# How do I find the integral intcos(x)ln(sin(x))dx ?

Jul 30, 2014

$= \ln \left(\sin \left(x\right)\right) \left(\sin \left(x\right) - 1\right) + c$, where $c$ is a constant

Explanation

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

let's $\sin \left(x\right) = t$, $\implies$ $\cos \left(x\right) \mathrm{dx} = \mathrm{dt}$

$= \int \ln \left(t\right) \cdot 1 \mathrm{dt}$

Using Integration by Parts,

$\int \left(I\right) \left(I I\right) \mathrm{dx} = \left(I\right) \int \left(I I\right) \mathrm{dx} - \int \left(\left(I\right) ' \int \left(I I\right) \mathrm{dx}\right) \mathrm{dx}$

where $\left(I\right)$ and $\left(I I\right)$ are functions of $x$, and $\left(I\right)$ represents which will be differentiated and $\left(I I\right)$ will be integrated subsequently in the above formula

Similarly following for the problem,

$= \ln \left(t\right) \int 1 \cdot \mathrm{dt} - \int \left(\left(\ln t\right) ' \int \mathrm{dt}\right) \mathrm{dt}$

$= t \ln \left(t\right) - \int \frac{1}{t} \cdot t \mathrm{dt}$

$= t \ln \left(t\right) - t + c$, where $c$ is a constant

$= t \left(\ln \left(t\right) - 1\right) + c$, where $c$ is a constant

$= \sin \left(x\right) \left(\ln \sin \left(x\right) - 1\right) + c$, where $c$ is a constant