What is the simplest way to compute :$\int^{\pi}_0\bigl(\frac{\sin(x)}{5-4\cos(x)}\bigr)^2dx=\frac{\pi}{24}$

Question

I am trying to compute the following integral.

$$\int^{\pi}_0\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx$$

My attempt was to expand the integrand and make use of the standard change of variables $t =\tan x/2$. But it turns out to be a lengthy and exhausting computations. I can put all details here since it is not pleasant. at the end I got the answer:

$$\int^{\pi}_0\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx =\frac{\pi}{24}$$ Maybe it is not correct so do not trust this result at 100%

I would like to know if there is an easiest way or trick that quickly leads to the answer?

Or if you want to avoid complex analysis you can simple use the t-formulas: put t = tan(x/2) — , Nov 09 '17 at 22:08
@Math_QED I have tried that trust me the out come is unpleasant — Guy Fsone, Nov 09 '17 at 22:14
Essentially, you have just asked the same question twice in a row: https://math.stackexchange.com/questions/2398037/calculate-int-pi-pi-bigg-sum-n-1-infty-frac-sinnx2n-bigg2dx?noredirect=1&lq=1 — Jack D'Aurizio, Nov 09 '17 at 22:51

score 7 · Accepted Answer · edited Nov 09 '17 at 23:24

Let $z=e^{ix}$ and then \begin{eqnarray} &&\int^{\pi}_0\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx\\ &=&\frac12\int^{\pi}_{-\pi}\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx\\ &=&\frac12\int_{|z|=1}\bigg(\frac{\frac{z-\frac1z}{2i}}{5-4\frac{z+\frac1z}{2}}\bigg)^2\frac{dz}{iz}\\ &=&\frac12\int_{|z|=1}\bigg(\frac{z^2-1}{2i(5z-2z^2-2)}\bigg)^2\frac{dz}{iz}\\ &=&-\frac1{8i}\int_{|z|=1}\bigg(\frac{z^2-1}{2z^2-5z+2}\bigg)^2\frac{dz}{z}\\ &=&-\frac1{8i}\int_{|z|=1}\frac{(z^2-1)^2}{z(2z-1)^2(z-2)^2}dz\\ &=&-\frac1{32i}\int_{|z|=1}\frac{(z^2-1)^2}{z(z-\frac{1}{2})^2(z-2)^2}dz\\ &=&-\frac1{32i}\cdot2\pi i\left(\text{Res}\bigg(\frac{(z^2-1)^2}{z(z-\frac{1}{2})^2(z-2)^2},z=0\bigg)+\text{Res}\bigg(\frac{(z^2-1)^2}{z(z-\frac{1}{2})^2(z-2)^2},z=\frac12\bigg)\right)\\ &=&-\frac\pi{16}(1-\frac{5}{3})\\ &=&\frac{\pi}{24}. \end{eqnarray}

Waoo this is really fast thanks I was right – Guy Fsone Nov 09 '17 at 22:04 — Guy Fsone, Nov 09 '17 at 22:04

score 4 · Answer 2 · answered Nov 09 '17 at 22:58

Here you seemed pretty sure of your (fine) proof of $$ \sum_{n\geq 1}\frac{\sin(nx)}{2^n} = \frac{2\sin x}{5-4\cos x} \tag{A}$$ so I wonder how you can be uncertain of the mentioned equality. Given $(A)$, by Parseval's identity

$$ \int_{0}^{\pi}\left(\frac{\sin x}{5-4\cos x}\right)^2\,dx =\frac{1}{8}\sum_{n\geq 1}\frac{\pi}{4^n}=\frac{\pi}{24}\tag{B}$$ nice and easy.

Paramanand Singh · Answer 3 · 2017-11-10T05:29:18.377

The following approach does not use anything beyond the material available in a first course in calculus.

