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Question


π/20cosx1+sinxdX=

A
log2
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B
loge
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C
12 log3
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D
0
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Solution

The correct option is B log2

π20cosx1+sinxdx
we know that
π20cosx1+sinxdx=π20d(sinx)1+sinx
π20d(sinx)1+sinx=log(1+sinx)π20
=log(1+sinπ2)log(1+sin0)
=log(2)
=log2


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