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?