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Question

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

∫0πxf(sin⁡x)dx=\int _ { 0 } ^ { \pi } x f ( \sin x ) d x =

A

π∫0πf(sin⁡x)dx\pi \int _ { 0 } ^ { \pi } f ( \sin x ) d x

B

π2∫0πf(sin⁡x)dx\frac { \pi } { 2 } \int _ { 0 } ^ { \pi } f ( \sin x ) d x

C

π2∫0π/2f(sin⁡x)dx\frac { \pi } { 2 } \int _ { 0 } ^ { \pi / 2 } f ( \sin x ) d x

D

None of these

Answer

π2∫0πf(sin⁡x)dx\frac { \pi } { 2 } \int _ { 0 } ^ { \pi } f ( \sin x ) d x

Explanation

Solution

Since f(a−x)=f(x)f ( a - x ) = f ( x ).