# How do you integrate int cosxln(sinx) using integration by parts?

Nov 22, 2016

It is the same as $\int \ln t \mathrm{dt}$

#### Explanation:

Let $u = \ln \left(\sin x\right)$ and $\mathrm{dv} = \cos x \mathrm{dx}$.

This makes $\mathrm{du} = \frac{1}{\sin} x \cos x \mathrm{dx}$ and $v = \sin x$

$u v - v \mathrm{du} = \sin x \ln \left(\sin x\right) - \int \sin x \left(\frac{1}{\sin} x \cos x\right) \mathrm{dx}$

$= \sin x \ln \left(\sin x\right) - \int \cos x \mathrm{dx}$

$= \sin x \ln \left(\sin x\right) - \sin x + C$.

Note if you know $\int \ln t \mathrm{dt}$

If you know that $\int \ln t \mathrm{dt} = t \ln t - t + C$, then you can simply substitue $t = \sin x$.