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Question: If \(\int_{0}^{\pi}{xf(\sin x)dx = k\int_{0}^{\pi}{f(\sin x)dx}}\), then the value of k will be...

If ∫0πxf(sin⁡x)dx=k∫0πf(sin⁡x)dx\int_{0}^{\pi}{xf(\sin x)dx = k\int_{0}^{\pi}{f(\sin x)dx}}, then the value of k will be

A

π

B

π/2

C

π/4

D

1

Answer

π/2

Explanation

Solution

Given, ∫0πxf(sin⁡x)dx=k∫0πf(sin⁡x)dx\int_{0}^{\pi}{xf(\sin x)dx = k\int_{0}^{\pi}{f(\sin x)dx}}

⇒ ∫0π(π−x)f(sin⁡(π−x))dx=k∫0πf(sin⁡(π−x))dx\int_{0}^{\pi}{(\pi - x)f(\sin(\pi - x)})dx = k\int_{0}^{\pi}{f(\sin(\pi - x))dx}

⇒ π∫0πf(sin⁡x)dx−∫0πxf(sin⁡x)dx=k∫0πf(sin⁡x)dx\pi\int_{0}^{\pi}{f(\sin x)dx} - \int_{0}^{\pi}xf(\sin x)dx = k\int_{0}^{\pi}{f(\sin x)dx}⇒

π∫0πf(sin⁡x)dx−2k∫0πf(sin⁡x)dx=0\pi\int_{0}^{\pi}{f(\sin x)dx - 2k}\int_{0}^{\pi}{f(\sin x)dx = 0} ⇒ (π−2k)∫0πf(sin⁡x)dx=0(\pi - 2k)\int_{0}^{\pi}{f(\sin x)dx = 0}

∴ π−2k=0⇒k=π⥂/⥂2\pi - 2k = 0 \Rightarrow k = \pi ⥂ / ⥂ 2.