Computing the value of $\int_0^\frac\pi2\frac{\sin^{2m-1}\theta\cos^{2n-1}\theta}{(a\sin^2\theta+b\cos^2\theta)^{m+n}}\,d\theta$ I have been trying to simplify the following integral that is given to prove the following.$$\int_0^\frac{\pi}{2}\frac{\sin^{2m-1}\theta \cos^{2n-1}\theta}{(a\sin^2\theta+b\cos^2\theta)^{m+n}}\,d\theta=\frac{1}{2}\frac{\Gamma(m)\Gamma(n)}{a^mb^n\Gamma(m+n)}.$$
How can I do the substitution here in the denominator so as to proceed and get the desired gamma function(or simply the beta function) on the R.H.S?
I have tried assuming $a\sin^2\theta=\sin^2t$ and the same $b\cos^2\theta=\cos^2t$ in the denominator and then doing the rest of the calculations but at last, ended up with a total mess and couldn't proceed further. What can I do to get the desired result?
All I know here is that $$\mathrm{B}(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=2\int_0^\frac{\pi}{2}\sin^{2m-1}\theta\cos^{2n-1}\theta\,d\theta$$
Please guide me.
Thanks in advance.

The first thing I should do is to let $x=\tan(\theta)$ which makes
$$\int_0^\frac{\pi}{2}\frac{\sin^{2m-1}\theta \cos^{2n-1}\theta}{(a\sin^2\theta+b\cos^2\theta)^{m+n}}d\theta=\int_0^\infty x^{2 m-1} \left(a x^2+b\right)^{-(m+n)}\,dx$$If you know about hypergeometric functions, the problem is simple since
$$\int x^{2 m-1} \left(a x^2+b\right)^{-(m+n)}\,dx=\frac {x^{2m}}{2m b^{m+n}} \, _2F_1\left(m,m+n;m+1;-\frac{a x^2}{b}\right)$$ Otherwise, let $x=\sqrt y$ and $\beta=\frac a b$ to get
$$\frac 12 b^{-(m+n)}\int_0^\infty  y^{m+1} (1+\beta y)^{-(m+n)}\,dy=\frac{\beta ^{-m} \Gamma (m) \Gamma (n) b^{-(m+n)}}{2 \Gamma (m+n)}$$ Replace $\beta$ by $\frac a b$ and you are done.
For the last integral, have a look at formula $3.194.3$  in "Table of Integrals, Series, and Products"
(seventh edition) by I.S. Gradshteyn and I.M. Ryzhik for the expression in terms of beta fuctions.
