Alternative Derivation of sin integral

saybrook1
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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
 
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saybrook1 said:

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##?
 
saybrook1 said:

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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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