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 eiθ = cosθ + isinθ, sine can be expressed as sinθ = (eiθ- e-iθ) / 2i. This can make the integration easier as the indefinite integral of ekx = (1/k)ekx and the indefinite integral of e-kx = (-1/k)e-kx
Thus ∫sinx dx = ∫(eix- e-ix) / 2i dx = (1/2i)[ ∫eixdx - ∫e-ix dx] = (1/2i)[eix/i + e-ix/i] + C = [-(eix + e-ix) / 2] + C.
Now just as sine can be expressed using complex numbers so can cosine such that cosθ = (eiθ + e-iθ) / 2.
Thus ∫sinx dx = [-(eix + e-ix) / 2] + C = - cosx + C
LC
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