# The beta function integration

• Aug 31st 2010, 05:13 AM
The beta function integration
the integral from 0 to pi/4 of (cot2theta)^1/3

This is suppose to somehow transform into the form of cos^(a)theta.sin^(b)theta
where a and b are constants

but i've tried so many times to put it in this form but the 1/3 is just making things difficult for me
• Aug 31st 2010, 11:11 AM
Opalg
Quote:

The definition of $\displaystyle B(x,y)$ is $\displaystyle \displaystyle B(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}dt$, the limits being from t=0 to t=1. The given integral goes from $\displaystyle \theta=0$ to $\displaystyle \theta=\pi/4$. So you need a change of variable which achieves that change of limits. I suggest making the substitution $\displaystyle t = \sin^2(2\theta)$, which has that effect. Then $\displaystyle dt = 4\sin(2\theta)\cos(2\theta)\,d\theta = 4t^{1/2}(1-t)^{1/2}d\theta$, and the integral becomes
\displaystyle \begin{aligned}\displaystyle\int_0^{\pi/4}\frac{(\cos(2\theta))^{1/3}}{(\sin(2\theta))^{1/3}}d\theta &= \int_0^1\frac{(1-t)^{1/6}}{t^{1/6}}\,\frac{dt}{4t^{1/2}(1-t)^{1/2}} \\ &= \frac14\int_0^1t^{-2/3}(1-t)^{-1/3}\,dt \\ &=\tfrac14 B\bigl(\tfrac13,\tfrac23\bigr).\end{aligned}