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Question

Evaluate 3π/4π/4xsinx1+sinxdx

A
π24(21)π
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B
π24(2+1)π
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C
π22(21)π
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D
π22(2+1)π
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Solution

The correct option is A π24(21)π
Let I=3π/4π/4xsinx1+sinxdx=3π/4π/4(πx)sin(πx)1+sin(πx)dx

=3π/4π/4(πx)sinx1+sinxdx=π3π/4π/4sinx1+sinxdx3π/4π/4xsinx1+sinxdx

=π3π/4π/4(1+sinx1)1+sinxdxI

2I=π3π/4π/4(111+sinx)dx

I=π2[x]3π/4π/4π23π/4π/41sinxcos2xdx

=π24π2[tanxsecx]3π/4π/4=π24π2[(11)(22)]

I=π24+π2(222)=π24(21)π

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