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.
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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First step: Multiply numerator and denominator by

$\ds{\sec^{2m + 2n}\pars{\theta} =
\sec^{2m - 1}\pars{\theta}\
\sec^{2n - 1}\pars{\theta}\
\sec^{2}\pars{\theta}}$.
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\pi/2}
{\sin^{2m - 1}\pars{\theta}\cos^{2n - 1}\pars{\theta} \over \bracks{a\sin^{2}\pars{\theta} + b\cos^{2}\pars{\theta}}^{m + n}}
\,\dd\theta}
\\[5mm] = &
\int_{0}^{\pi/2}
{\tan^{2m - 1}\pars{\theta} \over
\bracks{a\tan^{2}\pars{\theta} + b}^{m + n}}
\sec^{2}\pars{\theta}\,\dd\theta
\\[5mm] \stackrel{x\ =\ \tan\pars{\theta}}{=}\,\,\,&
\int_{0}^{\infty}{x^{2m - 1} \over
\pars{ax^{2} + b}^{m + n}}\,\dd x
\\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,&
{1 \over 2}\int_{0}^{\infty}{x^{m - 1} \over
\pars{ax + b}^{m + n}}\,\dd x
\\[5mm] \stackrel{ax/b\ \mapsto\ x}{=}\,\,\,&
{1 \over 2a^{m}b^{n}}\
\underbrace{\int_{0}^{\infty}{x^{m - 1} \over \pars{x + 1}^{m + n}}\,\dd x}_{\ds{\on{B}\pars{m,n}}}
\end{align}
See this link.
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\pi/2}
{\sin^{2m - 1}\pars{\theta}\cos^{2n - 1}\pars{\theta} \over \bracks{a\sin^{2}\pars{\theta} + b\cos^{2}\pars{\theta}}^{m + n}}
\,\dd\theta}
\\[5mm] = &\
\bbx{{1 \over 2a^{m}b^{n}}\,{\Gamma\pars{m}\Gamma\pars{n}
\over \Gamma\pars{m + n}}} \\ &
\end{align}
