Based on beta function, show that $\int_{-\pi/2}^{\pi/2} (\cos{\theta})^{n-1} d{\theta} = β(n/2, 1/2)$ I am trying to solve this problem but seems that I am stuck. In order to show that $\int_{-\pi/2}^{\pi/2} (\cos{\theta})^{n-1} d{\theta} = β(n/2, 1/2)$.
I understand that beta function follows β(a, b)= $\int_0^1 x^{a-1}(1-x)^{b-1} dx $, and it can also be expressed as β(a, b)= Γ(a)Γ(b)/Γ(a+b).
I have tried to solve the problem from both ways, I got Γ(1/2)=$\sqrt{\pi}$, and also tried the option of letting $\sqrt{x}=cos{\theta}$. But I can't process further to find the solution.
So please help, thank you.
 A: Put $x=\cos \theta$ then $dx = -\sin \theta d \theta = -(1-x^2)^{\frac 12}d \theta$. However, before doing this we must ensure that the substitution is $1-1$ , so we use the fact that $\cos$ is an even function, to write $\int_{- \frac \pi 2}^{\frac \pi 2} (\cos \theta)^n d \theta = 2\int_{0}^{\frac \pi 2} (\cos \theta)^n d \theta $. Therefore, we get :
$$
\int_{0}^{\frac \pi 2} (\cos \theta)^n d \theta = \int_0^1 x^n(1-x^2)^{- \frac 12} dx
$$
now put $y = x^2$, so that $dy = 2xdx $ so $dx = \frac{dy}{2 \sqrt y}$. Now we get :
$$
\int_0^1 x^n(1-x^2)^{- \frac 12} dx = \frac 12\int_0^1 y^{\frac n2} (1-y)^{- \frac 12} y^{- \frac 12}dy\\ = 
\frac 12\int_0^1 y^{\frac {n+1}2 - 1} (1-y)^{- \frac 12}dy = \frac{\beta(\frac{n+1}{2},\frac 12)}{2}
$$
this $2$ cancels out with the $2$ we got from the even break up at the start, leading to the final answer being $$
\int_{- \frac \pi 2}^{\frac \pi 2} (\cos \theta)^n d \theta = \beta\left(\frac{n+1}{2},\frac 12\right)
$$
Your problem is solved by changing $n$ to $n-1$ in this argument. Also, since you have apparently attempted this approach but failed, it is possible that some of the steps need to be elaborated further, so kindly let me know.
