Easy Calculus 3 : Imaginary

$\displaystyle I = \int i^x~ dx \\ \displaystyle I = \dfrac{i^x}{\ln(i)} \text{; since } \int a^x = \dfrac{a^x}{\ln(a)}\\ \displaystyle \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \displaystyle e^{ix} = \cos(x) + i\sin(x) \\ \displaystyle e^{i\dfrac{\pi}{2}} = \cos(\dfrac{\pi}{2}) + i\sin(\dfrac{\pi}{2}) \\ \displaystyle e^{i\dfrac{\pi}{2}} = 0 + i ( 1) = i \\ \displaystyle \ln(e^{i\dfrac{\pi}{2}}) = \ln(i) \\ \displaystyle \ln(i) = i\dfrac{\pi}{2} ~ \ln(e) = i\dfrac{\pi}{2} \\ \displaystyle \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \displaystyle I = \dfrac{i^x}{i\dfrac{\pi}{2}} = \dfrac{2i^x}{i\pi} = \dfrac{2i^x}{\pi} \cdot \dfrac{1}{i} = \dfrac{2i^x}{\pi} \cdot (-i) \\ \displaystyle I = -\dfrac{2i^{x+1}}{\pi} \\ \displaystyle \text{Now, } \\ \displaystyle \int_0^2 i^x~ dx = -\dfrac{2i^{x+1}}{\pi} ~\Big|^2_0 \\ \displaystyle = -\dfrac{2i^{2+1}}{\pi} - ( -\dfrac{2i^{0+1}}{\pi} ) \\ \displaystyle = -\dfrac{2i^{3}}{\pi} - ( -\dfrac{2i^{1}}{\pi} ) \\ \displaystyle = \dfrac{2i}{\pi} + \dfrac{2i}{\pi} \\ \displaystyle = \dfrac{4i}{\pi} \\ \displaystyle = 4\dfrac{i}{\pi} \\ \displaystyle \boxed{\therefore A = 4}$

$i=cos(\pi/2)+ i sin(\pi/2)$

So, $i^x=(cos(\pi/2)+ i sin(\pi/2))^x$

Applying De-Moivre's Theorem:

$i^x=cos(\pi x/2 ) + i sin(\pi x/2)$

Now it is to integrate the answer is $4 i/\pi$

Relevant wiki: Euler's Formula

$\begin{aligned} I & = \int_0^2 i^x \ dx & \small \color{#3D99F6}{\text{By Euler's formula: }e^{i\theta}=\cos \theta + i \sin \theta} \\ & = \int_0^2 e^{\frac {\pi x}2 i} \ dx \\ & = \frac {2 e^{\frac {\pi x}2}i}{\pi i} \bigg|_0^2 \\ & = \frac 2{\pi i} \left(e^{\pi i} - e^0 \right) \\ & = \frac 2{\pi i} \left(-1 - 1 \right) \\ & = \frac {4i}\pi \end{aligned}$

$\implies A = \boxed{4}$