Question

Integrate : ∫log(sinx)cosxdx

Answer 1

$\textbf{Answer :}$

Now, ∫ log(sinx) cosx dx

Let, sinx = z

Then, cosx dx = dz

So, ∫ log(sinx) cosx dx

= ∫ logz dz

= logz ∫dz - ∫ {d/dz (z) × ∫dz}dz

= z logz - ∫ z dz

= z logz - z²/2 + c, where c is integral constant

= sinx log(sinx) - (sin²x)/2 + c

Formula :

∫ uv dx

= u ∫ v dx - ∫ (du/dx × ∫ v dx) dx

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