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Question

π0 xf(sin x)dx=

A
π2π20f(sin x)dx
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B
ππ20f(cos x)dx
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C
ππ0f(cos x)dx
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D
ππ0f(sin x)dx
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Solution

The correct option is B ππ20f(cos x)dx
I=π0 xf(sin x)dx=π0(πx)f[sin(πx)]dx=π0(πx)f(sin x)dx=ππ0f(sin x)dxI
2I=ππ0 f(sin x)dx=2ππ20 f(sin x)dx
I=ππ20f(sin x)dx=ππ20f[sin(π2x)]dx=ππ20f(cos x)dx

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