Note that $$\frac{d} {dx} \frac{1}{5-4\cos x} =-\frac{4\sin x} {(5-4\cos x) ^{2}}$$ and hence using integration by parts we have $$I=\left.-\frac{\sin x}{4(5-4\cos x)}\right|_{x=0}^{x=\pi}+\frac{1}{4}\int_{0}^{\pi}\frac{\cos x} {5-4\cos x}\, dx$$ The first term vanishes and the second term can be written as $$-\frac{1}{16}\int_{0}^{\pi}\left(1-\frac{5}{5-4\cos x} \right) \, dx$$ and this evaluates to $$-\frac{\pi} {16}+\frac{5}{16}\frac{\pi}{3}=\frac{\pi}{24}$$ Here we have used the formula $$\int_{0}^{\pi}\frac{dx}{a + b\cos x} =\frac{\pi} {\sqrt{a^{2}-b^{2}}},\, a>|b|$$ which is (not so) easily proved via the substitution $$(a+b\cos x) (a-b\cos y) =a^{2}-b^{2}$$

score 2 · Answer 4 · answered Nov 09 '17 at 22:58

A slight improvement in the complex method - doing it this way you only have to calculate one residue instead of two. Noting that $$\int_{-\pi}^\pi \frac{\sin2x}{(5-4\cos x)^2}\,dx=0$$ because the integrand is odd, your integral $I$ is given by $$\eqalign{I &=\frac12\int_{-\pi}^\pi \frac{\sin^2x}{(5-4\cos x)^2}\,dx\cr &=\frac14\int_{-\pi}^\pi \frac{1-\cos2x}{(5-4\cos x)^2}\,dx\cr &=\frac14\int_{-\pi}^\pi \frac{1-\cos2x-i\sin2x}{(5-4\cos x)^2}\,dx\cr &=\frac14\int_{\rm unit\ circle} \frac{1-z^2}{(5-2(z+\frac1z))^2}\,\frac{dz}{iz}\cr &=\frac1{4i}\int_{\rm uc} \frac{z(1-z^2)}{(2z^2-5z+2)^2}\,dz\cr &=\frac\pi2\,{\rm Res}\Bigl(\frac{z(1-z^2)}{(2z^2-5z+2)^2},z=\frac12\Bigr)\cr}$$ and I'll leave you to show that the residue is $\frac1{12}$.

score 2 · Answer 5 · answered Nov 10 '17 at 08:06

Tha tangent half-angle substitution is not so bad since using $t=\tan(\frac x2)$ $$I=\int\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx=\int\frac{8 t^2}{\left(t^2+1\right) \left(9 t^2+1\right)^2}\,dt$$ Using partial fractions $$\frac{8 t^2}{\left(t^2+1\right) \left(9 t^2+1\right)^2}=\frac{9}{8 \left(9 t^2+1\right)}-\frac{1}{\left(9 t^2+1\right)^2}-\frac{1}{8 \left(t^2+1\right)}$$ The second term needs to be integrated by parts. So, we have $$\int\frac{dt}{\left(9 t^2+1\right)}=\frac{1}{3} \tan ^{-1}(3 t)$$ $$\int\frac{dt}{\left(9 t^2+1\right)^2}=\frac{t}{2 \left(9 t^2+1\right)}+\frac{1}{6} \tan ^{-1}(3 t)$$ $$\int\frac{dt}{ \left(t^2+1\right)}=\tan ^{-1}( t)$$ making $$I=-\frac{t}{2 \left(9 t^2+1\right)}-\frac{1}{8} \tan ^{-1}(t)+\frac{5}{24} \tan ^{-1}(3 t)$$ Integrating between $0$ and $\infty$, the first term is $0$ and you are left with $$\frac 5 {24}\frac \pi 2-\frac 1 {8}\frac \pi 2=\frac \pi {24}$$

What is the simplest way to compute :$\int^{\pi}_0\bigl(\frac{\sin(x)}{5-4\cos(x)}\bigr)^2dx=\frac{\pi}{24}$

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