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

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

A

−(π2)log⁡2- \left( \frac { \pi } { 2 } \right) \log 2

B

πlog⁡12\pi \log \frac { 1 } { 2 }

C

−πlog⁡12- \pi \log \frac { 1 } { 2 }

D

π2log⁡2\frac { \pi } { 2 } \log 2

Answer

−(π2)log⁡2- \left( \frac { \pi } { 2 } \right) \log 2

Explanation

Solution

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

⇒ 2I=∫0π/2log⁡sin⁡xcos⁡xdx=∫0π/2log⁡sin⁡2xdx−∫0π/2log⁡2dx2 I = \int _ { 0 } ^ { \pi / 2 } \log \sin x \cos x d x = \int _ { 0 } ^ { \pi / 2 } \log \sin 2 x d x - \int _ { 0 } ^ { \pi / 2 } \log 2 d x

=12∫0πlog⁡sin⁡tdt−π2log⁡2= \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } \log \sin t d t - \frac { \pi } { 2 } \log 2 , (Putting 2x=t2 x = t )

=12⋅2∫0π/2log⁡sin⁡tdt−π2log⁡2= \frac { 1 } { 2 } \cdot 2 \int _ { 0 } ^ { \pi / 2 } \log \sin t d t - \frac { \pi } { 2 } \log 2

⇒2I=I−π2log⁡2⇒I=−π2log⁡2\Rightarrow 2 I = I - \frac { \pi } { 2 } \log 2 \Rightarrow I = \frac { - \pi } { 2 } \log 2 ,{∵∫abf(x)dx=∫abf(t)dt}\left\{ \because \int _ { a } ^ { b } f ( x ) d x = \int _ { a } ^ { b } f ( t ) d t \right\}.