Inter 2nd Year Maths Exercise 7c Solutions

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7c will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7c

Integrals Exercise 7c Solutions

I. Find the integral of the functions

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

Question 2.
sin 3x cos 4x
Solution:
∫sin 3x cos 4xdx
We know that,
sin A cos B = \(\frac{1}{2}\)[(sin(A + B) + sin(A – B)]
= \(\frac{1}{2}\)∫(sin(3x + 4x) + sin(3x – 4x) dx
= \(\frac{1}{2}\)∫(sin 7x + sin(-x)) dx = \(\frac{1}{2}\)∫(sin 7x – sin x) dx
= \(\frac{1}{2}\)∫sin 7xdx – \(\frac{1}{2}\)∫sin x dx [∵ ∫ sin x dx = -cos x + C]
= \(\frac{1}{2}\) \(\left[\frac{-\cos 7 x}{7}-(-\cos x)\right]\) + C
= \(\frac{-\cos 7 x}{14}\) + \(\frac{\cos x}{2}\) + C

Question 3.
sin 4x sin 8x
Solution:
∫sin 4x sin 8x dx = \(\frac{1}{2}\) ∫ 2 sin 8x sin 4x dx
= \(\frac{1}{2}\) ∫(cos 4x – cos 12x) dx
∵ 2 Sin A Sin B = cos(A – B) – cos(A + B)
= \(\frac{1}{2}\left\{\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right\}\) + C [∵ ∫ cos x dx = sin x + C]

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

Question 5.
\(\frac{\cos x}{1+\cos x}\)
Solution:

Question 6.
Find the integral of \(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\)
Solution:
∫ \(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ \(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}\) dx + ∫ \(\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ \(\frac{\sin x}{\sin ^2 x \cos ^2 x}\) dx + ∫ \(\frac{\cos x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ tan x . sec x dx + ∫ cot x . cosec x dx
= sec x – cosec x + C
∵ ∫ sec x tan x dx = sec x + C, ∫ cosec x cot x dx = -cosec x + C

Question 7.
\(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\)
Solution:
∫ \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) dx = ∫ \(\frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x}\) dx [∵ cos 2x = 1 – 2 sin²x]
= ∫ \(\frac{1}{\cos ^2 x}\) dx = ∫ sec² x dx
= tan x + C [∵ ∫ sec² x dx = tan x + C]

Question 8.
\(\frac{\cos 2 x}{(\cos x+\sin x)^2}\)
Solution:

Question 9.
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 + cos^-1 t = \(\frac{\pi}{2}\) for |t| ≤ 1]
= \(\int\left(\frac{\pi}{2}-x\right) \) dx = \(\frac{\pi}{2}\) x – \(\frac{x^2}{2}\) + C [∵∫ dx = x + C; ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]

Question 10.
\(\frac{\sin ^2 x}{1+\cos x}\)
Solution:
∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{\left(1-\cos ^2 x\right)}{(1+\cos x)}\) dx
= ∫ \(\frac{(1+\cos x)(1-\cos x)}{(1+\cos x)}\) dx [∵ sin²x = 1 – cos²x]
= ∫ (1 – cos x) dx = ∫ 1 dx – ∫ cos x dx
= x – sin x + C [∵∫ dx = x +C, ∫ cos x dx = sin x + C]

II. Find the integral of functions

Question 1.
cos 2x cos 4x cos 6x
Solution:
∫ cos 2x cos 4x cos 6x dx
= ∫cos 2x[\(\frac{1}{2}\)[cos(4x + 6x) + cos(4x – 6x)]] dx
[∵ 2 Cos A cos B = cos(A + B) + cos(A – B)]
= \(\frac{1}{2}\) ∫ [cos 2x cos 10x + cos 2x cos(-2x) dx
= \(\frac{1}{2}\)∫\(\frac{1}{2}\)[cos 2x cos 10x + cos² 2x] dx [∵ cos(-θ) = cos θ]
= \(\frac{1}{4}\)∫(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 [∵ ∫ cos x dx = sin x + C, ∫ sin x dx = -cosx + C]

Question 2.
sin x sin 2x sin 3x
Solution:
∫ sin x sin 2x sin 3x dx
= \(\frac{1}{2}\) ∫ (2 sin 3x sin x) sin 2x dx
[∵ Sin A sin B = cos(A – B) – cos(A + B)]
= \(\frac{1}{2}\) ∫ {cos 2x – cos 4x}sin 2x dx
= \(\frac{1}{2}\) ∫ {2 sin 2x cos 2x – 2 cos 4x sin 2x}dx
= \(\frac{1}{4}\) ∫ {(sin 4x + sin 0) – (sin 6x – sin 2x)}dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – 8)
2 cos A sin B = sin (A + B) – sin (A – B)]
= \(\frac{1}{4}\) ∫ (sin 4x – sin 6x + sin 2x) dx
= \(\frac{1}{4}\left\{\frac{-\cos 4 x}{4}-\frac{(-\cos 6 x)}{6}+\frac{(-\cos 2 x)}{2}\right\}\) [∵ ∫ sin ax dx = – \(\frac{\cos a x}{a}\)]
= \(\frac{1}{4}\left\{\frac{-\cos 4 x}{4}-\frac{(-\cos 6 x)}{6}+\frac{(-\cos 2 x)}{2}\right\}\) + C

