Question

$\int \frac{x^{2}dx}{(x\sin x+\cos x{)}^2}$ is equal to

• A. $\frac{\sin x+\cos x}{x\sin x+\cos x}$
• B. $\frac{x\sin x-\cos x}{x\sin x+\cos x}+c$
• C. $\frac{\sin x-\cos x}{x\sin x+\cos x}x$
• D. None of these

Answer 1

Answer: (D)

None of these

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

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

$=\int \frac{x\cdot \cos x}{(x\sin x+\cos x)2}\cdot \frac{x}{\cos x}\cdot dx$

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

On integrating by parts,

$=\frac{x}{\cos x}\cdot \left(\frac{-1}{x\sin x+\cos x}\right)$

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

$=\frac{-x}{\cos x(x\sin x+\cos x)}+\int se{c}^{2}xdx$

$=\frac{-x}{\cos x(x\sin x+\cos x)}+\tan x+C$.