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!
 A: Here are a few shortcuts. With $t =\sin x$
\begin{equation}
I=\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx=36\int_0^1 \frac {1}{(t^4+t^2+1)^2}dt\tag1
\end{equation}
Note
$$\left(\frac{t-t^3}{t^4+t^2+1}\right)’
= \frac{t^6+5}{(t^4+t^2+1)^2}-\frac{4}{t^4+t^2+1}
$$
Integrate both sides over $(0,\infty)$
\begin{align}
\int_0^\infty \frac{4}{t^4+t^2+1}dt &= \int_0^\infty \overset{t\to 1/t} {\frac{t^6}{(t^4+t^2+1)^2} }dt
 + \int_0^\infty \frac{5}{(t^4+t^2+1)^2}dt\\
&= \int_0^1\frac{6}{(t^4+t^2+1)^2}dt +\int_1^\infty \frac{6}{(t^4+t^2+1)^2}dt\tag2
\end{align}
Similarly, integrate over $(0,1)$
\begin{align}
\int_0^1\frac{4}{t^4+t^2+1}dt=\int_0^1\frac{5}{(t^4+t^2+1)^2}dt +\int_1^\infty \frac{1}{(t^4+t^2+1)^2}dt \tag3
\end{align}
Combine (2) and (3) to obtain
\begin{align}
\int_0^1\frac{dt}{(t^4+t^2+1)^2}= \int_0^1\frac{dt}{t^4+t^2+1}
-\frac14 \int_0^\infty \frac{dt}{t^4+t^2+1}= \frac14\ln3+\frac{\pi}{6\sqrt3}
\end{align}
Plug into (1)
$$I= 9\ln3+ 2\sqrt3{\pi}$$

P.S. The last integral is carried out as follows
$$\int \frac{1}{t^4+t^2+1}dt= \frac12 \int\frac{1-t^2}{t^4+t^2+1}dt 
 + \frac12 \int\frac{1+t^2}{t^4+t^2+1}dt \\
= \frac12 \int\frac{d(t+\frac1t)}{(t+\frac1t)^2-1}
 + \frac12 \int\frac{d(t-\frac1t)}{(t-\frac1t)^2+3}
$$
A: 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.
