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How do I integrate log (sinx)?
Ah, I see. So we'd have log(sin(x)) = log(1/cos x), or -log(cosx). Then we can substitute this into our integral expression to get ∫log(sin(x)) dx = -∫log(cos(x)) dx.
Exactly right! And using the formula for integrating log(x) that we discussed earlier, we can simplify this further to get ∫log(sin(x)) dx = x(log(sin(x)) + log(cos(x))) + C.