<s>[INST] Compute $\int^{\pi/2}_0 \frac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$ I have tried solving this for about an hour and will probably resort to head banging in some time:
$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$
I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:
$$\int ^{\infty}_{0} \dfrac{1+t^2}{(a^2 + b^2 t^2)^2}dt$$
This has become unmanageable. Neither splitting the numerator, nor Partial fraction (taking t^2 = z and applying partial fraction) seems to work. 
 [/INST] Say that $ a, b >0 .$ From the substitution $t \mapsto \dfrac{a}{b}t$ followed by $ t \mapsto \dfrac{1}{t}$ we get
$$
I = \int_{0}^{\infty}\dfrac{1+t^{2}}{(a^{2}+b^{2}t^{2})^{2}}\, dt = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}\dfrac{b^2+a^2t^{2}}{(1+ t^{2})^{2}}\, dt
$$
and
$$
I = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}\dfrac{b^2t^2+a^2}{(1+ t^{2})^{2}}\, dt .
$$
Consequently
$$
2I = \dfrac{a^2+b^2}{a^3b^3}\int_{0}^{\infty}\dfrac{1}{1+t^2}\, dt = \dfrac{\pi}{2}\dfrac{a^2+b^2}{a^3b^3}
$$
and
$$
I = \dfrac{\pi}{4}\dfrac{a^2+b^2}{a^3b^3}.
$$
</s>