Find $\int \frac{2\cos x}{(1-\sin x)\left(2-\cos^{2}x\right)}\,dx$

$-\log(1-\sin x)+\dfrac{1}{2}\log\left(1+\sin^{2}x\right)+\tan^{-1}(\sin x)+C$

CBSE Class 12 · 2019 · Indefinite IntegralsLong Answer

Find

\int \frac{2 c o s x}{( 1 - s i n x ) ( 2 - c o s ^{2} x )} d x

Answer: $- lo g (1 - sin x) + \frac{1}{2} lo g (1 + sin^{2} x) + tan^{- 1} (sin x) + C$

Note

2 - cos^{2} x = 2 - (1 - sin^{2} x) = 1 + sin^{2} x

, so the integral is

\int \frac{2 cos x}{( 1 - sin x ) ( 1 + sin ^{2} x )} d x

. Put

t = sin x

d t = cos x d x

\int \frac{2 d t}{( 1 - t ) ( 1 + t ^{2} )} .

With

\frac{2}{( 1 - t ) ( 1 + t ^{2} )} = \frac{1}{1 - t} + \frac{t + 1}{1 + t ^{2}}

- lo g ∣1 - t ∣ + \frac{1}{2} lo g (1 + t^{2}) + tan^{- 1} t + C .

Restoring

t = sin x

- lo g (1 - sin x) + \frac{1}{2} lo g (1 + sin^{2} x) + tan^{- 1} (sin x) + C

CBSE Class 12 Board · 2019 · Subjective