If $\int_{0}^{\pi} x f(\sin x) d x=A \int_{0}^{\pi / 2} f(\sin x) d x$, then $A$ is

JEE Main · 2004 · 24 Apr 2004 · Offline · Definite Integrals

\int_{0}^{π} x f (sin x) d x = A \int_{0}^{π /2} f (sin x) d x

, then

A

Answer: (B)

Apply the reflection

x \mapsto π - x

\int_{0}^{π} x f (sin x) d x

; since

sin (π - x) = sin x

, the integral also equals

\int_{0}^{π} (π - x) f (sin x) d x

. Adding the two forms,

2 I = π \int_{0}^{π} f (sin x) d x

. Because

f (sin x)

is symmetric about

x = \frac{π}{2}

\int_{0}^{π} f (sin x) d x = 2 \int_{0}^{π /2} f (sin x) d x

, so

I = π \int_{0}^{π /2} f (sin x) d x

, giving

A = π

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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