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.
$\int\limits_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}\ dx$.

Well let's use Weierstrass and partial fractions ig and evaluate the definite with FTC II.
Using Weierstrass with the tangent half angle substitution, our integral becomes
$$\int \frac{1}{(3 + 5 \cos x)^2}\ dx = \int \frac{1}{\left(3+5\left(\frac{1-t^2}{1+t^2}\right)\right)^2}\cdot \frac{2dt}{1+t^2} = \frac12\int {t^2 + 1\over(t - 2)^2 (t + 2)^2}dt$$
Perform partial fraction decomposition by setting it up like this
$$ {t^2 + 1\over2(t - 2)^2 (t + 2)^2}dt = \frac{A}{2(t-2)} + \frac{B}{2(t-2)^2} + \frac{C}{2(t+2)} + \frac{D}{2(t+2)^2}$$
Then just multiply by the denominator, match powers, and solve the system of equations for the unknown. We get
$$ {t^2 + 1\over2(t - 2)^2 (t + 2)^2}dt = \frac{3}{64(t-2)} + \frac{5}{32(t-2)^2} + \frac{-3}{64(t+2)} + \frac{5}{32(t+2)^2}$$
Now integrate termwise (log for single power, 1/(whatever) for double power) and simplify to get
$$\int \frac{3}{64(t-2)} + \frac{5}{32(t-2)^2} + \frac{-3}{64(t+2)} + \frac{5}{32(t+2)^2} dt $$$$= {-5t\over 16(t^2-4)}  + {3\over 64}\ln(2-t) -\frac3{64}\ln(t+2)+C$$
The bounds transform as follows $\tan\left({\frac\pi2\over2}\right)=1$ and the lower one remains $0$ so
$${-5t\over 16(t^2-4)}  + {3\over 64}\ln(2-t) -\frac3{64}\ln(t+2)\Big|^1_0 = \frac5{48} - {3\ln(3)\over64}$$
