Integration

**dwsmith** · March 18th 2011, 09:01 PM

$\displaystyle\frac{4L^2}{\pi^2(1+2m)^2}\int_0^L\si n^2\left(\frac{\pi x(1+2m)}{L}\right)dx$

$\displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2} \int_0^L\left(1-\cos\left(\frac{2\pi x(1+2m)}{L}\right)\right)dx$

$\displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2} \left[x-\frac{L}{2\pi(1+2m)}\sin\left(\frac{2\pi x(1+2m)}{L}\right)\right]_0^L=\frac{2L^3}{\pi^2(1+2m)^2}$

Correct?

**CaptainBlack** · March 19th 2011, 01:28 AM

Originally Posted by dwsmith

$\displaystyle\frac{4L^2}{\pi^2(1+2m)^2}\int_0^L\si n^2\left(\frac{\pi x(1+2m)}{L}\right)dx$

$\displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2} \int_0^L\left(1-\cos\left(\frac{2\pi x(1+2m)}{L}\right)\right)dx$

$\displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2} \left[x-\frac{L}{2\pi(1+2m)}\sin\left(\frac{2\pi x(1+2m)}{L}\right)\right]_0^L=\frac{2L^3}{\pi^2(1+2m)^2}$

Correct?

Yes, since the integral is L/2 (assuming m and integer)

CB

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