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Q.

Let n∈N,∫02πx⋅sin2nxsin2nx+cos2nxdx=

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a

0

b

π22

c

π2

d

2π2

answer is C.

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Detailed Solution

I=∫02πx⋅sin2nxsin2nx+cos2nxdx,x→2π−x I=∫02π(2π−x)sin2nxsin2nx+cos2nxdx2I=2π∫02πsin2nxsin2nx+cos2nxdx       =4π∫0πsin2nxsin2nx+cos2nx dx2I=8π∫0π2sin2nsin2nx+cos2nxdx,x→π2−x2I=8π∫0π2cos2nsin2nx+cos2nxdx4I=8π∫0π21⋅dx=4π2→I=π2
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