Inetgral

use property 5 of definite integral

\hspace{-20}$Let $\bf{\mathbb{I}=\int_{0}^{\pi}x\cdot \cot xdx = \int_{0}^{\frac{\pi}{2}}x\cdot \cot xdx+\int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx}$\\\\\\ Bcz $\bf{x\cdot \cot x>0\;\in \left[0,\frac{\pi}{2}\right)}$ and $\bf{x\cdot \cot x\leq 0\;\in \left[\frac{\pi}{2},\pi\right)}$\\\\\\ Now Let $\bf{\mathbb{J}=\int_{0}^{\frac{\pi}{2}}x\cdot \cot xdx}$ and $\bf{\mathbb{K}=\int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx}$\\\\\\ $\bf{\underline{\underline{For\;\; Calculation \;\;of\;\; $\bf{\mathbb{J::}}}}}$\\\\\\ $Let $\bf{\mathbb{J}=\int_{0}^{\frac{\pi}{2}}x\cdot \cot xdx = \left[x\cdot \ln(\sin x)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx}$\\\\\\ So $\bf{\mathbb{J}=0+\frac{\pi}{2}\ln(2)\;\;\;....}$\\\\\\ Bcz value of $\bf{\int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\ln(2)}$\\\\\\ Left as a Exercise.......\\\\\\ $\bf{\underline{\underline{For\;\; Calculation \;\;of\;\; $\bf{\mathbb{K::}}}}}$\\\\\\

\hspace{-20}$Let $\bf{\mathbb{K}=\int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx\rightarrow -\infty}$\\\\\\ Bcz Using camparasion Test.....\\\\\\ $\bf{\Rightarrow \int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx<\int_{\frac{\pi}{2}}^{\pi}\cot x dx\rightarrow \infty}$\\\\\\ Bcz In $\bf{x\in \left[\frac{\pi}{2},\pi\right)\;\;,x\cdot \cot x\leq 0}$\\\\\\ So $\bf{\mathbb{R.H.S}}$ of Integral Diverges to $\bf{-\infty}$\\\\\\ So $\bf{\mathbb{L.H.S}}$ of Integral also Diverges to $\bf{-\infty}$\\\\\\ So $\bf{\int_{0}^{\pi}x\cdot \cot xdx=\int_{0}^{\frac{\pi}{2}}x\cdot \cot xdx+\int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx \rightarrow -\infty}$