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Q.

∫0π2n dx1+Cotnnx=

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a

π2n

b

π4n

c

π8n

d

πn

answer is B.

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Detailed Solution

Let I=∫0π/2nSinnnxSinnnx+Cosnnxdx I=∫0π/2nCosn nxCosn nx+Sinnnxdx       since ∫fxdx=∫fa-xdx2I=∫0π/2n1dx=π2nI=π4n
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