Find $\int \frac{\sec x}{1 + \text{cosec } x} dx$.

$\frac{1}{4} \ln \left| \frac{1 + \sin x}{1 - \sin x} \right| + \frac{1}{2(1 + \sin x)} + C$

The correct answer is Option (2) → $\frac{1}{4} \ln \left| \frac{1 + \sin x}{1 - \sin x} \right| + \frac{1}{2(1 + \sin x)} + C$

$I = \int \frac{\sec x}{1 + \text{cosec }x} dx$

$ = \int \frac{\sin x}{\cos x(1 + \sin x)} dx$

$ = \int \frac{\sin x \cos x}{(1 + \sin x)^2 (1 - \sin x)} dx$

$ = \int \frac{t}{(1 + t)^2 (1 - t)} dt$    [\sin x = t\,or\,\cos x\, dx = dt]$

Let $\frac{t}{(1+t)^2(1-t)} = \frac{A}{1+t} + \frac{B}{(1+t)^2} + \frac{C}{1-t}$

Or $t = A(1+t)(1-t) + B(1-t) + C(1+t)^2$  (an identity)

Put $t = -1$: $-1 = 2B ⇒B = -\frac{1}{2}$

Put $t = 1$: $1 = 4C ⇒C = \frac{1}{4}$

Put $t = 0$: $0 = A + B + C ⇒A = \frac{1}{4}$

Therefore the required integral

$ = \frac{1}{4} \int \frac{1}{1+t} dt - \frac{1}{2} \int \frac{1}{(1+t)^2} dt + \frac{1}{4} \int \frac{1}{1-t} dt$

$ = \frac{1}{4} \log |1+t| + \frac{-1}{2} \cdot \frac{-1}{1+t} - \frac{1}{4} \log |1-t| + c$

$ = \frac{1}{4} \log |1+\sin x| + \frac{1}{2} \cdot \frac{1}{1+\sin x} - \frac{1}{4} \log |1-\sin x| + c$

$ = \frac{1}{4} \log \left| \frac{1 + \sin x}{1 - \sin x} \right| + \frac{1}{2(1 + \sin x)} + c$