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01sin1x1x2dx

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a
π2
b
π24
c
π28
d
π22

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detailed solution

Correct option is C

Put sin−1⁡x=t⇒dt=11−x2dxL.L:t=sin−1⁡(0)=0U.L:t=sin−1⁡(1)=π2∫01 sin−1⁡x1−x2dx=∫0π/2 tdt=t220π/2=π28


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