10
$\begingroup$

How to evaluate the following integral $$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$

For integrating I took $\cos^{2}x$ outside and applied integration by parts.

Given answer is $\dfrac{\pi}{4ab^{2}(a+b)}$. But I am not getting the answer.

$\endgroup$
  • 1
    $\begingroup$ There are a lot of simplifications you can make on this integral to ease your workload a little bit. For instance, $\sin x\cos x = \frac12\sin(2x)$, and you should be able to rewrite your demoninator as $(m+n\cos(2x))^2$ for suitable $m$ and $n$. Once you have that, take $y=2x$. After that, you can integrate by parts pulling out your $y$ term; the integral in just $\sin y$ and $\cos y$ can be done easily with another substitution (if $z=\cos y$, $dz=\sin y dy$). $\endgroup$ – Steven Stadnicki Dec 15 '14 at 3:55
  • $\begingroup$ Try D.U.I.S. Seems work. $\endgroup$ – Venus Dec 15 '14 at 3:59
9
$\begingroup$

The case $a^2=b^2$ being simple, let's just consider, by symmetry, the case $a>b>0$.

Observe that $$ \partial _x \left(\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right)=2(a^2-b^2)\frac{\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2} $$ then, integrating by parts, you may write $$ \begin{align} I(a,b)&=\int_0^{\pi/2}\frac{x\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2} dx\\\\ &=\frac{x}{2(a^2-b^2)}\left.\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right|_{0}^{\pi/2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{dx}{a^2 \cos^2x+b^2 \sin^2 x}\\\\ &= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{1}{\left(a^2 +b^2 \large{\frac{\sin^2 x}{\cos^2x}}\right)}\frac{1}{\cos^2 x}dx\\\\ &= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{1}{\left(a^2 +b^2 \tan ^2x\right)}(\tan x)'dx\\\\ &= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\frac{1}{ab}\left.\arctan \left(\frac ba \tan x \right)\right|_{0}^{\pi/2}\\\\ &= \frac{\pi}{4(a^2-b^2)b^2}-\frac{\pi}{4(a^2-b^2)}\frac{1}{ab}\\\\ &= \frac{\pi}{4(a+b)ab^2}. \end{align} $$

$\endgroup$
  • $\begingroup$ The solution must be $\displaystyle{\pi\over 4\left(\,\left\vert\, a\,\right\vert + \left\vert\, b\,\right\vert\,\right)\left\vert\, a\,\right\vert b^{2}}$. $\endgroup$ – Felix Marin Dec 15 '14 at 17:52
  • $\begingroup$ @FelixMarin Thank you for your expression, it reduces to the answer I've given when $a>b>0$, which is the case I've explicitly considered. $\endgroup$ – Olivier Oloa Dec 15 '14 at 19:03
  • $\begingroup$ You are welcome. Thanks. $\endgroup$ – Felix Marin Dec 15 '14 at 19:05
3
$\begingroup$

Integrating by parts,

$$\int\frac{x\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$

$$=x\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx-\int\left[\frac{dx}{dx}\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx\right]dx$$

Now $a^2\cos^2x+b^2\sin^2x=u\implies2(b^2-a^2)\sin x\cos x\ dx=du$

For $\displaystyle\int\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\int\frac{\sec^2x\ dx}{a^2+b^2\tan^2x},$

set $b\tan x=a\tan y$

$\endgroup$
3
$\begingroup$

let $x=\dfrac{t}{2}$, we have $$I=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{\dfrac{t}{2}\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}}{\left(a^2\sin^2{\dfrac{t}{2}}+b^2\cos^2{\dfrac{t}{2}}\right)^2}dt=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{t\sin{t}}{[(a^2+b^2)+(a^2-b^2)\cos{t}]^2}dt$$ So \begin{align*}I&=-\dfrac{1}{2(a^2-b^2)}\int_{0}^{\pi}t\;d\left(\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}\right)\\ &=-\dfrac{1}{2(a^2-b^2)}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}\Bigg|_{0}^{\pi}+\int_{0}^{\pi}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}dt\\ &=-\dfrac{\pi}{4b^2(a^2-b^2)}+\int_{0}^{\pi}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}dt \end{align*}

$\endgroup$
1
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{\pi/2}{x\sin\pars{x}\cos\pars{x}\over \bracks{a^{2}\cos^{2}\pars{x} + b^{2}\sin^{2}\pars{x}}^{2}}\,\dd x} \\[5mm]&=\ \overbrace{\int_{0}^{\pi/2}{x\sin\pars{2x}/2\over\braces{% a^{2}\bracks{1 + \cos\pars{2x}}/2 + b^{2}\bracks{1 - \cos\pars{2x}}/2}^{2}}\,\dd x} ^{\dsc{2x}\ \ds{\mapsto}\ \dsc{x}} \\[5mm]&=\half\int_{0}^{\pi}{x\sin\pars{x}\over\bracks{% a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}}^{2}}\,\dd x \\[5mm]&={1 \over 2\pars{a^{2} - b^{2}}}\int_{x\ =\ 0}^{x\ =\ \pi}x\, \dd\bracks{1 \over a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}} \\[1cm]&={1 \over 2\pars{a^{2} - b^{2}}}\,\left. {x \over a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}} \right\vert_{x\ =\ 0}^{x\ =\ \pi} \\[5mm]&-{1 \over 2\pars{a^{2} - b^{2}}}\ \underbrace{\int_{0}^{\pi} {\dd x \over a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}}} _{\dsc{t}\ \ds{=}\ \dsc{\tan\pars{x \over 2}}} \\[1cm]&={\pi \over 4b^{2}\pars{a^{2} - b^{2}}} -{1 \over 2\pars{a^{2} - b^{2}}}\ \overbrace{\int_{0}^{\infty} {\dd x \over b^{2}t^{2} + a^{2}}}^{\ds{=}\ \dsc{\pi \over 2\verts{ab}}} ={\pi \over 4\verts{b}\pars{a^{2} - b^{2}}} \pars{{1 \over \verts{b}} - {1 \over \verts{a}}} \\[5mm]&=\color{#66f}{\large% {\pi \over 4\verts{a}b^{2}\pars{\verts{a} + \verts{b}}}} \end{align}

$\endgroup$
1
$\begingroup$

You can use this property : $$\int_a^b f(x)\hspace{1mm}dx = \int_a^b f(a+b-x)\hspace{1mm}dx$$

To prove this property : Substitute $a+b-x = u$

Let us have $$I = \int_0^{\pi/2}\dfrac{x\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (1)$$

After applying the property, you will get

$$I = \int_0^{\pi/2}\dfrac{(\pi/2-x)\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (2)$$

After adding the two equation, you get

$$I =\dfrac{\pi}{4} \int_0^{\pi/2}\dfrac{\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx $$

Now you will substitute $a^2\cos^2x+b^2\sin^2x = u$

To get $$I = \dfrac{\pi}{8(b^2-a^2)}\int_{a^2}^{b^2} \dfrac{1}{u^2}\hspace{1mm}du$$

I hope you can Integrate from here

$\endgroup$
  • $\begingroup$ But $\cos\left(\frac{\pi}{2}-x\right)=\sin x$, so the denominator ... $\endgroup$ – Venus Dec 31 '14 at 12:29
  • $\begingroup$ I made a mistake, my method won't work $\endgroup$ – stoic Jan 1 '15 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.