Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^{2}(a+b)}$.
But I am not getting the answer.

Integrating by parts,
$$\int\frac{x\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
$$=x\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx-\int\left[\frac{dx}{dx}\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx\right]dx$$
Now $a^2\cos^2x+b^2\sin^2x=u\implies2(b^2-a^2)\sin x\cos x\ dx=du$
For $\displaystyle\int\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\int\frac{\sec^2x\ dx}{a^2+b^2\tan^2x},$
set $b\tan x=a\tan y$
