# How do you integrate int ln(sint)cost by integration by parts method?

Jan 15, 2017

$\int \ln \left(\sin t\right) \cos t \mathrm{dt} = \sin t \left(\ln \left(\sin t\right) - 1\right) + C$

#### Explanation:

First let $w = \sin t$, implying that $\mathrm{dw} = \cos t \mathrm{dt}$. This will help simplify the problem before we attempt integration by parts. Through this substitution:

$I = \int \ln \left(\sin t\right) \cos t \mathrm{dt} = \int \ln \left(w\right) \mathrm{dw}$

Now we should apply integration by parts. Let:

$\left\{\begin{matrix}u = \ln \left(w\right) \text{ "=>" "du=1/wdw \\ dv=dw" "=>" } v = w\end{matrix}\right.$

Thus:

$I = w \ln \left(w\right) - \int w \frac{1}{w} \mathrm{dw} = w \ln \left(w\right) - \int \mathrm{dw}$

Integrating and factoring:

$I = w \ln \left(w\right) - w = w \left(\ln \left(w\right) - 1\right)$

From $w = \sin t$:

$I = \sin t \left(\ln \left(\sin t\right) - 1\right) + C$