Question

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

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

Hint:

Note that \(\frac{d}{dx}(x\sin x+\cos x)=x\cos x\); write the integrand as \(\left(x\sec x\right)\cdot\frac{x\cos x}{(x\sin x+\cos x)^2}\) and integrate by parts.

Answer 1

The Correct Option is D

Let \(D=x\sin x+\cos x\); then \(\frac{dD}{dx}=x\cos x\) and \(\int\frac{x\cos x}{D^2}dx=-\frac{1}{D}\).
\(\int\frac{x^2}{D^2}dx=\int (x\sec x)\cdot\frac{x\cos x}{D^2}dx=-\frac{x\sec x}{D}+\int\frac{\sec x+x\sec x\tan x}{D}dx\).
Now \(\sec x+x\sec x\tan x=\frac{\cos x+x\sin x}{\cos^2x}=\frac{D}{\cos^2 x}\), so the last integral is \(\int\sec^2x\,dx=\tan x\).
Hence the integral \(=\tan x-\frac{x\sec x}{x\sin x+\cos x}+C\).
Answer: (4) \(\tan x-\frac{x\sec x}{x\sin x+\cos x}+C\)