$$ \int \frac{1}{ (a^2 \cos ^2 x + b ^ 2 \sin ^2x) ^2} \ dx $$
So this is the question .
The solution given in book is to divide numerator and denominator by $\cos ^4x$ and then substitute $\tan x = t$ in the resulting integrand.
Other way of doing this was to substitute $b \tan x = a \tan t$.
So I was thinking is not there any other way to solve this as it seems to a complicated problem as the given methods are very lengthy while solving.
Any simpler\shorter method anyone could think of ?
Note that \begin{align} I(a,b)&=\int \frac{1}{ a^2\cos ^2 x + b^2 \sin ^2x}dx\\ &=\int \frac{d(\tan x)}{ a^2+b^2\tan ^2 x } =\frac1{ab}\tan^{-1}\bigg({\frac ba}\tan x \bigg) \end{align} and \begin{align} &\int \frac{1}{ (a^2\cos ^2 x + b^2 \sin ^2x)^2}dx = -\frac12\left(\frac{I’_a}a+\frac{I’_b}b\right)\\ =&\ \frac1{2ab}\left(\frac1{a^2}+\frac1{b^2}\right)\tan^{-1}\left(\frac ba\tan x \right) + \frac12\left(\frac1{a^2}-\frac1{b^2}\right)\frac{\tan x}{b^2\tan^2x+a^2} \end{align}