Integration $\int_0^{\pi/2} \frac{dx}{(3 + 5 \cos x)^2}$ I had tried to solve this integral; using the substitution $\tan(x/2) =t$, and $\cos x= \frac{1-t^2}{1+t^2}$. But after making terms in $t$, I am not able to integrate further as numerator contains quadratic and denominator contains biquadratic.
$\int\limits_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}\ dx$.

An alternative approach is to integrate both sides of
$$\left(\frac{5\sin x}{3 + 5 \cos x}\right)’
= \frac{16}{(3 + 5 \cos x)^2}+\frac3{3 + 5 \cos x}$$
to simply the integral
\begin{align}
\int_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}dx
=  \frac{5}{48}
-\frac1{16}\int_0^{\pi/2} \frac3{3 + 5 \cos x}dx
\end{align}
