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Question

Evaluate : 10sin1(x1xx1x2)dx,0x1

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Solution

We have
E=10sin1(x1xx1x2)dx

E=10sin1x10sin1(x)dx

Now,
I=10sin1(x)dx

I=10sin1(1x)dx

I=10cos1(x)dx

2I=10π2dx

2I=π2

I=π4

10sin1x=xsin1x+122x1x2dx

=[xsin1x]10+[1x2]10

=π21

Hence,
E=π21π4=π41

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