ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.9

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.9

Evaluate the following (1 to 4) integrals:

Question 1.
(i) ∫ \(\frac{d x}{2+\cos x}\)
(ii) ∫ \(\frac{d x}{5+4 \cos x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{2+\cos x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
and cos x = \(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{1-t^2}{1+t^2}\)

(ii) Let I = ∫ \(\frac{d x}{5+4 \cos x}\)

Question 2.
(i) ∫ \(\frac{d x}{1+2 \cos x}\)
(ii) ∫ \(\frac{d x}{4 \cos x-1}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{1+2 \cos x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
and cos x = \(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{1-t^2}{1+t^2}\)

(ii) Let I = ∫ \(\frac{d x}{4 \cos x-1}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt

Question 3.
(i) ∫ \(\frac{d x}{4+5 \sin x}\)
(ii) ∫ \(\frac{d x}{1-\sin x+\cos x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{4+5 \sin x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt

(ii) Let I = ∫ \(\frac{d x}{1-\sin x+\cos x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt

= – log |1 – tan x/2| + c

Question 4.
(i) ∫ \(\frac{d x}{1-\tan x}\) (NCERT)
(ii) ∫ \(\frac{d x}{1+\cot x}\) (NCERT)
Solution:

(i) Let I = ∫ \(\frac{d x}{1-\tan x}\)
= ∫ \(\frac{\cos x d x}{\cos x-\sin x}\)
Let cos x = l (cos x – sin x) + m \(\frac{d}{d x}\) (cos x – sin x)
cos x = l (cos x – sin x) + m (- sin x – cos x)
Comparing the coefficients of sin x and cos x on both sides,
we have 0 = – l – m
and 1 = l – m
⇒ m = \(-\frac{1}{2}\) ; l = \(\frac{1}{2}\)
∴ I = ∫ \(\frac{l(\cos x-\sin x)+m(-\sin x-\cos x)}{\cos x-\sin x}\)
= ∫ l dx + m \(\frac{-\sin x-\cos x}{\cos x-\sin x}\) dx
= lx + m ∫ \(\frac{d t}{t}\)
where t = cos x – sinx
⇒ dt = (- sin x – cos x) dx
= \(\frac{x}{2}\) – \(\frac{1}{2}\) log |cos x – sin x| + c

(ii) Let I = ∫ \(\frac{d x}{1+\cot x}\)
= ∫ \(\frac{\sin x d x}{\sin x+\cos x}\)
Let Numerator = l (denominator + m \(\frac{d}{d x}\) (derivative of denominator)
⇒ sin x = l (sin x + cos x) + m \(\frac{d}{d x}\) (sin x + cos x)
⇒ sin x = l (sin x + cos x) + m (cos x – sin x)
Equating the coefficients of sin x and cos x on both sides,
we have 1 = l – m
and 0 = l + m
On solving the eqns ; we have
⇒ l = \(\frac{1}{2}\) ; m = \(-\frac{1}{2}\)
∴ I = ∫ \(\frac{l(\sin x+\cos x)+m(\cos x-\sin x)}{\sin x+\cos x}\)
= ∫ l dx + m ∫ \(\frac{(\cos x-\sin x) d x}{\sin x+\cos x}\) dx
= \(\frac{1}{2}\) x – \(\frac{1}{2}\) log |sin x + cos x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

Question 5.
If ∫ \(\frac{5 \tan x}{\tan x-2}\) dx = ax + b log |sin x – 2 cos x| + C, then find the values of a and b.
Solution:
Let I = ∫ \(\frac{5 \tan x}{\tan x-2}\) dx
= ∫ \(\frac{5 \frac{\sin x}{\cos x} d x}{\frac{\sin x}{\cos x}-2}\)
= ∫ \(\frac{5 \sin x d x}{\sin x-2 \cos x}\)
Let 5 sin x = l (sin x – 2 cos x) + m \(\frac{d}{d x}\) (sin x – 2 cos x)
⇒ 5 sin x = l (sin x – 2 cos x) + m (cos x + 2 sin x) ………… (1)
comparing the coefficients of sin x and cos x on both sides ; we have
5 = l + 2m …………(2)
0 = – 2l + m ……………(3)
On solving eqn. (2) and (3) ; we have
m = 2 and l = 1
∴ I = ∫ \(\frac{5 \sin x d x}{\sin x-2 \cos x}\)
= ∫ \(\frac{l(\sin x-2 \cos x)+m(\cos x+2 \sin x) d x}{\sin x-2 \cos x}\)
= l ∫ dx + m ∫ \(\frac{(\cos x+2 \sin x) d x}{\sin x-2 \cos x}\) dx
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= lx + m log |sin x – 2 cos x| + C
= x + 2 log |sin x – 2 cos x| + C ……………. (1)
Also, Given I = ∫ \(\frac{5 \tan x d x}{\tan x-2}\)
= ax + b log |sin x – 2 cos x| + C …………….(2)
∴ From (1) and (2) ; we have
a = 1 ; b = 2.