Question

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

• A. $ \frac{sin\,x+cos\,x}{x\,sin\,x+cos\,x}+C $
• B. $ \frac{x\,sin\,x-cos\,x}{x\,sin\,x+cos\,x}+C $
• C. $ \frac{sin\,x-cos\,x}{x\,sin\,x+cos\,x}+C $
• D. None of the above

Answer 1

Answer: (C)

Since, $\frac{d}{d x}(x \sin x+\cos x)=x \cos x$
$\therefore I=\int \frac{x^{2} d x}{(x \sin x+\cos x)^{2}}$
$=\int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x+\cos x)^{2}} d x$
$=\frac{x}{\cos x} \cdot\left(\frac{-1}{x \sin x+\cos x}\right)$
$-\int \frac{\cos x-x(-\sin x)}{\cos ^{2} x} \cdot \frac{-1}{(x \sin x+\cos x)} d x$
$=\frac{-x}{\cos x(x \sin x+\cos x)}+\int \sec ^{2} d x$
$=\frac{-x}{\cos x(x \sin x+\cos x)}+\tan x+C$
$=\frac{-x+\sin x(x \sin x+\cos x)}{\cos x(x \sin x+\cos x)}+C$
$=\frac{-x \cos ^{2} x+\sin x \cdot \cos x}{\cos x(x \sin x+\cos x)}+C$
$=\left(\frac{\sin x-x \cos x}{\cos x+x \sin x}\right)+C$