integrating complex functions I need to integrate $$\int\limits_0^{\frac{\pi}{2}} e^{t+it} dt$$. This is equal to $\int\limits_0^{\frac{\pi}{2}} e^t \cdot e^{it} dt$ However I'm not sure if I can substitute $e^{it}=\cos(t)+i\sin(t)$ and then solve the integral.  
 A: Why not just using the primitive method?
\begin{align*}
\int_{0}^{\pi/2}e^{t+it}dt&=\int_{0}^{\pi/2}e^{(1+i)t}dt\\
&=\dfrac{1}{1+i}e^{(1+i)t}\bigg|_{t=0}^{t=\pi/2}\\
&=\dfrac{1}{1+i}(e^{(1+i)\pi/2}-1)\\
&=\dfrac{1}{1+i}(e^{\pi/2}i-1).
\end{align*}
A: The complex number is just like any constant:
$$I=\int_0^{\frac{\pi}{2}} e^{t+it} dt=\int_0^{\frac{\pi}{2}} e^{t(1+i)} dt=\int_0^{\frac{\pi}{2}} e^{Ct} dt$$
A: it can be done the way you suggested, however. first evaluate these two integrals:
$$
I_1 = \int_0^\frac{\pi}2 e^t \cos t \, dt = \bigg[e^t \cos t \bigg]_0^\frac{\pi}2 + \int_0^\frac{\pi}2 e^t \sin t \, dt = -1 + I_2
$$
where
$$
I_2 = \int_0^\frac{\pi}2 e^t \sin t \, dt = \bigg[e^t \sin t \bigg]_0^\frac{\pi}2 - \int_0^\frac{\pi}2 e^t \cos t \,dt = e^{\frac{\pi}2} - I_1
$$
so now we have
$$
2I_1 = e^{\frac{\pi}2} -1
$$
and
$$
2I_2 = e^{\frac{\pi}2} +1
$$
now all you need to do is to compute:
$$
I_1+iI_2 = \frac{(1+i)}2  e^{\frac{\pi}2} + \frac{i-1}2
$$
this differs slightly from the value given in a previous answer, because user284331 has slipped in the erroneous assumption that $e^{i\frac{\pi}2} = 1$, whereas that should, of course be $i$. (i make happy assumptions like this all the time ;-)
A: Because $e^{it} = \cos(t) + i \sin(t)$ is true, you can substitute $\cos(t) + i \sin(t)$ anywhere where you have $e^{it}$ and vice versa. 
Thus you do indeed have
$$\int_0^{\frac{\pi}{2}} e^t \cdot e^{it} dt
= \int_0^{\frac{\pi}{2}} e^t \cdot \left( \cos(t) + i \sin(t) \right) dt
$$
