Question

$\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{6}}{\int }}\ln \left(\sin x\right)dx-\underset{-\ln 2}{\overset{\frac{1}{2}\ln \left(\frac{3}{4}\right)}{\int }}{\sin }^{-1}{e}^{x}dx$ is

• A. $\frac{\pi }{6}\ln \left(\frac{2}{3}\right)$
• B. $\frac{\pi }{6}\ln \left(\frac{3}{2}\right)$
• C. $\frac{\pi }{3}\ln \left(\frac{3}{4}\right)$
• D. $\pi \ln \left(\frac{4}{3}\right)$

Answer 1

The correct option is A $\frac{\pi }{6}\ln \left(\frac{2}{3}\right)$

$Letf\left(x\right)=\ln \left(\sin x\right)\Rightarrow f^{-1}x=\sin \left(e^x\right)$
$\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{6}}{\int }}\ln \left(\sin x\right)dx-\underset{-\ln 2}{\overset{\frac{1}{2}\ln \left(\frac{3}{4}\right)}{\int }}{\sin }^{-1}{e}^{x}dx$
$=\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{6}}{\int }}f\left(x\right)dx-\underset{-\ln 2}{\overset{\frac{1}{2}\ln \left(\frac{3}{4}\right)}{\int }}{f}^{-1}\left(x\right)dx$
$Let{f}^{-1}\left(x\right)=t\Rightarrow x=f\left(t\right)\Rightarrow dx=f^{\prime }\left(t\right)dt$
$\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{6}}{\int }}f\left(x\right)dx-\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}t.f^{\prime }\left(t\right)dt$
$=\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{6}}{\int }}f\left(x\right)dx-[[tf\left(t\right){]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}-\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}f\left(t\right)dt]$
$\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{6}}{\int }}f\left(x\right)dx-[tf\left(t\right){]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}-\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{6}}{\int }}f\left(t\right)dt$
$=-\frac{\pi }{6}\ln 2-\frac{\pi }{6}\ln \frac{3}{4}=\frac{\pi }{6}\ln \left(\frac{2}{3}\right)$