# Math Help - Definite Integral

1. ## Definite Integral

I'm stumped by the following:

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

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

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

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

So now I can write:

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

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

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

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

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

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

I'm stumped by the following:

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

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

$\frac{du}{dt}=\frac{2\pi}{\textcolor{red}{T}}$
...

3. Thanks, I didn't catch that. That definitely simplified things. It probably wouldn't have happened if I had used something other than a T as the constant. I got it confused with the variable t.

I'm stumped by the following:

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

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

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

I think this is simpler than substitution and stuff.

Tonio

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

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

So now I can write:

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

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

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

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

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

[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
.