Find ##\int_0^π \sin^ n x dx ##

[chwala]

TL;DR Summary

I am looking at this definite integral; i will go slow on it with the whole intention being to understand the steps and how to apply. I will attempt to use the reduction method and i also noted that this can be approached using Beta and Gamma relationship. Its Time to upscale my intellect.

$$\int_0^π \sin^ n x dx = \int_0^π \sin^ {n-1} x\sin x dx$$

Letting $u=\sin^{n-1} x$ and $v^{'}= \sin x$ then,

$$\int_0^π \sin^ {n-1} x\sin x dx = [-\sin^ {n-1} x⋅\cos x]_0^π+ \int_0^π (n-1)\sin^{n-2} x ⋅ \cos^2 x
dx$$

$$=\int_0^π (n-1)\sin^{n-2} x ⋅ \cos^2 x dx $$

$$=\int_0^π (n-1)\sin^{n-2} x ⋅ (1- \sin^2 x) dx $$

Letting
$$I_n = \int_0^π \sin^ n x dx $$

$$I_n = (n-1)\int_0^π \left(\sin^{n-2} x - \sin^n x \right)dx $$

$$I_n = (n-1)(I_{n-2} - I_n)$$

$$nI_n = (n-1)I_{n-2}$$

Now up to this point it is quite clear. The next thing for me to check on is on conditions where $n$ is even or odd. Most importantly to understand the beta and gamma approach that forms the basis of this post.

Cheers Man!

 

[anuttarasammyak]

Your result with obvious
$$I_0=\pi, I_1=2$$
will give $I_n$.

 

[chwala]

Now i need to understand this $$\int_0^{0.5π} \sin^n dx = \dfrac{1}{2} β \left(\dfrac {n+1}{2}, \dfrac{1}{2}\right)$$

Let me check on the literature...