If ∫0πx f(Sinx)dx=A ∫0π/2f(sinx)dx then A=

∫0πx f(Sinx)dx= = ∫0π/2π-x fsin(π-x)dx =∫0π/2π-x fsinxdx =π∫0πf(sinx)-∫0πx f(sinx)dx =π∫0πf(sinx)-I 2I=π∫0πf(sinx)dx I=π22∫0π/2f(sinx)dx I=π∫0π/2 f(sinx)dx A=π

If ∫0πx f(Sinx)dx=A ∫0π/2f(sinx)dx then A=

∫0πx f(Sinx)dx= = ∫0π/2π-x fsin(π-x)dx =∫0π/2π-x fsinxdx =π∫0πf(sinx)-∫0πx f(sinx)dx =π∫0πf(sinx)-I 2I=π∫0πf(sinx)dx I=π22∫0π/2f(sinx)dx I=π∫0π/2 f(sinx)dx A=π

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If $\int_{0}^{π} x f (S i n x) d x = A \int_{0}^{π / 2} f (\sin x) d x t h e n A =$

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$π$

$π / 4$

$2 π$

answer is D.

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Detailed Solution

$\int_{0}^{π} x f (S i n x) d x = = \int_{0}^{π / 2} (π - x) f (\sin (π-x))dx = \int_{0}^{π / 2} (π - x) f (\sin x) d x = π \int_{0}^{π} f (\sin x) - \int_{0}^{π} x f (\sin x) d x = π \int_{0}^{π} f (\sin x) - I 2 I = π \int_{0}^{π} f (\sin x) d x I = \frac{π}{2} (2 \int_{0}^{π / 2} f (\sin x) d x) I = π \int_{0}^{π / 2} f (\sin x) d x A = π$