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Question

The value of ππ(1x2)sinxcos2xdx is

A
0
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B
ππ33
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C
2ππ3
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D
722π3
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Solution

The correct option is A 0
Let I=ππ(1x2)sinxcos2xdx
Using baf(x)dx=baf(a+bx)dx
I=ππ(1(x)2sinlx)cos2(x)dx

=ππ(1x2)sinxcos2xdx=I

2I=0I=0

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