Originally Posted by

**adkinsjr** I'm stumped by the following:

$\displaystyle I=\int_{0}^{\frac{T}{2}}sin(\frac{2\pi t}{T}-\alpha)dt$

I thought this would be pretty simple, I set $\displaystyle u=\frac{2\pi t}{T}-\alpha $.

Why not directly? $\displaystyle \int\limits_0^{\frac{T}{2}}\sin\!\!\left(\frac{2\p i t}{T}-\alpha\right)dt=\frac{T}{2\pi}\,\int\limits_0^{\fr ac{T}{2}}\left(\frac{2\pi}{T}\,dt\right)\sin\!\!\l eft(\frac{2\pi t}{T}-\alpha\right)$ $\displaystyle =-\frac{T}{2\pi}\cos\!\!\left(\frac{2\pi t}{T}-\alpha\right)_0^{T\slash 2}$ $\displaystyle =-\frac{T}{2\pi}\left[\cos(\pi-\alpha)-\cos\alpha\right]$ $\displaystyle =\frac{T}{\pi}\cos\alpha$

I think this is simpler than substitution and stuff.

Tonio

$\displaystyle \frac{du}{dt}=\frac{2\pi}{t}$

$\displaystyle dt=\frac{tdu}{2\pi}$

So now I can write:

$\displaystyle I=\frac{1}{2\pi}\int_{0}^{\frac{T}{2}}tsin(u)du$

$\displaystyle t=\frac{T(u+\alpha)}{2\pi}$

$\displaystyle I=\frac{1}{2\pi}\int_{0}^{\frac{T}{2}}\frac{T(u+\a lpha)}{2\pi}sin(u)du$

$\displaystyle =\frac{T}{4\pi^2}\int_{0}^\frac{T}{2}(u+\alpha)sin (u)du$

$\displaystyle =\frac{T}{4\pi^2}\left(\int_{0}^{\frac{T}{2}}usin( u)du+\alpha \int_{0}^{\frac{T}{2}}sin(u)du\right)$

[color=red]I know how to proceed from here using elementary methods of integration, but it seems as if I have made this more complicated than it should have been. I'm really bad at integration, so if anyone could point out a more obvious solution, or where I went wrong, that would be great.

Thanks