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Q.

∫0π2xsinxcosxcos4x+sin4x=

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a

π24

b

π28

c

π216

d

π22

answer is C.

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Detailed Solution

I=∫0π2xsinxcosxcos4x+sin4xdx,x→π2−x I=∫0π2(π2−x)cosxsinxcos4x+sin4xdx 2I=π2∫0π2cosxsinxcos4x+sin4xdx,tan2x=t       =π4∫0∞dt1+t2=π4⋅π2=π28→I=π216
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