin²(2x + 5) = ∫ \(\frac{1-\cos (4 x+10)}{2}\)dx
= \(\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x=\frac{1}{2} x-\frac{1}{2}\left(\frac{\sin (4 x+10)}{4}\right)+C=\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+C\)

Question 2.
Find the integral of sin 3xcos 4x
Solution:
I = ∫sin 3x cos 4xdx = \(\frac{1}{2}\)∫2sin3x cos4x dx
= \(\frac{1}{2}\)∫(sin(3x + 4x) + sin(3x – 4x)dx [∵ 2sin A cos B = sin(A + B) + sin(A – B)]
= \(\frac{1}{2}\)∫(sin 7x + sin(-x)) dx = \(\frac{1}{2}\)∫(sin 7x – sin x) dx [∵ sin(θ) = -sinθ]
= \(\frac{1}{2}\)[∫sin 7xdx – ∫sinxdx = \(\frac{1}{2}\) \(\left[\frac{-\cos 7 x}{7}-(-\cos x)\right]\) + c
= \(\frac{-1}{14}\)cos 7x + \(\frac{1}{2}\)cos x + C

Question 3.
Find the integral of sin4x sin8x
Solution:
We know that SinA SinB = \(\frac{1}{2}\)[cos(A – B) – cos(A + B)]
∴ ∫sin4x sin 8x dx = ∫[\(\frac{1}{2}\)cos(4x – 8x) – \(\frac{1}{2}\)cos(4x + 8x)] dx
= \(\frac{1}{2}\)∫(cos(-4x) – cos12x) dx = \(\frac{1}{2}\)∫(cos 4x – cos 12x) dx = \(\frac{1}{2}\left[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right]\)

Question 4.
Find the integral of \(\frac{1-\cos x}{1+\cos x}\)
Solution:

Question 5.
Find the integral of \(\frac{\cos x}{1+\cos x}\)
Solution:

Question 6.
Find the integral of \(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\)
Solution:
\(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}=\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x}=\frac{\sin x}{\cos ^2 x}+\frac{\cos x}{\sin ^2 x}\) = tan xsec x + cot xcosec x
∴ \(\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x} d x\) = ∫(tan x sec x + cot x cosec x) dx = sec x – cosecx + C

Question 7.
Find the integral of \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\)
Solution:

Question 8.
Find the integral of \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\)
Solution:
Let I = \(\int \frac{\cos 2 x}{(\cos x+\sin x)^2} d x=\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x\)

Put cos x + sin x = t ⇒ (-sinx + cosx)dx = dt
∴ From (i), I = \(\int \frac{d t}{t}\) = log|t| + C = log|cos x + sinx| + C

Question 9.
Find the integral of sin^-1(cos x)
Solution:
\(\int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} \sin \left(\frac{\pi}{2}-x\right) d x\) [∵ sin^-1 sin θ = θ]
\(=\int\left(\frac{\pi}{2}-x\right) d x=\int \frac{\pi}{2} d x-\int x d x=\frac{\pi}{2} \int 1 d x-\int x^1 d x=\frac{\pi}{2} x-\frac{x^2}{2}+c\)

Question 10.
Find the integral of \(\frac{\sin ^2 x}{1+\cos x}\)
Solution:

II.

Question 1.
Find the integral of cos 2x cos 4x cos 6x
Solution:
We know that Cos A cos B = \(\frac{1}{2}\)[cos(A + B) + cos(A – B)]
∫cos 2x(cos 4x cos 6x) dx = ∫cos 2x[\(\frac{1}{2}\)[cos(4x + 6x) + cos(4x – 6x)]] dx
= ∫\(\frac{1}{2}\)[cos 2x cos 10x + cos 2x cos(-2x) dx = \(\frac{1}{2}\)∫\(\frac{1}{2}\)[cos 2x cos 10x + cos² 2x] dx
= \(\frac{1}{2}\)∫[(\(\frac{1}{2}\)cos(2x + 10x) + \(\frac{1}{2}\)cos(2x – 10x)) + (\(\frac{1+\cos 4 x}{2}\))] dx
= \(\frac{1}{4}\)∫(cos 12x + cos8x + 1 + cos4x) dx = \(\frac{1}{4}\left[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}\right]\) + C

Question 2.
Find the integral of sin x sin 2x sin 3x
Solution:
We know that Sin A sin B = \(\frac{1}{2}\)[cos(A – B) – cos(A + B)]

Question 3.
Find the integral of sin³(2x + 1)
Solution:
Let I = ∫sin³(2x + 1) = ∫sin²(2x + 1) sin(2x + 1) dx
= ∫(1 – cos²(2x + 1)) sin(2x + 1) dx
Put cos(2x + 1) = t ⇒ -2 sin(2x + 1) dx = dt ⇒ sin(2x + 1) dx = \(\frac{-\mathrm{dt}}{2}\)
∴ \(I=\frac{-1}{2} \int\left(1-t^2\right) d t=\frac{-1}{2}\left[t-\frac{t^3}{3}\right]=\frac{-1}{2}\left[\cos (2 x+1)-\frac{\cos ^3(2 x+1)}{3}\right]\)
\(=\frac{-\cos (2 x+1)}{2}+\frac{\cos ^3(2 x+1)}{6}+C=\frac{\cos ^3(2 x+1)}{6}-\frac{\cos (2 x+1)}{2}++C\)

