3
$\begingroup$

$$\int\limits_{0}^{\pi/2}{\frac{\theta \left( 1+\sin^{2}\theta \right)\cos \theta }{\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta }$$

Attempt: I tried making the substitution $u = \sin(\theta)$ so that the integral could become easier to split.

I then ended up with http://www.wolframalpha.com/input/?i=integrate+arcsin(x)(1%2Bx%5E2)%2F((1%2B3x%5E2)(3%2Bx%5E2))+from+0+to+1 and tried to use a from of partial fractions (using a system of equations) to arrive at the answer.

I somehow got $A = -\pi/16$ and $B = 3\pi/16$, from the system $3A + B = 0$ and $A + B = \pi/8$. But my answer was off by a bit, which is why I'm curious what to do next. Any help would be appreciated!

$\endgroup$
  • $\begingroup$ The extra $\theta$ in the numerator is not a typo, correct? $\endgroup$ – abiessu Aug 15 '16 at 21:06
  • $\begingroup$ @abiessu correct, it indeed is not a typo. :) $\endgroup$ – Why Do You Care Aug 15 '16 at 21:07
  • $\begingroup$ Have you tried using the tangent half-angle substitution? $\endgroup$ – Mark Viola Aug 15 '16 at 21:30
  • $\begingroup$ @Dr.MV Yes and it is way, way too bashy/tedious to use. Also, it's ineffective because of the stray theta term. You end up with the product of rationals times arctan and you can't do much from there. $\endgroup$ – Why Do You Care Aug 15 '16 at 21:39
  • $\begingroup$ Integrate by parts with $u=\theta$ and $$v=\int \frac{(1+\sin^2 (\theta))\,\cos( \theta)}{(1+3\sin^2(\theta))(3+ \sin^2(\theta ))}\, d\theta$$ $\endgroup$ – Mark Viola Aug 15 '16 at 21:48
1
$\begingroup$

It is not complete answer, but hopefully will help to approach this task.

First of all, let us generalize integral: $$ I(b)=\int\limits_{0}^{\pi/2}{\frac{\arccos\left(b\cos(\theta)\right) \left( 1+\sin^{2}\theta \right)\cos \theta }{\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta },\quad b\in[0,1] $$ For $b=1$ it appears to be initial one.

Next, derivate with respect to $b$. (Change order of integration/derivation). $$ \frac{\partial I }{\partial b}=-\int\limits_{0}^{\pi/2}{\frac{ \left( 1+\sin^{2}\theta \right)\cos^{2} \theta }{\sqrt{1-b^2\cos^{2}(\theta)}\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta } $$ Notice that argument is an even function, thus $$ \int\limits_{0}^{\pi/2}=\int\limits_{-\pi/2}^{0}\to\frac{\partial I }{\partial b}=-\frac{1}{2}\int\limits_{-\pi/2}^{\pi/2} $$

Make shift $\theta\to\theta+\pi/2$. (Change $\sin^2(\theta)\to\cos^2(\theta),$ $\cos^2(\theta)\to\sin^2(\theta)$)

$$ \frac{\partial I }{\partial b}=-\frac{1}{2}\int\limits_{0}^{\pi}{\frac{ \left( 1+\cos^{2}\theta \right)\sin^{2} \theta }{\sqrt{1-b^2\sin^{2}(\theta)}\left( 1+3\cos^{2}\theta \right)\left( 3+\cos^{2}\theta \right)}d\theta } $$ Once again, notice that shift $\theta\to\theta+\pi$ don't change the integral, thus we can write $$ \frac{\partial I }{\partial b}=-\frac{1}{4}\int\limits_{0}^{2\pi}{\frac{ \left( 1+\cos^{2}\theta \right)\sin^{2} \theta }{\sqrt{1-b^2\sin^{2}(\theta)}\left( 1+3\cos^{2}\theta \right)\left( 3+\cos^{2}\theta \right)}d\theta } $$ At this point we can use theory of residuals. Corresponding contour integral after some simplifications $$ \frac{\partial I }{\partial b}=\frac{i}{6b}\int\limits_{|z|=1}\frac{(z^4+6z^2+1)(z^2-1)^2}{\sqrt{-z^4-1+z^2\left(2+\frac{4}{b^2}\right)}(z^4+\frac{10}{3}z^2+1)(z^4+14z^2+1)}dz $$ Next step should be calculating of residuals, but I stubmle here, having $0$ everywhere (probably, miscalculations).

Anyway, idea of further solution is to integrate obtained expression over $b$, calculate constant of integration from $I(0)=\int\limits_{0}^{\pi/2}{\frac{\left( 1+\sin^{2}\theta \right)\cos \theta }{\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta }$, which is super easy to compute, and then just take value of $I(1)$.

$\endgroup$
  • $\begingroup$ Beautiful Answer! $\endgroup$ – Why Do You Care Aug 20 '16 at 5:57

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.