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Question: \(\int _ { 0 } ^ { \pi } \log \sin ^ { 2 } x d x =\)...

∫0πlog⁡sin⁡2xdx=\int _ { 0 } ^ { \pi } \log \sin ^ { 2 } x d x =

A

2πlog⁡e(12)2 \pi \log _ { e } \left( \frac { 1 } { 2 } \right)

B

πlog⁡e2+c\pi \log _ { e } 2 + c

C

π2log⁡e(12)+c\frac { \pi } { 2 } \log _ { e } \left( \frac { 1 } { 2 } \right) + c

D

None of these

Answer

2πlog⁡e(12)2 \pi \log _ { e } \left( \frac { 1 } { 2 } \right)

Explanation

Solution

∫0π2log⁡sin⁡xdx=2∫02π2log⁡sin⁡xdx=4∫0π/2log⁡sin⁡xdx\int _ { 0 } ^ { \pi } 2 \log \sin x d x = 2 \int _ { 0 } ^ { 2 \frac { \pi } { 2 } } \log \sin x d x = 4 \int _ { 0 } ^ { \pi / 2 } \log \sin x d x

=4×(−π2log⁡2)=−2πlog⁡e2=2πlog⁡e(12)= 4 \times \left( - \frac { \pi } { 2 } \log 2 \right) = - 2 \pi \log _ { e } 2 = 2 \pi \log _ { e } \left( \frac { 1 } { 2 } \right).