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Question

Integral π/40sinx+cosx3+sin2x is equal to ?

A
log 2
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B
log 3
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C
14log13
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D
14log 3
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Solution

The correct option is B 14log13
I=π/40sinx+cosx4(1sin2x)dx

=π/40sinx+cosx4(sinxcosx)2dx
sinxcosx=t
(cosx+sinx).dx=dt
When x=0sin0cosθ=01=1
When x=π4sinπ4cosπ4=0

=01dt4t2

=12×2[ln2+t2t]01

=14[ln|1|ln13]

=14ln[13]

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