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Question

Evaluate the integral
10sin1xdx

A
π
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B
π
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C
π/4
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D
π/3
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Solution

The correct option is B π/4

10sin1(x)dx

Let x=tx=t2

dx=2t dt

10sin1(t)2t dt

=210sin1(t)tdt

=2[sin1(t)t22]10210t221t2dt

=π2+10=t21t2dt

Let t=sinθ

=π2+π/20sin2θcosθcosθdθ

=π2π/20sin2θdθ

=π2(12×π2) (using walli’s formula)

=π4

10sin1(x)dx=π4.


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