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Question

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

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

A

π\pi

B

π2\frac { \pi } { 2 }

C

π2−12\frac { \pi } { 2 } - \frac { 1 } { 2 }

D

π−1\pi - 1

Answer

π2\frac { \pi } { 2 }

Explanation

Solution

∫−π/2π/2sin⁡2xdx=2∫0π/2sin⁡2xdx=2Γ(32)Γ(12)2Γ(2+22)=π2\int _ { - \pi / 2 } ^ { \pi / 2 } \sin ^ { 2 } x d x = 2 \int _ { 0 } ^ { \pi / 2 } \sin ^ { 2 } x d x = 2 \frac { \Gamma \left( \frac { 3 } { 2 } \right) \Gamma \left( \frac { 1 } { 2 } \right) } { 2 \Gamma \left( \frac { 2 + 2 } { 2 } \right) } = \frac { \pi } { 2 }.