How to evaluate the following integral $\int \frac{\sin^2 x}{(x\cos x-\sin x)^2} dx$?

Question

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.

Answer 1

$$\int\frac{\sin^{2}(x)}{(x\cos(x)-\sin(x))^{2}}dx=\int\frac{\tan^{2}(x)}{(x-\tan(x))^{2}}dx$$ Now try the substitution $u=x-\tan(x)$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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}