Thread: Integration

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

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

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

Correct?

Originally Posted by dwsmith

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

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

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

Correct?

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

CB