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### 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

1 year ago

Answered by Lucile, an IB Maths tutor with MyTutor

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