Solving Integration Problem: \int_0^a x Sin^2\left(\frac{nx\pi}{a}\right) dx

[dangsy]

Homework Statement

\int_0^a x Sin^2\left(\frac{nx\pi}{a}\right) dx

The Attempt at a Solution

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b f'(x) g(x)\, dx\

\left| x \left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right] \right|^{a}_{0}- \int_0^a \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}

a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \int_0^a \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}

a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a} \int_0^a nx\pi-\frac{Sin\left(2nx\pi\right)}{2}

a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ \int_0^a nx\pi-\int_0^a\frac{Sin\left(2nx\pi\right)}{2}\right]

a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi\int_0^a x- \frac{1}{2}\int_0^a Sin\left(2nx\pi\right)\right]


a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi \left | x\right |^{a}_{0} - \frac{1}{2} \left| Sin\left(2nx\pi\right)\right|^{a}_{0}\right]

a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi a - \frac{1}{2} Sin\left(2na\pi\right)\right]

a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \left[ \frac{n\pi}{2} - \frac{1}{4a} Sin\left(2na\pi\right)\right]

Sin\left(2n\pi\right) = 0
Sin\left(2na\pi\right) = 0

a \left[ \frac{n\pi}{2}\right] - \left[ \frac{n\pi}{2} \right]

I'm kinda stuck here...

I know the answer is \frac{a}{2}

but I'm not sure how to get to it...maybe an integration mistake?

 

[sutupidmath]

first of all you are claiming that

\frac{d}{dx}\left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right]=Sin^2\left(\frac{nx\pi}{a}\right) which is not true

 

[ad absurdum]

You need to use the identity \cos(2x) = 1 - 2\sin^2(x) to turn \sin^2(x) into a sum of things you know how to integrate (either directly or using integration by parts).