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Q.

∫0πcos2xlnsinxdx=

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a

−π

b

−π2

c

−π4

d

π

answer is B.

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Detailed Solution

I=∫0πcos2xlnsinxdx=2∫0π2cos2xlnsinxdx                                           =2[sin2x2lnsinx|0π2−∫0π2sin2x2cotxdx]                                           =−2∫0π2cos2xdx=−2⋅12⋅π2=−π2
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∫0πcos2xlnsinxdx=