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B
π∫π20f(cosx)dx
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C
π∫π0f(cosx)dx
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D
π∫π0f(sinx)dx
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Solution
The correct option is Bπ∫π20f(cosx)dx I=∫π0xf(sinx)dx=∫π0(π−x)f[sin(π−x)]dx=∫π0(π−x)f(sinx)dx=π∫π0f(sinx)dx−I ⇒2I=π∫π0f(sinx)dx=2π∫π20f(sinx)dx ⇒I=π∫π20f(sinx)dx=π∫π20f[sin(π2−x)]dx=π∫π20f(cosx)dx