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Question

π0 xf(sin x)dx is equal to

A
ππ0 f(sin x)dx
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B
π2π20 f(sin x)dx
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C
ππ20 f(cos x)dx
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D
πx0 f(cos x)dx
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Solution

The correct option is C ππ20 f(cos x)dx
Let I=π0 xf(sin x)dx(1)
I=π0(πx)f(sin(πx))dx
I=π0(πx)f(sin x)dx(2)
Adding (1) and (2)
2I=π0 πf(sin x)dx
=2ππ20f(sin x)dx
I=ππ20f[sin(π2x)]dx
=ππ20 f(cosx)dx

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