Question

If $\int \cot x\cdot \ln \left(\sin x\right)dx=f\left(x\right)+C$, then the number of solution(s) of the equation $f\left(x\right)=0$ in $[0,2\pi ]$, is
(where $C$ is constant of integration)

Answer 1

$I=\int \cot x\cdot \ln \left(\sin x\right)dx$
Putting $\ln \left(\sin x\right)=y\Rightarrow \frac{1}{\sin x}\times \cos xdx=dy$
$\Rightarrow \cot xdx=dy$
$\Rightarrow I=\int ydy=\frac{y^2}{2}+C=\frac{\left(\ln \right(\sin x){)}^2}{2}+C$
$\Rightarrow f\left(x\right)=\frac{\left(\ln \right(\sin x){)}^2}{2}=0$
$\Rightarrow \ln \left(\sin x\right)=0\Rightarrow \sin x=1$
$\Rightarrow x=\frac{\pi }{2}\{\because x\in [0,2\pi \left]\right\}$
So, only one solution​​