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I know this integral exists - I've plugged it into Mathematica which spit out some answer. But what I would really like to know is what techniques could be useful in calculating the integral $$\int \frac{1}{(2\sin x + 3\sin 2x - 2)^2+1}dx$$ by hand.

Edit: There have been put forth concerns that this question lacks 'context'. This is somewhat surprising considering one can literally search for 'Calculate integral for' and have returned 10 if not 20 completely similar questions without any sort of 'context' to them.

To those concerned - this equation is a one specific instance of a family of equations that arise from the analysis of certain types of complexity classes.

Edit II: Previous attempts at solving integrals of this type have included integration by parts which didn't yield anything of much use.

C Shreve
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  • What are the bounds? – Jacky Chong Oct 28 '16 at 01:24
  • Well I was looking for the indefinite integral...but if it helps the discussion, the lower bound would be 0 and the upper bound 2Pi. – C Shreve Oct 28 '16 at 04:30
  • Please edit your question to add additional context. What is the source of the integral? Why is it of interest? Of the infinite number of integrals that could be posed, why focus on this one? Information of this sort makes the question more useful for others. – Carl Mummert Oct 29 '16 at 18:27
  • @TheGreatDuck - none of these similar questions have been closed. Indeed - see all the 'Related' questions. None of these require 'context' or have been put on 'hold'. So why this one? – C Shreve Oct 31 '16 at 05:10
  • @TheGreatDuck - pretty sure this question was seen. http://math.stackexchange.com/questions/1164183/how-to-evaluate-i-int-limits-0-pi-2-fracx-log-sinx-sinx-dx?rq=1 Bonus points: No context. – C Shreve Oct 31 '16 at 05:14
  • @TheGreatDuck - copy that. FWIW - I've added additional context to my question. – C Shreve Oct 31 '16 at 05:17
  • Let us continue this discussion in chat. – C Shreve Oct 31 '16 at 18:50
  • @TheGreatDuck: for this kind of problem, the source and motivation are more important IMO than the asker's attempts. Even if someone has no idea how to solve an integral, they can explain its source and significance. Cf. http://meta.math.stackexchange.com/a/9960/630 – Carl Mummert Nov 01 '16 at 11:43
  • @C Shreve: " 10 if not 20 completely similar questions without any sort of 'context' to them." - that is exactly the problem that we are trying to address. It is challenging on a decentralized site to uphold high quality standards on all questions. But the expectation is that questions should go beyond merely stating an integral to be solved. See "How to ask a good question" here on meta: http://meta.math.stackexchange.com/q/9959/630 – Carl Mummert Nov 01 '16 at 11:45

1 Answers1

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Considering $$I=\int \frac{dx}{(2\sin (x) + 3\sin (2x) - 2)^2+1}=\int \frac{dx}{(2\sin( x) + 6\sin(x)\cos(x) - 2)^2+1}$$ use the tangent half angle substitution. This lead to $$I=\int \frac{2 \left(t^2+1\right)^3 }{ 5 t^8+32 t^7+84 t^6-226 t^4-96 t^3+276 t^2-64 t+5}\,dt$$ The hard part is the denominator but, fortunately, it does not show any real root. So $$\frac 52I=\int \frac{ \left(t^2+1\right)^3 }{(t^2+a^2)(t^2+b^2)(t^2+c^2)(t^2+d^2)}\,dt$$ Now, using partial fraction decomposition makes the integrand to be $$\frac A{t^2+a^2}+\frac B{t^2+b^2}+\frac C{t^2+c^2}+\frac D{t^2+d^2}$$ where $$A=\frac{(a^2-1)^3}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(a^2-d^2\right)}$$ $$B=\frac{(b^2-1)^3}{\left(b^2-a^2\right) \left(b^2-c^2\right) \left(b^2-d^2\right)}$$ $$C=\frac{(c^2-1)^3}{\left(c^2-a^2\right) \left(c^2-b^2\right) \left(c^2-d^2\right)}$$ $$D=\frac{(d^2-1)^3}{\left(d^2-a^2\right) \left(d^2-b^2\right) \left(d^2-c^2\right)}$$ All of the above make $$\frac 52I=\frac{A }{a}\tan ^{-1}\left(\frac{t}{a}\right)+\frac{B }{b}\tan ^{-1}\left(\frac{t}{b}\right)+\frac{C }{c}\tan ^{-1}\left(\frac{t}{c}\right)+\frac{D }{d}\tan ^{-1}\left(\frac{t}{d}\right)$$ Numerically, we should find $a^2\approx 0.0206317$, $b^2\approx 1.00839$, $c^2\approx 3.75827$, $d^2\approx 12.7893$.

Claude Leibovici
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  • @TheGreatDuck. I think that this question is interesting even if it suffers lack of context. If you try with any CAS, the result is just a monster. I thought that it could be interesting to show what a modest human being could do without any of these tools. Cheers. – Claude Leibovici Oct 31 '16 at 05:09
  • @Claude Leibovici: And to be quite clear - the monstrosity that Mathematica spits out indicates that this can be done mechanically. Which lead to the question - how would a human approach this problem. I must confess - complicated integration was not my forte in university. Thank you for your input. – C Shreve Oct 31 '16 at 05:19
  • @CShreve. To me, this problem was a challenge and I wanted to see what a human being could do by hand. Where I have been tricky is when I had to look at the polynomial in $t^8$ for which I used a solver for the roots. Once I dentified that all roots are complex, then it became quite easy to make it in a formal way. – Claude Leibovici Oct 31 '16 at 05:27