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Question

π20|sinxcosx|dx is equal to

A
0
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B
2(21)
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C
21
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D
2(2+1)
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Solution

The correct option is B 2(21)
π20|sinxcosx|dx=π40(sinxcosx) dx+π2π4(sinxcosx) dx=[cosx+sinx]π40[cosx+sinx]π2π4=[12+1210][0+11212]=222=2(21)

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