Indefinite integral $\int(\sin x)^2/(x\cdot\cos x-\sin x)^2\,dx$ How to find the antiderivative of $\dfrac{(\sin x)^2}{(x\cdot\cos x-\sin x)^2}$? I know that series theory can solve the problem. But I wish to solve the problem by the antiderivative way. The numerator and denominator divide the $x^2$ at the same time, don't they? 
 A: Try to imagine what kind of form it has to be.
The answer is 
$$\int \frac{{\sin^2 x}}{{(x \cos x-\sin x)}^2}dx=\frac{\cos x}{x \cos x-\sin x}+C.$$
You can know this is true if you differenciate the right hand side.
A: 
The numerator and denominator divide the $x^2$ at the same time, don't
  they?

No.
Divide both the numerator and denominator by $\cos ^{2}x$ 
\begin{equation*}
I=\int \frac{\sin ^{2}x}{(x\cos x-\sin x)^{2}}\,dx=\int \frac{\tan ^{2}x}{
(x-\tan x)^{2}}dx.
\end{equation*}
Notice that the numerator ($\tan ^{2}x$) is apart the sign the derivative of 
$x-\tan x$ in the denominator, which suggests the use of the substitution $
u=x-\tan x$. So $du=-\tan ^{2}x\,dx$ and $I$ becomes
\begin{equation*}
I=\int \frac{-1}{u^{2}}\,du=\frac{1}{u}+C=\frac{1}{x-\tan x}+C.
\end{equation*}
A: As $\displaystyle \frac{d(x\cos x-\sin x)}{dx}=-x\sin x,\int\frac{-x\sin x}{(x\cos x-\sin x)^2}dx=-\frac1{x\cos x-\sin x}$
Integrating by parts, 
$$\int\frac{\sin^2x}{(x\cos x-\sin x)^2}dx=\int \frac{(-x\sin x)}{(x\cos x-\sin x)^2}\cdot\frac{-\sin x}x dx$$
$$=\frac{-\sin x}x\int\frac{(-x\sin x)}{(x\cos x-\sin x)^2}dx-\int\left(\frac{d\left(\frac{-\sin x}x\right)}{dx}\cdot\int \frac{(-x\sin x)}{(x\cos x-\sin x)^2}dx\right)dx$$
$$=\frac{\sin x}x\cdot\frac1{x\cos x-\sin x}+\int\left(\frac{x\cos x-\sin x}{x^2}\cdot\frac1{x\cos x-\sin x}\right)dx$$
$$=\frac{\sin x}x\frac1{x\cos x-\sin x}+\int\frac{dx}{x^2}$$
Can you take it from here?
