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.

You can use this property : $$\int_a^b f(x)\hspace{1mm}dx = \int_a^b f(a+b-x)\hspace{1mm}dx$$
To prove this property : Substitute $a+b-x = u$
Let us have $$I = \int_0^{\pi/2}\dfrac{x\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (1)$$
After applying the property, you will get  
$$I = \int_0^{\pi/2}\dfrac{(\pi/2-x)\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (2)$$
After adding the two equation, you get
$$I =\dfrac{\pi}{4} \int_0^{\pi/2}\dfrac{\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx $$
Now you will substitute $a^2\cos^2x+b^2\sin^2x = u$
To get $$I = \dfrac{\pi}{8(b^2-a^2)}\int_{a^2}^{b^2} \dfrac{1}{u^2}\hspace{1mm}du$$
I hope you can Integrate from here
