definite integral problems by s.k.goyal.help needed.

\hspace{-16}\bf{(2)\;::\; }$ Given $\bf{\int_{\infty}^{0}\frac{x\cdot e^{-x}}{\sqrt{1-e^{-2x}}}dx = -\int_{0}^{\infty}\frac{x\cdot e^{-x}}{\sqrt{1-e^{-2x}}}dx}$\\\\\\ Let $\bf{e^{-x}=\sin \phi\;,}$ Then $\bf{-e^{-x}dx = \cos \phi d\phi\Rightarrow dx = -\frac{\cos \phi}{\sin \phi}d\phi}$\\\\\\ and Changing Limits......\\\\\\ So Limit is $\bf{=-\int_{0}^{\frac{\pi}{2}}\ln(\sin \phi)d\phi = \frac{\pi}{2}\ln(2)}$\\\\\\ Here $\bf{\int_{0}^{\frac{\pi}{2}}\ln (\sin \phi)d\phi}$ is a Standard Definite Integral.......\\\\\\ Which is $\bf{= -\frac{\pi}{2}\ln(2)}$ and You can calculate easily.....\\\\\\