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Question

Evaluate the integral
10(sin1x)21x2dx

A
π324
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B
π2
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C
π2
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D
0
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Solution

The correct option is A π324

10(sin1x)21x2dx

(sin1x)21x2dx=(sin1x)d(sin1x)

=(sin1x)33+c

10(sin1x)21x2dx=[(sin1x)33+c]10

=((sin11)33+c)((sin10)33+c)

=13(π2)3

=π324


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