1. ## Beta Integral

$\int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta$ = $\frac{\beta(m,n)}{2a^mb^n}$

I started from R.H.S.
i.e.

$\frac{\beta(m,n)}{2a^mb^n}$ = $\frac{1}{2a^mb^n} \int_{0}^{1} x^{m-1}(1-x)^{n-1}dx$

and then put $x=sin^2\theta$
but i couldn't reach at the result..

I also tried from LHS, but then also i couldn't.

LHS = $\int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta$

= $\int_{0}^{\frac{\pi}{2}} \frac{({\cos}^{2}\theta)^m ({\sin}^{2}\theta)^n (sin\theta cos\theta)}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta$

= $\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{(1-{\sin}^{2}\theta )^m({\sin}^{2}\theta)^n (2sin\theta cos\theta)}{(a +(b-a)\sin^2 \theta)^{m+n}}d\theta$

then I tried to put $x=sin^2\theta$, but...

so pl help..

2. omg
what is this :/

Scary integral

3. Originally Posted by kjchauhan

$\int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta$ = $\frac{\beta(m,n)}{2a^mb^n}$

I started from R.H.S.
i.e.

$\frac{\beta(m,n)}{2a^mb^n}$ = $\frac{1}{2a^mb^n} \int_{0}^{1} x^{m-1}(1-x)^{n-1}dx$

and then put $x=sin^2\theta$
but i couldn't reach at the result..

I also tried from LHS, but then also i couldn't.

LHS = $\int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta$

= $\int_{0}^{\frac{\pi}{2}} \frac{({\cos}^{2}\theta)^m ({\sin}^{2}\theta)^n (sin\theta cos\theta)}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta$

= $\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{(1-{\sin}^{2}\theta )^m({\sin}^{2}\theta)^n (2sin\theta cos\theta)}{(a +(b-a)\sin^2 \theta)^{m+n}}d\theta$

then I tried to put $x=sin^2\theta$, but...

so pl help..
How much do you know before I type this all up?

4. Hello,

I'd suggest you use this formula of the beta function :

$\beta(m,n)=\int_0^{\pi/2} \cos^{2m-1}\theta \sin^{2n-1}\theta ~d\theta$

5. Substitute

$(b/a) \tan^2 \theta = \tan^2(x)$