The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to (where $C$ is constant of integration):

(B) $\tan x-\frac{x \sec x}{x \sin x+\cos x}+C$

(A) $\tan x+\frac{x \sec x}{x \sin x+\cos x}+C$

(C) $\sec x+\frac{x \tan x}{x \sin x+\cos x}+C$

(D) $\sec x-\frac{x \tan x}{x \sin x+\cos x}+C$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Indefinite Integrals

The integral

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

is equal to (where

C

is constant of integration):

Answer: (B)

Write the integrand as

(\frac{x}{x sin x + cos x})^{2} = \frac{x}{cos x} \cdot \frac{x cos x}{( x sin x + cos x ) ^{2}}

. Since

\frac{d}{d x} (x sin x + cos x) = x cos x

, the second factor integrates to

- \frac{1}{x sin x + cos x}

, so integrating by parts with

u = \frac{x}{cos x}

gives

- \frac{x sec x}{x sin x + cos x} + \int \frac{cos x + x sin x}{cos ^{2} x} \cdot \frac{1}{x sin x + cos x} d x

. The remaining integral collapses to

\int sec^{2} x d x = tan x

, leaving

tan x - \frac{x sec x}{x sin x + cos x} + C

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option