<s>[INST] How to solve $\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx$ 
\begin{equation}
\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx
\end{equation}

This is a MCQ question and there are 5 options to choose which are "A.$2\sqrt{3}\pi+9\ln 3$, B.$2\sqrt{7}\pi+8\ln 3$ , C.$2\sqrt{3}\pi+8\ln 3$ , D.$2\sqrt{2}\pi+2\ln 3$, E.Other solution."
How do you solve this integral?
This appears in MCQ test, so I think there should be a trick to solve this without using too much force. The test have 30 questions and 2 hours, so each question should be finished under 4mn ( I doubted it though). Here is my solution that I spend around 3h to solve it. ( Sorry for not using latex and sorry for the inconveniences )Photo
I appreciate any solution tricks. Thanks!
 [/INST] Tou have done a good work and obtained the correct result.
Trying on my side, using as you did $s=\sin(x)$, we ned with
$$\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx=36 \int_{0}^1  \frac{ds}{ \left(s^4+s^2+1\right)^2}$$
$$s^4+s^2+1=\left(s^2-s+1\right) \left(s^2+s+1\right)$$ Using partial fraction decomposition
$$\frac{1}{ \left(s^4+s^2+1\right)^2}=\frac{1-s}{2 \left(s^2-s+1\right)}+\frac{s+1}{2 \left(s^2+s+1\right)}-$$ $$\frac{s}{4
   \left(s^2-s+1\right)^2}+\frac{s}{4 \left(s^2+s+1\right)^2}$$ The first and the second are not difficult
$$I_1=\int\frac{1-s}{ s^2-s+1}\,ds=-\frac 12\Bigg[\int \frac{2s-1}{ s^2-s+1}ds -\int\frac{ds}{ s^2-s+1}\Bigg]$$
$$I_2=\int\frac{s+1}{s^2+s+1}ds=\frac 12\Bigg[\int \frac{2s+1}{ s^2+s+1}ds -\int\frac{ds}{ s^2+s+1}\Bigg]$$
Now, for the third and fourth antiderivatives which look like
$$J=\int \frac s{(s^2+as+1)^2} ds=\int \frac s{(s-r_1)^2(s-r_2)^2} ds$$ partial fraction decomposition again
$$\frac s{(s-r_1)^2(s-r_2)^2}=\frac{r_1}{(r_2-r_1)^2 (s-r_1)^2}+\frac{r_1+r_2}{(r_2-r_1)^3 (s-r_1)}-$$ $$\frac{r_1+r_2}{(r_2-r_1)^3
   (s-r_2)}+\frac{r_2}{(r_2-r_1)^2 (s-r_2)^2}$$
This is a pure nightmare !
At the end, after recombining evrything, the antiderivative is
$$\frac{6 s\left(1-s^2\right)}{s^4+s^2+1}+4 \sqrt{3} \tan ^{-1}\left(\frac{16 s \left(1-s^2\right)}{3 \sqrt{3}}\right)+9 \log \left(\frac{s^2+s+1}{s^2-s+1}\right)$$ and, for the definite integral, your good result
$$2 \sqrt{3} \pi +9 \log (3)$$
All this work took me more than one hour. Be sure that I have been looking for tricks but ... no one came to my mind.
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