How to evaluate the following integral $\int \frac{\sin^2 x}{(x\cos x-\sin x)^2} dx$? From the above I can only get:$$\int \frac{1}{(x\cot x-1)^2} dx$$
Then I have no ideas. Is there anyone helping me? Any tips are useful.
 A: $$\int\frac{\sin^{2}(x)}{(x\cos(x)-\sin(x))^{2}}dx=\int\frac{\tan^{2}(x)}{(x-\tan(x))^{2}}dx$$
Now try the substitution $u=x-\tan(x)$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
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The integrand resembles the formula $\ds{\pars{\fermi \over {\rm g}}'
={\fermi'{\rm g} - {\rm g}'\fermi\over {\rm g}^{2}}}$. Then we 'rebuild' the formula to check if that sort is true: 

\begin{align}
&\int{\sin^{2}\pars{x} \over \bracks{x\cos\pars{x} - \sin\pars{x}}^{2}}\,\dd x
\\[3mm]&=\int{-\sin\pars{x}\bracks{x\cos\pars{x} - \sin\pars{x}}
   -\bracks{\cos\pars{x} - x\sin\pars{x} - \cos\pars{x}}\cos\pars{x}
   \over \bracks{x\cos\pars{x} - \sin\pars{x}}^{2}}\,\dd x
\\[3mm]&=\int\totald{}{x}\bracks{\cos\pars{x} \over x\cos\pars{x} - \sin\pars{x}}
\,\dd x = {\cos\pars{x} \over x\cos\pars{x} - \sin\pars{x}} + \mbox{a constant}
\end{align}

