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Question

120sin1x(1x2)32dx=

A
π4+12log 2
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B
π412log 2
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C
π2+log 2
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D
π2log 2
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Solution

The correct option is B π412log 2
I=120sin1x(1x2)32dx
Put sin1x=t11x2dx=dt and x=sin t
Also t=0 to π4as x=0 to 12
I=π40 t.sec2 t dt=π412log 2

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