Question 4.
Find the integral of sin³x cos³ x
Solution:
Let I = ∫sin³x cos³x dx = ∫cos³ xsin²x sin x dx = ∫cos³x(1 – cos² x) sin x dx
Put cos x = t ⇒ -sin x dx = dt
∴ I = \(-\int t^3\left(1-t^2\right) d t=-\int\left(t^3-t^5\right) d t=-\left[\frac{t^4}{4}-\frac{t^6}{6}\right]+C\)
\(=-\left[\frac{\cos ^4 x}{4}-\frac{\cos ^6 x}{6}\right]+C=\frac{\cos ^6 x}{6}-\frac{\cos ^4 x}{4}+C\)

Question 5.
Find the integral of sin⁴x.
Solution:
We have sin⁴ x = sin² x sin² x = \(\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)=\frac{1}{4}(1-\cos 2 x)^2\)

Question 6.
Find the integral of cos⁴ x .
Solution:
We have cos ⁴2x = (cos ² 2x)² = \(\left(\frac{1+\cos 4 x}{2}\right)^2\)

Question 7.
Find the integral of \(\frac{\cos 2 x-\cos 2 u}{\cos x-\cos \alpha}\)
Solution:

Question 8.
Find the integral of \(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Solution:
We have \(\frac{\cos x-\sin x}{1+\sin 2 x}=\frac{\cos x-\sin x}{\left(\sin ^2 x+\cos ^2 x\right)+2 \sin x \cos x}\) [∵ sin² x + cos² x = 1; sin 2x = 2sin x cosx ]
= \(=\frac{\cos x-\sin x}{(\sin x+\cos x)^2}\)
Put sin x + cos x = t ⇒ (cos x – sin x) dx = dt
∴ \(\int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^2} d x\)
\(=\int \frac{d t}{t^2}=\int t^{-2} d t=-t^{-1}+C=-\frac{1}{t}+C=\frac{-1}{\sin x+\cos x}+C\)

Question 9.
Find the integral of tan³2x sec 2x
Solution:
Let I = ∫tan³2xsec 2x dx = ∫tan² 2x tan 2x sec 2x dx
= ∫(sec² 2x – 1) sec 2x tan ex dx [∵ tan² θ = sec² θ – 1]
= \(\frac{1}{2}\)∫(sec² 2x – 1)(2 sec 2x tan 2x) dx ……(i)
Put sec 2x = t ⇒ sec 2x tan 2x \(\frac{d}{dx}\)(2x) = \(\frac{dt}{dx}\) ⇒ 2 sec 2x tan 2x dx = dt
∴ From (i), I = \(\frac{1}{2} \int\left(\mathrm{t}^2-1\right) \mathrm{dt}=\frac{1}{2}\left(\int \mathrm{t}^2 \mathrm{dt}-\int 1 \mathrm{dt}\right)\)
= \(\frac{1}{2}\left(\frac{t^3}{3}-t\right)+c=\frac{1}{6} t^3-\frac{1}{2} t+c=\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c\)   [∵ t = sec 2x]

Question 10.
Find the integral of tan⁴ x.
Solution:
∫tan⁴ x dx = ∫tan² x tan² x dx = ∫tan² x(sec² x – 1) dx
= ∫(tan² x sec² x – tan² x) dx = ∫tan² x sec² dx – ∫tan² dx
= ∫tan² x sec² x dx – ∫(sec²x – 1)dx
= ∫tan² x sec²xdx – ∫sec²dx + 1dx
= ∫tan² x sec² x dx – tan x + x + C₁ …….(i)
For this integral, put tan x = t, ⇒ sec²x dx = dt
∴ \(\int \tan ^2 x \sec ^2 x d x=\int t^2 d t=\frac{t^3}{3}+C_2=\frac{\tan ^3 x}{3}+C_2\)
From (i) I = \(\frac{\tan ^3 x}{3}\) – tan x + x + C, where C = C₁ + C₂

Question 11.
Find the integral of \(\frac{1}{\sin x \cos ^3 x}\)
Solution:

Question 12.
Find the integral of \(\frac{1}{\cos (x-a) \cos (x-b)}\)
Solution:

Question 13.
Find the integral of \(\frac{\sin x}{\sin (x-a)}\)
Solution:
Put x – a = t ⇒ x = t + a ⇒ dx = dt
∴ \(\int \frac{\sin x}{\sin (x-a)} d x=\int \frac{\sin (t+a)}{\sin t} d t=\int \frac{\sin t \cos a+\cos t \sin a}{\sin t} d t=\int(\cos a+\cot t \sin a) d t\)
= tcos a + sin a log|sin t| + C₁ = (x – 1)cos a + sin alog|sin(x – a)| + C₁
= x cos a + sin a log|sin(x – a)| – a cos a + C₁ = sin a log|sin(x – a)| + x cos a + C

Question 14.
Find the integral of \(\frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}\)
Solution:

Question 15.
Find the integral of \(\frac{1}{\cos (x+a) \cos (x+b)}\)
Solution:
Given integral is \(\frac{1}{\cos (x+a) \cos (x+b)}\) Multiplying and dividing by sin(a – b), we get