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Question

The value of π20log(sinx)dx is equal to?

A
π20log(cosx)dx
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B
π2log2
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C
Both (a) & (b)
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D
None of these
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Solution

The correct option is D Both (a) & (b)
By integration property
a0f(x)dx=a0f(ax)dx
I=π/20logsin(π/2x)dx=π/20logcosxdx
2I=π/20(logsinx+logcosx)dx
2I=π/20log(sinxcosx)dx.
=π/20logsin2x2dx.
=π/20logsin2xdxπ/20log2dx.
Substitute 2x=t in the 1st. 2dx=dt and limits become 0 to π.
2I=12π0logsintdt[xlog2]π/20
Now apply Prop. Vi in 1st. f(2ax)=f(x).
2I=12.2π/20logsintdtπ2log2
or 2I=π/20logsinxdx(π/2)log2, by prop. Ior 2I=I+π2log12
I=π/20logsinxdx=π/20logcosxdx=π2log12

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