Calculus is Awesome!

A couple of changes of variable gives $\begin{array}{rcl} \displaystyle M \; = \; \int_0^{\frac12\pi} \frac{\ln(\sin x)}{\sin x - 1}\,dx & = & \displaystyle \int_0^{\frac12\pi} \frac{\ln(\cos y)}{\cos y - 1}\,dy \; = \; -\tfrac12\int_0^{\frac12\pi}\ln(\cos y) \mathrm{cosec}^2\tfrac12y\,dy \\ & = & \displaystyle-\int_1^\infty \ln\left(\frac{u^2-1}{u^2+1}\right)\,du \; = \; \int_1^\infty \ln\left(\frac{u^2+1}{u^2-1}\right)\,du \end{array}$ using first $y= \tfrac12\pi - x$ and then $u = \cot\tfrac12y$ . Since this integral is improper, being both infinite and with a singularity at $u=1$ , we shall first integrate from $a > 1$ to $X$ , and take limits. Integrating by parts gives $\begin{array}{rcl} \displaystyle \int_a^X \ln\left(\frac{u^2+1}{u^2-1}\right)\,du & = & \left[u \ln\left(\frac{u^2+1}{u^2-1}\right)\right]_a^X + 4\int_a^X \frac{u^2}{u^4-1}\,du \\ & = & \left[u\ln\left(\frac{u^2+1}{u^2-1}\right)\right]_a^X + \int_a^X \left(\frac{2}{u^2+1} + \frac{1}{u-1} - \frac{1}{u+1}\right)\,du \\ & = & \left[ u\ln \left(\frac{u^2+1}{u^2-1}\right) + 2\tan^{-1}u + \ln\left(\frac{u-1}{u+1}\right)\right]_a^X \end{array}$ Now $X\ln\left(\frac{X^2+1}{X^2-1}\right) + 2\tan^{-1}X + \ln\left(\frac{X-1}{X+1}\right) \; = \; X\ln\left(\frac{1 + X^{-2}}{1 - X^{-2}}\right) + 2\tan^{-1}X + \ln\left(\frac{1 - X^{-1}}{1 + X^{-1}}\right)$ which tends to $0 + \pi + 0 = \pi$ as $X \to \infty$ , and hence $\begin{array}{rcl} \displaystyle \int_a^\infty \ln\left(\frac{u^2+1}{u^2-1}\right)\,du & = & \displaystyle\pi - a\ln\left(\frac{a^2+1}{a^2-1}\right) - 2\tan^{-1}a - \ln\left(\frac{a-1}{a+1}\right) \\ & = & \displaystyle \pi - 2\tan^{-1}a - a\ln(a^2+1) + (a+1)\ln(a+1) + (a-1)\ln(a-1) \end{array}$ for any $a > 1$ ; this tends to $M \; = \; \pi - 2\tan^{-1}1 - \ln2 + 2\ln 2 + 0 \; = \; \tfrac12\pi + \ln 2$ as $a \to 1$ . The answer is $\lfloor 10M \rfloor \,=\, \lfloor 5\pi + 10\ln2\rfloor \,=\, \boxed{22}$ .

Note1:

$\begin{aligned} &\int \frac{1}{\sin(x)-1} {\color{#D61F06} \cdot\frac{\sin(x)+1}{\sin(x)+1}}dx\\ &=\int \frac{\sin(x)}{-\cos^2(x)}+\frac{1}{-\cos^2(x)}\\ &= -\left ( \int \sec(x)\tan(x) +\int sec^2(x) \right ) \\ &= -\left ( \sec(x)+\tan(x) \right ) + C \end{aligned}$

Note2:

$\begin{aligned} &\int \csc(x){\color{#D61F06} \cdot \frac{\csc(x)+\cot(x)}{\csc(x)+\cot(x)}}dx\\ &= - \int \frac{-\csc^2(x)-\csc(x)\cot(x)}{\csc(x)+\cot(x)}dx\\ &= -\ln\left ( \csc(x)+\cot(x) \right ) + C \end{aligned}$

Solution:

$\begin{aligned} &u=\ln(\sin(x)) && dv=\frac{dx}{\sin(x)-1} \\ &du=\cot(x) && v= -\left ( \sec(x)+\tan(x) \right ) \end{aligned}$

$\begin{aligned} & \int{\frac{\ln(\sin(x))}{\sin(x)-1}}dx &&= -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x) \right )+\int{\cot(x)\cdot \left ( \sec(x)+\tan(x) \right )dx} \\ &&& = -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x) \right )+\int{(\csc(x)+1)dx} \\ &&& = -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x) \right )- \ln\left ( \csc(x)+\cot(x) \right )+x \\ &&& = -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x) \right )- \ln\left ( 1+\cos(x) \right )+\ln(sin(x))+x \\ &&& = -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x)-1 \right )- \ln\left ( 1+\cos(x) \right )+x \\ & \int_0^{\pi/2}{\frac{\ln(\sin(x))}{\sin(x)-1}}dx &&= \left. -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x)-1 \right )- \ln\left ( 1+\cos(x) \right )+x \right |_0^{\pi/2} \\ & x\rightarrow \pi/2 && -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x)-1 \right )\rightarrow 0 \\ & x\rightarrow 0 && -\ln(\sin(x))\cdot \left ( \sec(x)+\tan(x)-1 \right )\rightarrow 0 \\ & \int_0^{\pi/2}{\frac{\ln(\sin(x))}{\sin(x)-1}}dx &&=\frac{\pi}{2}+\ln(2)\approx 2.26 \end{aligned}$