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Question

The value of π40xsin2x dx is:

A
π24π+8 64
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B
π2+4π836
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C
π22π+864
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D
π2+8π+632
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Solution

The correct option is A π24π+8 64
Let I=π40xsin2xdx
=π40x2(1cos2x)dx
Using integrating by parts, we have
I=[x2(xsin2x2)]π40π4012(xsin2x2)dx=π8(π412)[x24+cos2x8]π40
=(π232π16)(π26418)=π24π+864

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