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Question

The value of π2π2sin2x1+2xdx is :

A
4π
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B
π4
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C
π8
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D
π2
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Solution

The correct option is B π4
Using,
I=baf(x)dx=baf(abx)dx

I2=π/2π/2sin2x1+2xdx

Remember I1=aaf(x)dx

I1=a0[f(x)+f(x)]dx

I2=π/2π/2sin2x1+2xdx=π/20(sin2x1+2x+sin2(x)1+2x)dx

I2=π/20sin2x(1+2x1+2x)dx

I2=π/20sin2xdx .....(1)

I2=π/20sin2(π2x)dx=π/20cos2x dx .....(2)

(1) + (2)

I2=π/201.dx

I2=π/22

I2=π4


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