Partial Integration $ \int \frac{x\cos x}{\sin^3x}dx $

Question

The problem:

$$ \int \frac{x\cos x}{\sin^3x}dx $$

Can someone give me a hint on how to solve this without using cosecant?

The solution provided is:

$$ -\frac{1}{2}\left(\frac{x}{\sin^2x}+\cot x\right)\:+\:C $$

Answer 1

HINT:

$$\int x\cdot\frac{\cos x}{\sin^3x}dx=x\int\frac{\cos x}{\sin^3x}dx-\int\left(\frac{dx}{dx}\int\frac{\cos x}{\sin^3x}dx\right)dx$$

For $\int\dfrac{\cos x}{\sin^3x}dx$ write $\sin x=u$

Answer 2

$$\begin{gathered} I = \int {\frac{{x\cos x}} {{{{\sin }^3}x}}dx} = \int {\frac{{x\cos x}} {{\sin x}}.\frac{1} {{{{\sin }^2}x}}dx} = - \int {\frac{{x\cos x}} {{\sin x}}.d\left( {\cot x} \right)} \hfill \\ = - \frac{{x\cos x}} {{\sin x}}.\cot x + \int {\cot xd\left( {\frac{{x\cos x}} {{\sin x}}} \right)} \hfill \\ = - \frac{{x{{\cos }^2}x}} {{{{\sin }^2}x}} + \int {\cot x.\frac{{\left( {\cos x - x\sin x} \right)\sin x - \left( {x\cos x} \right)\cos x}} {{{{\sin }^2}x}}dx} \hfill \\ = - x{\cot ^2}x + \int {\cot x.\frac{{\sin x\cos x - x}} {{{{\sin }^2}x}}dx} \hfill \\ = - x{\cot ^2}x + \int {\frac{{\cos x}} {{\sin x}}.\left( {\frac{{\cos x}} {{\sin x}} - \frac{x} {{{{\sin }^2}x}}} \right)dx} \hfill \\ = - x{\cot ^2}x + \int {{{\cot }^2}xdx} - I \hfill \\ \end{gathered} $$ So, $$2I = - x{\cot ^2}x + \int {{{\cot }^2}xdx} \Rightarrow I = \frac{1} {2}\left( { - x{{\cot }^2}x + \int {{{\cot }^2}xdx} } \right).$$ But $$\int {{{\cot }^2}xdx} = \int {\left( {\frac{1} {{{{\sin }^2}x}} - 1} \right)dx} = - \cot x - x + C.$$ So, $$I = \frac{1}{2}\left( { - x{{\cot }^2}x - \cot x - x + C} \right).$$ Finally, $$I = \frac{1} {2}\left[ { - x\left( {1 + {{\cot }^2}x} \right) - \cot x + C} \right] = - \frac{1} {2}\left[ {x\left( {1 + {{\cot }^2}x} \right) + \cot x - C} \right] = - \frac{1} {2}\left( {\frac{x} {{{{\sin }^2}x}} + \cot x - C} \right)$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int\frac{x\cos\pars{x}}{\sin^{3}\pars{x}}\,\dd x} =-\,\half\int x\,\dd\bracks{\frac{1}{\sin^{2}\pars{x}}} =-\,\half\,\frac{x}{\sin^{2}\pars{x}} + \half\, \overbrace{\int\frac{\dd x}{\sin^{2}\pars{x}}}^{\dsc{-\cot\pars{x}}} \\[5mm]&=\color{#66f}{\large-\,\half\bracks{x\csc^{2}\pars{x} + \cot\pars{x}}} + \mbox{a constant} \end{align}