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Question

π/20log(sinx) dx=

A
(π2)log2
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B
πlog12
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C
πlog12
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D
12log2
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Solution

The correct option is A (π2)log2
I=π/20log(sinx) dx=π/20log(cosx) dx2I=π/20log(sinx cosx) dx=π/20log(sin2x) dxπ/20log(2) dx
=12π0log(sint) dtπ2log2 [Putting in first integral 2x=t]=122π/20logsintdtπ2log2 [2a0f(x)dx=2a0f(x)dx if f(2ax)=f(x)]2II=π2log2I=π2log2. [baf(x)dx=baf(t)dt]

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