Integrate it! Part- IV

We can take the indefinite integral using integration by parts.

$I=\displaystyle\int \ln (\sin x) \sin x \, \mathrm{d}x$

$=\displaystyle-\ln(\sin x)\cos x + \displaystyle\int\cos x \cot x \, \mathrm{d}x$

Now, the integral on the right side of this equation is

$\to \displaystyle\int \dfrac{\cos^2 x}{\sin x} \, \mathrm{d}x= \displaystyle\int \dfrac{1-\sin^2 x}{\sin x}\, \mathrm{d}x=\displaystyle\int \csc x - \sin x \, \mathrm{d}x=-\ln(\csc x + \cot x) + \cos x$

(disregarding the absolute value bars, since $x\geq 0\,$ .)

So our original indefinite integral is

$I= \displaystyle-\ln(\sin x)\cos x -\ln(\csc x + \cot x) + \cos x$

which after some rearranging and using a trigonometric identity we can write as

$\displaystyle I=(\cos x)(1-\ln(\sin x)) + \ln\left(\tan \frac{x}{2}\right)$ .

Now we evaluate this at the limits of integration.

$\displaystyle I\left(\frac{\pi}{2}\right)=0$ , but we must take a limit (and use L'Hôpital's rule and logarithmic differentiation) to find $\displaystyle I\left(0\right)$ .

$\displaystyle\lim_{x\to 0} \,\, \left[ \cos x - (\cos x)\ln(\sin x) + \ln\left(\tan \frac{x}{2}\right)\right]$

$\displaystyle= 1 + \displaystyle\lim_{x\to 0} \,\, \ln\left( \frac{\tan \frac{x}{2}}{(\sin x)^{\cos x} } \right)$

$\displaystyle= 1 + \ln \left( \displaystyle\lim_{x\to 0} \,\, \frac{\frac{1}{2}\sec^2 \frac{x}{2}}{(\sin x)^{\cos x} \left[(\cos x) (\cot x) - (\sin x) \ln(\sin x)\right] } \right)$ $\displaystyle = 1 + \ln \left( \displaystyle\lim_{x\to 0} \,\, \frac{ \frac{1}{2} \sec^2 \frac{x}{2}} {(\cos^2 x)(\sin x)^{\cos x \,-\,1}-\ln(\sin x)(\sin x)^{\cos x \,+\,1} } \right)$

$\displaystyle = 1+\ln\left(\frac{1}{2}\right)$

Finally, the definite integral in question is

$\displaystyle I\left(\frac{\pi}{2}\right)-I\left(0\right)=\displaystyle 0-\left(1+\ln\left(\frac{1}{2}\right)\right)=\displaystyle \ln (2) -1\approx \boxed{-0.3069}$

$\begin{aligned} I & = \int_0^\frac \pi 2 \sin x \ln (\sin x) \ dx & \small \color{#3D99F6} \text{Using integration by parts.} \\ & = - \cos x \ln (\sin x) \ \bigg|_0^\frac \pi 2 + \int_0^\frac \pi 2 \frac {\cos^2 x}{\sin x} dx \\ & = - \cos x \ln (\sin x) \ \bigg|_0^\frac \pi 2 + \int_0^\frac \pi 2 \frac {1-\sin^2 x}{\sin x} dx \\ & = - \cos x \ln (\sin x) \ \bigg|_0^\frac \pi 2 + \int_0^\frac \pi 2 \csc x\ dx - \int_0^\frac \pi 2 \sin x \ dx \\ & = - \cos x \ln (\sin x) - \ln({\color{#3D99F6}\cot x + \csc x}) + \cos x \ \bigg|_0^\frac \pi 2 & \small \color{#3D99F6} \text{Let }t = \tan \frac x2 \implies \sin x = \frac {2t}{1+t^2} \\ & = - \cos x \ln (\sin x) - \ln \left({\color{#3D99F6}\frac {1+t^2}{2t} + \frac {1-t^2}{2t}}\right) + \cos x \ \bigg|_0^\frac \pi 2 & \small \color{#3D99F6} \implies \cos x = \frac {1-t^2}{1+t^2} \\ & = - \cos x \ln (\sin x) - \ln \left(\frac 1t \right) + \cos x \ \bigg|_0^\frac \pi 2 \\ & = - \cos x \ln (\sin x) + \ln \left(\sin \frac x2\right) - \ln \left(\cos \frac x2\right)+ \cos x \ \bigg|_0^\frac \pi 2 \\ & = 0 + \lim_{x \to 0} \left(\cos x \ln (\sin x) - \ln \left(\sin \frac x2\right) + \ln \left(\cos \frac x2\right) - \cos x\right) \\ & = \lim_{x \to 0} \ln \left(\frac {\sin x}{\sin \frac x2}\right) + 0 - 1 \\ & = \lim_{x \to 0} \ln \left(\frac {2\sin \frac x2 \cos \frac x2}{\sin \frac x2}\right) - 1 \\ & = \ln 2 - 1 \approx \boxed{-0.307} \end{aligned}$