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Evaluate: $$ \int_{0}^{\pi / 2} \frac{x^{3} \cos x d x}{3 \sin x-\sin 3 x} $$

Here I can see that the denominator nicely converts into $4\sin^{3}{x}$ so I basically get $$ \int_{0}^{\pi / 2}\left(\frac{x}{\sin x}\right)^{3} \cos x\>{dx} $$ After that I tried substituting $\sin{x}$ as $u$ but that only complicates the problem further, leaving me with an inverse function to deal with. Also, the King's property is not useful here.

Can anyone provide an alternate approach to this question?

Quanto
  • 97,352
Shriom707
  • 323

5 Answers5

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Integrate by parts $$I= \int_{0}^{\pi / 2}\left(\frac{x}{\sin x}\right)^{3} d(\sin x ) =\frac{\pi^3}{8}+ 3I - 3\int_{0}^{\pi / 2}x^2\csc^2xdx$$ Integrate the last term by parts again

$$\int_{0}^{\pi / 2}x^2\csc^2x dx= 2\int_{0}^{\pi / 2}x\cot xdx =- 2\int_{0}^{\pi / 2}\ln\sin xdx=\pi\ln2 $$

Thus, $I= \frac{3\pi}2\ln2-\frac{\pi^3}{16}$ and

$$ \int_{0}^{\pi / 2} \frac{x^{3} \cos x d x}{3 \sin x-\sin 3 x} = \frac14 I= \frac{3\pi}8\ln2-\frac{\pi^3}{64} $$

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Note $$\begin{eqnarray*} \int_{0}^{\pi/2}\ln\sin x\,dx &=&\frac12\int_{0}^{\pi}\ln\sin x\,dx =\int_{0}^{\pi/2}\ln\sin(2t)dt\\&=&\frac\pi2\ln2+\int_{0}^{\pi/2}\ln\sin tdt+\int_{0}^{\pi/2}\ln\cos tdt\\&=&\frac\pi2\ln2+ 2 \int_{0}^{\pi/2}\ln\sin x dx = - \frac\pi2\ln2 \end{eqnarray*}$$

Quanto
  • 97,352
2

Let us do repeated integration by parts: $$\int x^3 f(x) dx=x^3 I^{(1)}(x)-3x^2I^{(2)}(x)+6xI^{(3)}(x)-6I^{(4)}(x)~~~~(1)$$ Here $I^{(k)}(x)$ denotes $k$th integration of $f(x)=\frac{\cos x}{\sin ^3 x}$, We have $$I^{(1)}(x)=-\frac{1}{2} \cot^2 x, I^{(2)}(x)=-\frac{1}{2}(-x-\cot x), I^{(3)}(x)=-\frac{1}{2}(-x^2/2-\ln(\sin x))$$ $$\implies I^{(4)}(x)=-\frac{1}{2}(-x^4/6-\int \ln \sin x) dx$$ Using $\int_{0}^{\pi/2} \ln\sin x dx=-\frac{\pi}{2} \ln 2$ and putting these expressions in (1), we get $$\int_{0}^{\pi/2} \frac{x^3 \cos x}{\sin^3 x}dx=\frac{3\pi}{2}\ln 2-\frac{\pi^3}{16}$$

Z Ahmed
  • 43,235
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\pi/2}{x^{3}\cos\pars{x} \over 3\sin\pars{x} - \sin\pars{3x}}\,\dd x} = {1 \over 4}\int_{0}^{\pi/2}x^{3}\,{\cos\pars{x} \over \sin^{3}\pars{x}}\,\dd x \\[5mm] = &\ -\,{1 \over 8}\int_{x\ =\ 0}^{x\ =\ \pi/2}x^{3}\,\dd\bracks{1 \over \sin^{2}\pars{x}} = -\,{\pi^{3} \over 64} + {3 \over 8} \color{#00f}{\bf\int_{0}^{\pi/2}{x^{2} \over \sin^{2}\pars{x}}\,\dd x} \label{1}\tag{1} \end{align}

\begin{align} &\color{#00f}{\bf\int_{0}^{\pi/2}{x^{2} \over \sin^{2}\pars{x}} \,\dd x} = \left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}{\bracks{-\ic\ln\pars{z}}^{\, 2} \over \bracks{\pars{z - 1/z}/\pars{2\ic}}^{\, 2}} {\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.4\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}{z\ln^{2}\pars{z} \over \pars{z^{2} - 1}^{\, 2}}\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ -\,4\,\Im\int_{1}^{0}{\ic y\,\bracks{\ln\pars{y} + \ic\pi/2}^{\, 2} \over \pars{\ic^{2}y^{2} - 1}^{\, 2}}\,\ic\,\dd y \\[5mm] = &\ -\,4\pi\int_{0}^{1}{y\ln\pars{y}\over \pars{1 + y^{2}}^{\, 2}}\,\dd y = -\,\pi\int_{0}^{1}{\ln\pars{y}\over \pars{1 + y}^{\, 2}}\,\dd y = \pi\ln\pars{2} \end{align} (\ref{1}) becomes $\ds{\ \bbox[#ffd,10px,border:1px solid navy]{{3 \over 8}\,\pi\ln\pars{2} - {\pi^{3} \over 64}}}$ $\ds{\ \approx\ 0.3321}$
\begin{align} \int_{0}^{1}{\ln\pars{y}\over \pars{1 + y}^{\, 2}}\,\dd y & = \lim_{\epsilon \to 0^{+}}\bracks{\ln\pars{\epsilon} + \int_{\epsilon}^{1}{\dd y \over y\pars{1 + y}}} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\bracks{\ln\pars{\epsilon} + \int_{\epsilon}^{1}{\dd y \over y} - \int_{\epsilon}^{1}{\dd y \over 1 + y}} = -\ln\pars{2} \end{align}
Felix Marin
  • 89,464
1

\begin{aligned} & \int_{0}^{\frac{x}{2}} \frac{x^{2} \cos x}{3 \sin x-\sin 3 x} d x \\ =& \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{3 \sin x-3 \sin x+4 \sin ^{3} x} \\ =& \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{\sin ^{3} x} d x \\ =&-\frac{1}{8} \int_{0}^{\frac{\pi}{2}} x^{3} d\left(\frac{1}{\sin ^{2} x}\right) \\ =&-\left[\frac{x^{3}}{8 \sin ^{2} x}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{8} \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x \\ =&-\frac{\pi^{3}}{64}+\frac{3}{8} \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x \\ =& \frac{3 \pi \ln 2}{8}-\frac{\pi^{3}}{64}\quad (\textrm{ Using } \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x=\pi \ln 2)\\ \end{aligned}

Lai
  • 20,421
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$$\begin{align*} & \int_0^{\tfrac\pi2} \frac{x^3 \cos x }{3 \sin x-\sin(3x)} \, dx \\ &= \frac14 \int_0^{\tfrac\pi2} \frac{x^3 \cos x }{\sin^3x} \, dx \\ &= \frac14 \int_0^{\tfrac\pi2} \frac{x^3 \sec^2x }{\tan^3x} \, dx \\ &= \frac14 \int_0^\infty \frac{\arctan^3y}{y^3} \, dy & x=\arctan y \\ &= \frac14 \left(\frac{3\pi}2\ln2 - \frac{\pi^3}{16}\right) = \boxed{\frac{3\pi}8\ln2-\frac{\pi^3}{64}} \end{align*}$$

where the last step follows from $\displaystyle\int_0^\infty\left(\frac{\arctan x}x\right)^n\,dx$ with $n=3$.

user170231
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