Question

If $\int \frac{x^2}{{(x\sin x+\cos x)}^2}dx=\frac{f\left(x\right)}{x\sin x+\cos x}+\tan x+c$, then

• A. $f\left(x\right)=\frac{x}{\cos x}$
• B. $f\left(x\right)=\frac{\cos x}{x}$
• C. $f\left(x\right)=-\frac{x}{\cos x}$
• D. none of these

Answer 1

Answer: (C)

$f\left(x\right)=-\frac{x}{\cos x}$

Substitute $\frac{1}{x\sin x+\cos x}=z\Rightarrow -\frac{1}{{(x\sin x+\cos x)}^2}dx=dz$

$\Rightarrow \frac{x\cos x}{{(x\sin x+\cos x)}^2}=-dz\Rightarrow \int \frac{x^2}{{(x\sin x+\cos x)}^2}dx=-z$

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

$=\frac{x}{\cos x}.(-z)-\int (-z)\left(\frac{\cos x.1+x\sin x}{{\cos }^{2}x}\right)dx$

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

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

$\therefore f\left(x\right)=-\frac{x}{\cos x}$