Evaluation of an integral.

Question

$$I_1=\int \frac{a^2\sin^2(x)+b^2\cos^2(x)}{a^4\sin^2(x)+b^4\cos^2(x)}dx$$

I tried writing it as $$\int \frac{a^2+\cos^2(x)(b^2-a^2)}{a^4+\cos^2(x)(b^4-a^4)}dx$$

But I don't know how to proceed. What's the procedure to evaluate this integral?

Answer 1

$$\frac{a^2u^2+b^2}{a^4u^2+b^4}=\frac{(a^2+b^2)(a^2u^2+b^2)}{(a^2+b^2)(a^4u^2+b^4)}=\cdots =\frac{1}{a^2+b^2}+\frac{a^2(u^2+1)}{(a^2+b^2)(\frac{a^4}{b^2}u^2+b^2)}.$$ So if $u=\tan{x}$ and integrate the previous you get $$\frac{x}{a^2+b^2}+\frac{a^2}{a^2+b^2}\int \frac{\tan^2{x}+1}{\frac{a^4}{b^2}\tan^2{x}+b^2} dx=$$ $$\frac{x}{a^2+b^2} +\frac{1}{a^2+b^2}\arctan(\frac{a^2}{b^2}\tan{x})+c$$