$\int^0_{-\pi} {i e^{2it}\sin(2t)}$ by parts?

Question

By integrating by parts the following:

$$\int^0_{-\pi} {i e^{2it}\sin(2t)}dt$$

with $$u=\sin(2t); v'=ie^{2it}$$ $$u'=2\cos(2t);v=\frac{1}{2}e^{2it}$$ I get:

$$\int^0_{-\pi} {ie^{2it}\sin(2t)}dt = \frac{1}{2} [e^{2it}\sin(2t)]_{-\pi}^0 - \int^0_{-\pi} {e^{2it}\cos(2t)}dt$$

The first term is equal to $0$. I integrate the second term the same way, with $$u=\cos(2t); v'=e^{2it}$$ $$u'=-2\sin(2t);v=-i\frac{1}{2}e^{2it}$$

$$\int^0_{-\pi} {ie^{2it}\sin(2t)}dt = i\frac{1}{2} [e^{2it}\cos(2t)]_{-\pi}^0 + \int^0_{-\pi} {ie^{2it}\sin(2t)}dt$$

I cannot proceed from here. Where did I go wrong, and how do I solve this?

Entering the integral in Wolfram Alpha gives me $-\frac{\pi}{2}$.

Answer 1

Hint: Writing $$\int^0_{-\pi} {i e^{2it}\color{red}{\sin(2t)}}dt=\int^0_{-\pi} {i e^{2it} \color{red}{\dfrac{e^{2it}-e^{-2it}}{2i}}}dt$$ is easier to work.

Answer 2

You don't need to integrate by parts. Just write $e^{2it}=\cos 2t + i\sin 2t$, then use $\cos 2t\sin 2t=\frac12 \sin 4t$ and $\sin^2 2t=\frac12(1-\cos 4t)$.

Edited to add: Nosrati's answer is even simpler.

Answer 3

Hint:

By parts,

$$\int e^{i2t}\sin 2t\,dt=\frac1{2i}e^{i2t}\sin 2t-\frac1i\int e^{i2t}\cos 2t\,dt \\=\frac1{2i}e^{i2t}\sin 2t-\frac1{2i^2}e^{i2t}\cos 2t+\frac1{i^2}\int e^{i2t}\sin 2t\,dt.$$