Question

• A. $\frac { \pi } { 2 } \log \frac { 1 } { 2 }$
• B. $\frac { \pi ^ { 2 } } { 2 } \log \frac { 1 } { 2 }$
• C. $\pi \log \frac { 1 } { 2 }$
• D. $\pi ^ { 2 } \log \frac { 1 } { 2 }$

Answer 1

Answer: (B)

$\frac { \pi ^ { 2 } } { 2 } \log \frac { 1 } { 2 }$

Answer 2

$I = \int _ { 0 } ^ { \pi } x \log \sin x d x$ …..(i)

= $\int _ { 0 } ^ { \pi } ( \pi - x ) \log \sin ( \pi - x ) d x$ …..(ii)

By adding (i) and (ii), we get

$2 I = \int _ { 0 } ^ { \pi } \pi \log \sin x d x \Rightarrow I = \frac { 2 \pi } { 2 } \int _ { 0 } ^ { \pi / 2 } \log \sin x d x$

$= \pi \left( \frac { \pi } { 2 } \log \frac { 1 } { 2 } \right) = \frac { \pi ^ { 2 } } { 2 } \log \frac { 1 } { 2 }$ .