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

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

CBSE Class 12 · 2018 · Indefinite IntegralsLong Answer

Find

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

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

Put

t = sin x

d t = cos x d x

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

Partial fractions give

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

, so

\int \frac{d t}{1 - t} + \frac{1}{2} \int \frac{2 t}{1 + t ^{2}} d t + \int \frac{d t}{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 · 2018 · Subjective