1. ## Help with integral

Hi, I'm having some trouble with the following integral:

$\displaystyle i \int sin(t) e^{it} dt$

Making $\displaystyle u=sin(t)$, then $\displaystyle du=cos(t)$
And if $\displaystyle dv=e^{it}$, then $\displaystyle v=-ie^{it}$

So

$\displaystyle i \int sin(t) e^{it} dt= i(-ie^{it}sin(t) - \int -ie^{it}cos(t) dt)= =e^{it}sin(t) -\int e^{it}cos(t) dt$

Now, for the integral of $\displaystyle e^{it}cos(t) dt$:

$\displaystyle u=cos(t), du=-sin(t)$
$\displaystyle dv=e^{it}, v=-ie^{it}$

So
$\displaystyle \int e^{it}cos(t)dt= -ie^{it}cos(t)- \int -ie^{it}(-sin(t))dt= -ie^{it}cos(t)- i\int e^{it}sin(t)dt$

So
$\displaystyle i \int sin(t) e^{it} dt= e^{it}sin(t) +ie^{it}cos(t)+ i\int e^{it}sin(t)dt$

So I guess integration by parts won't do it here. How can I solve it?

2. ## Re: Help with integral

convert i*e(it) to i*(cos(t) + i*sin(t)

If you want to use by parts, set i*e(it)dt = du and sin(t)

3. ## Re: Help with integral

Originally Posted by votan
convert i*e(it) to i*(cos(t) + i*sin(t)

If you want to use by parts, set i*e(it)dt = du and sin(t)
Thanks, I'll try that. Is there other way besides by parts?

Edit: Just figured it's better to express sin t as (1/2)(-ieit+ie-it) and the integral becomes so much easier.