Problem : $\int^{\pi/2}_0 \sin\theta \ln(\sin\theta)d\theta$ is equal to .... Problem : 
$\int^{\pi/2}_0 \sin\theta \ln(\sin\theta)d\theta$  is equal to 
(a) $\ln\frac{2}{e}$ 
(b) $-\ln\frac{2}{e}$  
(c) $\ln 2$ 
(d) $1$ 
My approach  : 
Let $I =\int^{\pi/2}_0 \sin\theta \ln(\sin\theta)d\theta$......(1)
Now using $\int^{\pi/2}_0 f(x)dx = \int^{\pi/2}_0 f(\pi/2-x)dx$
$\Rightarrow I = \int^{\pi/2}_0 \cos\theta \ln(\cos\theta)d\theta$.......(2)
Adding (1) and (2) we get 
$\Rightarrow 2I = \int^{\pi/2}_0 \ln\{(\sin\theta)^{\sin\theta} (\cos\theta)^{\cos\theta}\}d\theta$ [ Using property $m\log n = \log n^m$]
Now please suggest how to proceed further or some other alternate method , will be of great help thanks. 
 A: Even without doing the computations, just by looking at the options, we can conclude the answer should be $\ln(2/e)$, since all the other options are positive, whereas the integrand is always negative from $0$ to $\pi/2$, since $\ln(\sin(t)) \le 0$. A detailed computation is also given below.

We have
$$\sin(t) \ln(\sin(t)) = \dfrac12 \sin(t) \ln(\sin^2(t)) = \dfrac12 \sin(t) \ln(1-\cos^2(t))$$
Setting $\cos(t) = x$, we have $-\sin(t)dt = dx$. Hence, the integral becomes
\begin{align}
I & = \int_0^{\pi/2}\sin(t) \ln(\sin(t))dt = \dfrac12\int_0^1 \ln(1-x^2)dx = \dfrac12\int_0^1 \ln(1+x) dx + \dfrac12\int_0^1 \ln(1-x)dx\\
& = \ln(2/e)
\end{align}
A: You may try integration by parts by substituting $x=\sin \theta$:
$\int_0^1\dfrac{x}{\sqrt{1-x^2}}\ln(x)\operatorname{dx}=[\ln(x)\int \dfrac{x}{\sqrt{1-x^2}}\operatorname{dx}]_0^1-\int _0^1 \dfrac{1}{x}(\int \dfrac{x}{\sqrt{1-x^2}} \operatorname{dx})\operatorname{dx}$.
Check that $\int \dfrac{x}{\sqrt{1-x^2}} \operatorname{dx}=-\sqrt{1-x^2}$
A: Have when $t\in(0,\frac\pi2):$
$$\int\sin t\ln\sin t\,dt = -\int\ln\sin t\,d\cos t $$(by parts:)$$= -\cos t\ln\sin t + \int\frac{\cos^2 t}{\sin t} dt = -\cos t\ln\sin t +\int\frac{\cos^2t-1+1}{1-\cos^2t}\sin t dt $$$$ = -\dfrac12\cos t\ln(1-\cos^2t) - \int \left(-1+\dfrac12\dfrac1{1-\cos t}+\dfrac12\dfrac1{1+\cos t}\right)d\cos t $$$$= -\dfrac12\cos t(\ln(1-\cos t) + \ln(1+\cos t)) + \cos t + \dfrac12\ln(1-\cos t) - \dfrac12\ln(1+\cos t)+C $$$$ = \dfrac12(1-\cos t)(\ln(1-\cos t) + \ln(1+\cos t)) - \ln(1+\cos t) + \cos t + C $$$$= (1-\cos t)\ln\sin t - \ln(1+\cos t) + \cos t +C.$$
So:
$$\int_0^\frac\pi2\sin t\ln\sin t\,dt = \left((1-\cos t)\ln\sin t - \ln(1+\cos t) + \cos t\right)\biggr|_0^\frac\pi2 = \ln 2 -1 = \ln\dfrac2e.$$
