Integration of 1/(a^2sin^2(x)+b^2cos^2(x))^2)

[s0ft]

How do I integrate:
\int\dfrac{dx}{(a^2sin^2(x)+b^2cos^2(x))^2}
Multiplying and dividing by sec^4(x) doesn't work, neither does substituting tan.
Any pointers would be appreciated.

 

[Simon Bridge]

Use other trig identities to simplify the expression.

$$a^2\sin^2x + b^2\cos^2x = \cos^2x[a^2\sec^2x +(b^2-a^2)]\\ \qquad =a^2+(b^2-a^2)\cos^2x\\ \qquad = (a^2-b^2)\sin^2x + b^2$$

Also, consider similar triangles with the same angle x but different hypotenuses ... one has hypotenuse $a^2$ and the other $b^2$

Can you sub $u^2=a^2\sin^2x+b^2\cos^2x$

It's basically a matter of working through lots of things until you get a feel for the relationships.

 

[s0ft]

Thanks Simon, (a^2−b^2)sin^2(x)+b^2 did help move forward but now I'm stuck at \int \dfrac{1+t^2}{(a^2t^2+b^2)^2} dt And though I may be able to continue on from here trying this and that, isn't there a shorter way?

 

[s0ft]

The actual problem from where this stemmed out was this btw:
\int_0^\pi\dfrac{x dx}{(a^2sin^2(x)+b^2cos^2(x))^2}
If it could've been solved without of all this, please tell me how.

 

[Simon Bridge]

Thanks Simon, (a^2−b^2)sin^2(x)+b^2 did help move forward but now I'm stuck at \int \dfrac{1+t^2}{(a^2t^2+b^2)^2} dt And though I may be able to continue on from here trying this and that, isn't there a shorter way?

Have you tried partial fractions?