Question

# The value of π4∫0xsin2x 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 π2−4π+8 64Let I=π4∫0xsin2xdx =π4∫0x2(1−cos2x)dx Using integrating by parts, we have I=[x2(x−sin2x2)]π40−π4∫012(x−sin2x2)dx=π8(π4−12)−[x24+cos2x8]π40 =(π232−π16)−(π264−18)=π2−4π+864

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