How do I derive the indefinite integral of sine?

The integral of sine is pretty easy to remember as it is -cos + C. However you need to be able to prove this, without using the integral of cosine. This method uses sines exponential form.

As e^iθ​ = cosθ + isinθ, sine can be expressed as sinθ = (e^iθ​- e^-iθ​) / 2i. This can make the integration easier as the indefinite integral of e^kx = (1/k)e^kx and the indefinite integral of e^-kx = (-1/k)e^-kx

Thus ∫sinx dx = ∫(e^ix- e^-ix) / 2i dx = (1/2i)[ ∫e^ixdx - ∫e^-ix dx] = (1/2i)[e^ix/i + e^-ix​/i] + C =  [-(e^ix​ + e^-ix) / 2]  + C.

Now just as sine can be expressed using complex numbers so can cosine such that cosθ = (e^iθ​ + e^-iθ​) / 2.

Thus  ∫sinx dx = [-(e^ix​ + e^-ix) / 2]  + C = - cosx + C