Easy but conceptual

$\displaystyle \int \sin x \ln \left( \sin x \right) dx= \displaystyle \int \sin x \ln \left(\sqrt{1-\cos^{2} x}\right) dx$

$\frac {d \cos x}{d x } =- \sin x$

$d x =\frac{-d \cos x }{\sin x }$

substitute this to original integral which then becomes:

$-\ln\left( \sqrt{1-\cos^{2}x}\right) d\cos x$

antiderivative of this form is same as $\displaystyle \int \frac{-1}{2} \ln\left( 1-m^{2} \right) dm$

$\displaystyle \int \frac{-1}{2} \ln\left(\left( 1-m\right)\left( 1+m\right) \right)dm$

$\displaystyle \int \frac{-1}{2} \ln \left( 1-m\right) dm + \frac{-1}{2}\ln\left( 1+m\right) dm$

$\frac{1}{2} \left(\left( 1-m\right)\ln(1-m) +m - \left(1+m\right)\ln\left(1+m\right) +m \right)$

here m serves as $\cos x$ so the limits become $\cos \frac{\pi}{2} and \cos0$

evaluating it :

$\frac{1}{2} \left(\left( 1-0\right)\ln(1-0) +0 - \left(1+0\right)\ln\left(1+0\right) +0 \right)$ - $\frac{1}{2} \left(\left( 1-1\right)\ln(1-1) +1 - \left(1+1\right)\ln\left(1+1\right) +1 \right)$

$\frac{1}{2} \left(\left( 1\right)\ln(1) - \left(1\right)\ln\left(1\right) \right)$ - $\frac{1}{2} \left(\left( 0\right)\ln(0) +1 - \left(2\right)\ln\left(2\right) +1 \right)$

$\ln 2 - 1= \ln 2 -\ln e =\ln\frac{2}{e}$

First, we compute the indefinite integral: $\begin{aligned} \int\sin x\ln(\sin x)dx&=-\cos x\ln(\sin x)+\int\cos x\frac{\cos x}{\sin x}dx\\ &=-\cos x\ln(\sin x)+\int\frac{\cos^2x}{\sin x}dx\\ &=-\cos x\ln(\sin x)+\int\frac{1-\sin^2x}{\sin x}dx\\ &=-\cos x\ln(\sin x)+\int(\csc x-\sin x)dx\\ &=-\cos x\ln(\sin x)-\ln(\csc x+\cot x)+\cos x+C\\ &=-\cos x\ln(\sin x)+\ln(\sin x)-\ln(1+\cos x)+\cos x+C\\ &=(1-\cos x)\ln(\sin x)-\ln(1+\cos x)+\cos x+C \end{aligned}$ This is defined at $x=\frac{\pi}2$ , but not at $x=0$ , so we have to take a limit. $\begin{aligned} \int_0^{\frac{\pi}2} \sin x\ln(\sin x)dx&=\big((1-0)\ln(1)-\ln(1+0)+0\big)-\lim_{x\to0}\big((1-\cos x)\ln(\sin x)-\ln(1+\cos x)+\cos x\big)\\ &=0+\lim_{x\to0}\left(\frac{\ln(\sin x)}{1/(\cos x-1)}\right)+\ln(1+1)-1\\ &=\lim_{x\to0}\left(\frac{\sin x/\cos x}{\sin x/(\cos x-1)^2}\right)+\ln(2)-1\qquad\text{by L'Hôpital's rule}\\ &=\lim_{x\to0}\left(\frac{(\cos x-1)^2}{\cos x}\right)+\ln(2)-1\\ &=\frac{(1-1)^2}1+\ln(2)-1\\ &=\ln(2)-1 \end{aligned}$

$\begin{aligned} \int_0^{\frac12\pi}\sin x \ln(\sin x)\,dx & = \; \frac12 \frac{d}{du} \int_0^{\frac12\pi} \sin^{2u-1}x\,dx \; = \; \frac14 \frac{d}{du}B(u,\tfrac12) \Big|_{u=1} \\ & = \; \frac14B(u,\tfrac12)\big\{\psi(u) - \psi(u+\tfrac12)\big\}\Big|_{u=1} \; = \; \frac14B(1,\tfrac12)\big\{\psi(1)-\psi(\tfrac32)\big\} \; = \; \ln2 - 1 \; = \; \ln\left(\frac{2}{e}\right) \end{aligned}$ making the answer $\boxed{2}$ .

Guys can you post your solutions how you did it?