Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$

In this case, the following trick also works: Dividing both the numerator and the denominator by $\cos^4 x$, we can use the substitution $ t = \tan x$ to obtain

\begin{align*} \int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2} &= \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x)^2} \sec^2 x \, dx \\ &= \int_{0}^{\infty} \frac{1 + t^2}{(a^2 + b^2 t^2)^2} \, dt \\ &= \frac{1}{a^2}\int_{0}^{\infty} \left( \frac{1}{a^2 + b^2 t^2} + \frac{(a^2 - b^2) t^2}{(a^2 + b^2 t^2)^2} \right) \, dt. \end{align*}

The first one can be evaluated as follows: Let $bt = a \tan\varphi$. Then

$$ \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{1}{ab} \int_{0}^{\frac{\pi}{2}} d\varphi = \frac{\pi}{2ab}. $$

For the second one, we perform the integration by parts:

\begin{align*} \int_{0}^{\infty} \frac{t^2}{(a^2 + b^2 t^2)^2} \, dt &= \left[ - \frac{1}{b^2}\frac{1}{a^2 + b^2 t^2} \cdot \frac{t}{2} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{2b^2}\frac{dt}{a^2 + b^2 t^2} \\ &= \frac{1}{2b^2} \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} \\ &= \frac{\pi}{4ab^3}. \end{align*}

Putting together, the answer is

$$ \frac{(a^2 + b^2)\pi}{4(ab)^3}. $$

You want to integrate $$J = \int_0^{\pi/2} \dfrac{dx}{\left(a^2 +(b^2-a^2) \sin^2(x) \right)^2} = \dfrac1{a^4}\int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$$ where $c = \dfrac{b^2-a^2}{a^2}$. We now want to integrate $I = \displaystyle \int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$. From Taylor series, we have $$\dfrac1{(1+cy^2)^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k y^{2k}$$ Hence (swapping integral and infinite summation), we get that $$\int_0^{\pi/2}\dfrac{dx}{(1+c\sin^2(x))^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \int_0^{\pi/2}\sin^{2k}(x)dx$$ From here, we have $$\int_0^{\pi/2} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{\pi}{2^{2k+1}}$$ Hence, $$I = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \dbinom{2k}k \dfrac{\pi}{2^{2k+1}} = \dfrac{\pi}2 \sum_{k=0}^{\infty} (k+1) \dbinom{2k}k \left(-\dfrac{c}4\right)^k$$ Now from Taylor series, we have $$\sum_{k=0}^{\infty} (k+1) \dbinom{2k}k x^k = \dfrac{1+4x\sqrt{1-4x}-2x-\sqrt{1-4x}}{(1-4x)^{3/2}}$$ Hence, $$J = \dfrac{\pi}2 \left(\dfrac{1-c\sqrt{1+c}+c/2-\sqrt{1+c}}{a^4(1+c)^{3/2}} \right)$$ Now plug in the value of $c$ and get the value of the original integral $J$.