Prove $\int i\cos(\theta) \ d\theta$ = $ ( e^{i\theta} - e^{-i\theta}) /2$ Could $I\cdot\sin(\theta) + c$ also be a solution?
$e$ = Euler's constant 2.7...
$i$ = complex number or number of a second kind
 A: Euler's formula is a good hint. It does not really prove the problem.
You are better served with another definition.
$\cos(\theta)=\frac{\exp(\mathbb i \theta)+\exp(-\mathbb i \theta)}{2}$
You may easily verify that the given identity holds at least for $\theta=0$.
Where the numerator adds up for 2 and the identity is $1=1$ and so trivial.
The identity follows if the pair of $\{\sin(x), \cos(x)\}$ is via a matrix represented in the function basis $\{\exp(\mathbb i x), \exp(-\mathbb i x)\}$. This can be done without using a matrix just with coefficients. $x$ is real.
The coefficients can be calculated by using the Euler formula for both sine and cosine. Do not use the imaginary unit in the trigonometric functions under examination in this case.
If You integrate the exponential function just the constant coefficient has to be divided if it is not present. That is simply Your identity for real $\theta$.
The right-hand side of Your equation can be simplified further. If You divide by $\mathbb i$ then this is the usual sinus.
$\sin(\theta)=\frac{\exp(\mathbb i \theta)-\exp(-\mathbb i \theta)}{2 \mathbb i}$
With this identity or equation, the set of identities is complete. It is school knowledge that the sine and cosine can be differentiated to give each other and integrated. There is just a change of sign. And that is what is behind Your equation.
You ask for
$\int \cos (x) dx= sin(x)$
without a sign change. You just have to work around the representation change I described in detail to write it the required way.  Have luck.
A: sin instead of cos
Use Euler formula to directly get
$$ \int i\sin \theta \cdot d\theta  =  (e^{i\theta}−e^{−i\theta})/2.$$
