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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.

slamdunk
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1 Answers1

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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.

Sarvesh Ravichandran Iyer
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