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

Question 4.
sin³x cos³ x
Solution:
∫sin³x cos³x dx
Let cos x = t ⇒ -sin x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-\sin x}\)
∴ ∫ sin³ x . cos³x dx = ∫ sin³ x t³ \(\frac{d t}{-\sin x}\)
= -∫ sin³ x t³ dt = -∫ t³ (1 – cos² x)dt
= -∫ t³ (1 – t²)dt = – ∫ (t³ – t⁵) dt
= \(-\left(\frac{t^4}{4}-\frac{t^6}{6}\right)\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{\cos ^6 x}{6}-\frac{\cos ^4 x}{4}\) + C

Question 5.
sin⁴x.
Solution:
∫ sin⁴ x dx = ∫ (sin² x)² dx
= ∫ \(\left(\frac{1-\cos 2 x}{2}\right)^2\) dx [∵ sin² x = \(\frac{1-\cos 2 x}{2}\)]
= \(\frac{1}{4}\) ∫ (1 – cos 2x)² dx = \(\frac{1}{4}\) ∫ (1 + cos²2x – 2 cos 2x) dx
= \(\frac{1}{4}\) [∫ 1 dx + ∫ cos² 2x dx – 2∫ cos 2x dx]
= \(\frac{1}{4}\) [∫ 1 dx + ∫ \(\frac{(1+\cos 4 x)}{2}\) dx – 2 ∫ cos 2x dx] [cos²x = \(\frac{1+\cos 2 x}{2}\)]

Question 6.
cos⁴ x .
Solution:

Question 7.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Solution:
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx = ∫ \(\frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{(\cos x-\cos \alpha)}\) dx
= ∫ \(\frac{2 \cos ^2 x-1-2 \cos ^2 \alpha+1}{(\cos x-\cos \alpha)}\) dx [∵ cos 2x = 2 cos² – 1]
= ∫ \(\frac{2\left(\cos ^2 x-\cos ^2 \alpha\right)}{(\cos x-\cos \alpha)}\) dx = 2 ∫ \(\frac{(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)}\) dx
= 2 [∫ cos x dx + cos α ∫ 1 dx]
= 2[sin x + cos α . x] + C = 2[sin x + x cos α] + C
[∵ cos x dx = sin x + C, ∫ dx = x + C]

Question 8.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Solution:

Question 9.
tan³2x sec 2x
Solution:
∫tan³2x sec 2x dx
Let sec 2x = t

Question 10.
tan⁴ x.
Solution:
Let I = ∫ tan⁴ x dx = ∫ (tan²x)² dx
I = ∫ (tan² x) (tan² x) dx
= ∫ (sec²x – 1)(tan² x) dx
= ∫ sec² x tan² x dx – ∫ tan² x dx
= ∫ sec² x tan²x dx – ∫ [sec²x – 1] dx
= ∫ sec² x tan² x dx – [∫ sec² x dx – f1 dx]dx
Now, let I₁ = ∫ sec² x tan² x dx and
I₂ = ∫ sec² x dx – ∫ 1 dx
Then, I = I₁ – I₂ …………… (1)
Put tan x ⇒ sec² x = \(\frac{d t}{d x}\) ⇒ dx = \(\frac{d t}{\sec ^2 x}\)
∴ I₁ = ∫ sec² x t² \(\frac{d t}{\sec ^2 x}\) = ∫ t² dt
= \(\frac{t^3}{3}\) + C₁ = \(\frac{\tan ^3 x}{3}\) + C₁ [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
I₂ = ∫ sec² x dx – ∫ 1 dx = tan x – x + C₂
∴ Putting the values of I₁ and I₂ in (I), we get
I = \(\frac{\tan ^3 x}{3}\) + C₁ – (tan x – x) + C₂
⇒ I = \(\frac{\tan ^3 x}{3}\) – tan x + x + C (∵ C₁ + C₂ = C)

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

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

Question 13.
\(\frac{\sin x}{\sin (x-a)}\)
Solution:
∫ \(\frac{\sin x}{\sin (x-a)}\) dx = ∫ \(\frac{\sin \{(x-a)+a\}}{\sin (x-a)}\) dx
= ∫ \(\frac{\sin (x-a) \cos a+\cos (x-a) \sin a}{\sin (x-a)}\) dx
[∵ sin(A + B) = sin A cos B + cos A sin B]
= ∫ \(\frac{\sin (x-a) \cos a}{\sin (x-a)}\) dx + ∫ \(\frac{\cos (x-a) \sin a}{\sin (x-a)}\) dx
= cos a ∫ 1 dx + sin a ∫ cot(x – a) dx
= x cos a + sin a log|sin(x – a)| + C₁ [∵ ∫ cot x dx = log|sin x|]

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

Question 15.
\(\frac{1}{\cos (x+a) \cos (x+b)}\)
Solution: