Alternative Derivation of sin integral

[saybrook1]

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 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]

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.

 

[saybrook1]

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