Integration headache, please help.

Using Euler's formula
\[\int_0^{\pi} \sin t e^{it}dt=\int_0^{\pi} \sin t (\cos t+i \sin t)dt\]\[=\int_0^{\pi} \sin t \cos t dt+i \int_0^{\pi}\sin^2 tdt\]
\[=\int_0^{\pi}\frac{\sin 2t}{2}dt+i\int_0^{\pi}\frac{1-\cos 2t}{2}dt\]
\[=-\frac{\cos 2t}{4} |_0^{\pi}+\frac{i}{2}(t -\frac{\sin 2t }{2})|_0^{\pi}\]
\[=(-\frac{1}{4}+\frac{1}{4})+\frac{i}{2}[(\pi -0)-(0-0)]\]
\[=i\frac{\pi}{2}.\]
Here we have used the identities: 
\[\sin 2t =\sin t \cos t,    \sin^2 t= \frac{1-\cos t}{2}.\]