TITLE: Finding $\int^{\frac{\pi}{2}}_{0}\ln(\sin x)\cdot \sin xdx$
QUESTION [7 upvotes]: Finding $\displaystyle \int^{\frac{\pi}{2}}_{0}\ln(\sin x)\cdot \sin xdx$

What I try:-> Integration by parts 
assuming $\displaystyle I = \int\ln(\sin x)\cdot \sin xdx = -\ln(\sin x)\cdot \cos x+\int\frac{\cos^2 x}{\sin x}dx$
$\displaystyle I = -\ln(\sin x)\cdot \cos x+\int\frac{1-\sin^2 x}{\sin x}dx$
$ = -\ln(\sin x)\cos x+\ln\bigg(\tan\frac{x}{2}\bigg)-\cos x$
$ \displaystyle \int^{\frac{\pi}{2}}_{0}\ln(\sin x)\cos xdx = \bigg[-\ln(\sin x)\cos x+\ln\bigg(\tan\frac{x}{2}\bigg)-\cos x\bigg]\bigg|^{\frac{\pi}{2}}_{0}=-\ln(0)+\ln(0)$
but answer is $\ln(2/e)$
could some explain me why I have got wrong answer,thanks
also explain me How I solve it using double integral

REPLY [2 votes]: Here is an approach following along lines similar to your own answer. There is however a small subtlety used in the first integration by parts step.
On integrating by parts, we have
$$\int_0^{\frac{\pi}{2}} \sin x \ln (\sin x) \, dx = (1 - \cos x) \ln (\sin x) \Big{|}_0^{\pi/2} - \int_0^{\frac{\pi}{2}} (1 - \cos x) \cdot \frac{\cos x}{\sin x} \, dx.$$
Note the subtlety here. Having chosen $v' = \sin x$ we have used $v = 1 - \cos x$, that is, a non-zero constant of integration has been selected. Doing so means one has zero at the upper and lower limits of integration. 
Continuing, we have
\begin{align}
\int_0^{\frac{\pi}{2}} \sin x \ln (\sin x) \, dx &= \int_0^{\frac{\pi}{2}} \frac{-\cos x + \cos^2 x}{\sin x} \, dx\\
&= \int_0^{\frac{\pi}{2}} \frac{-\cos x + 1 - \sin^2 x}{\sin x} \, dx\\
&= \int_0^{\frac{\pi}{2}} \left [\text{cosec} \, x  - \cot x - \sin x \right ] \, dx\\
&= \left [-\ln (\text{cosec} \,x + \cot x) - \ln (\sin x) + \cos x \right ]_0^{\pi/2}\\
&= \left [-\ln (1 + \cos x) + \cos x \right ]_0^{\pi/2}\\
&= \ln 2 - 1,
\end{align}
as expected.