Question

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

• A. $\tan x-\frac{x\sec x}{x\sin x+\cos x}+C$
• B. $\sec x-\frac{x\tan x}{x\sin x+\cos x}+C$
• C. $\sec x+\frac{x\tan x}{x\sin x+\cos x}+C$
• D. $\tan x+\frac{x\sec x}{x\sin x+\cos x}+C$

Answer 1

The correct option is A $\tan x-\frac{x\sec x}{x\sin x+\cos x}+C$

$\int {\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx$
$\int \underset{}{x\sec x}.\frac{x\cos x}{(\underset{}{x\sin x+\cos x}{)}^2}dx$
Put $(x\sin x+\cos x)=t$
$\Rightarrow x\cos xdx=dt$
$=x\sec x\left(\frac{-1}{x\sin x+\cos x}\right)+\int \frac{\sec x+x\left(\sec x\right)\left(\tan x\right)}{(x\sin x+\cos x)}dx$
$=-\frac{x\sec x}{x\sin x+\cos x}+\int \frac{(\cos x+x\sin x)}{{\cos }^{2}x(x\sin x+\cos x)}dx$
$=-\frac{x\sec x}{x\sin x+\cos x}+\tan x+C$