I 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...

 

[pasmith]

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...

Substitute u = \sin^2 x, and use the definition <br /> \beta(p,q) = \int_0^1 u^{p-1}(1-u)^{q-1}\,du.

This is a special case of <br /> \int_0^{\pi/2} \cos^n x \sin^m x\,dx = \frac12 \beta\left( \frac{n+1}{2}, \frac{m + 1}{2}\right)

 

[Gavran]

$\begin{align} I_n&=\int_{0}^{\frac\pi2}sin^nx dx\nonumber\\ &=\frac12\int_{0}^{\frac\pi2}\frac {sin^nx}{sinx}\frac1{cosx}2\sin x\cos xdx\nonumber\\ &=\frac12\int_{0}^{\frac\pi2}(sin^2x)^\frac{n-1}2(1-sin^2x)^{-\frac12}2\sin x\cos xdx\nonumber\\ \end{align}$

$\begin{align} &u=sin^2x\nonumber\\ &du=2\sin x\cos x dx\nonumber\\ \end{align}$

$\begin{align} I_n&=\frac12\int_{0}^{1}u^\frac{n-1}2(1-u)^{-\frac12}du\nonumber\\ &=\frac12\beta(\frac{n+1}2,\frac12)\nonumber\\ \end{align}$

This is actually an integration by substitution method which gives a beta function as a result.

The final result is:
$$ \int_{0}^{\pi}sin^nx dx=\beta(\frac{n+1}2,\frac12) $$

 

[chwala]

Substitute u = \sin^2 x, and use the definition <br /> \beta(p,q) = \int_0^1 u^{p-1}(1-u)^{q-1}\,du.

This is a special case of <br /> \int_0^{\pi/2} \cos^n x \sin^m x\,dx = \frac12 \beta\left( \frac{n+1}{2}, \frac{m + 1}{2}\right)

aaaaah @pasmith now i see...

from,

$\int \sin^{2m-1} θ \cos ^{2n-1} θ dθ= \dfrac{1}{2} β (m,n)$ This is the key.

Noting that the Beta function and gamma function are related by,

$β(m,n) = \dfrac {Γ(m) Γ(n)}{Γ(m+n)}$

then in our case,by considering the integral we have,

$\cos^{2n-1} θ= 1$ implying that

$n=\dfrac{1}{2}$

and letting, $\sin^{2m-1} θ = \sin^n θ$
$2m-1=n$
$m=\dfrac{n+1}{2}$

from that point of view the problem can be handled...