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Q.

02πxsin2nxsin2nx+cos2nxdx,n>0, is equal to

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a

π

b

12π2

c

2π

d

π2

answer is C.

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Detailed Solution

Let I=02πxsin2nxsin2nx+cos2nxdx                                    …(i)

Using property IV, we have 

In=02π(2πx)sin2n(2πx)sin2n(2πx)+cos2n(2πx)dx In=02π(2πx)sin2nxsin2nx+cos2nxdx

Adding (i) and (ii), we get

2In=2π02πsin2nxsin2nx+cos2nxdx In=π02πsin2nxsin2nx+cos2nxdx In=2π0πsin2nxsin2nx+cos2nxdx        [Using property VII]  In=4π0π/2sin2nxsin2nx+cos2nxdx       [Using property VII]  In=4π×π4=π2 

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