# Alternative Derivation of sin integral

## Homework Statement

Hi guys; I'm just dealing with fourier series and they evaluate integrals such as ∫sin(nπx/L)dx from 0 to L as (L/nπ)[1-(-1)^n]. Can someone please tell me how to get to this conclusion or point me in the direction of a resource that will show me? Additionally I need to solve similar integrals like ∫sin^2(nπx/L)dx and ∫xsin(nπx/L) so I really need to know the technique. Thanks in advance.

## Homework Equations

∫sin(nπx/L)dx from 0 to L = (L/nπ)[1-(-1)^n]

## The Attempt at a Solution

I would traditionally evaluate ∫sin(nπx/L)dx from 0 to L as [L-Lcos(nπ)]/nπ

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Hi guys; I'm just dealing with fourier series and they evaluate integrals such as ∫sin(nπx/L)dx from 0 to L as (L/nπ)[1-(-1)^n]. Can someone please tell me how to get to this conclusion or point me in the direction of a resource that will show me? Additionally I need to solve similar integrals like ∫sin^2(nπx/L)dx and ∫xsin(nπx/L) so I really need to know the technique. Thanks in advance.

## Homework Equations

∫sin(nπx/L)dx from 0 to L = (L/nπ)[1-(-1)^n]

## The Attempt at a Solution

I would traditionally evaluate ∫sin(nπx/L)dx from 0 to L as [L-Lcos(nπ)]/nπ

So you have ##\frac{L}{n\pi}(1-\cos(n\pi))##. How does ##\cos(n\pi)## compare with ##(-1)^n##?

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Hi guys; I'm just dealing with fourier series and they evaluate integrals such as ∫sin(nπx/L)dx from 0 to L as (L/nπ)[1-(-1)^n]. Can someone please tell me how to get to this conclusion or point me in the direction of a resource that will show me? Additionally I need to solve similar integrals like ∫sin^2(nπx/L)dx and ∫xsin(nπx/L) so I really need to know the technique. Thanks in advance.

## Homework Equations

∫sin(nπx/L)dx from 0 to L = (L/nπ)[1-(-1)^n]

## The Attempt at a Solution

I would traditionally evaluate ∫sin(nπx/L)dx from 0 to L as [L-Lcos(nπ)]/nπ
If n is even, then cos(nπ) = 1.

If n is odd, then cos(nπ) = -1.

I see now! Thank you LCKurtz and SammyS, I appreciate the help.