I'm having trouble evaluating the integral
$$\int \frac{\ln(\sin x)}{\sin^2 x}\,\mathrm dx.$$
I tried $u$-substitution and integration by parts but they didn't work.
Answer
Here is a start. using integration by parts,
$$ \int u dv = u v - \int v du .$$
Let
$$ u=\ln(\sin(x)) \implies u'=\frac{\cos(x)}{\sin(x)}=\cot(x),\quad v=\int \frac{dx}{\sin^2 x}=-\cot(x). $$
Can you finish it now?
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