Question

If $ \int_{0}^{\lambda }{xf(\sin x)}dx=A\int_{0}^{\pi /2}{f(\sin x)}dx, $ then A is equal to

• A. $ 0 $
• B. $ \pi $
• C. $ \frac{\pi }{4} $
• D. $ 2\pi $
• E. $ 3\pi $

Answer 1

Answer: (B)

Let $ I=\int_{0}^{\pi }{x}f(\sin x)dx $ ...(i)
$=\int_{0}^{\pi }{(\pi -x)}f[\sin (\pi -x)]dx $
$ \Rightarrow $ $=\int_{0}^{\pi }{(\pi -x)}f(\sin x)dx $ ...(ii)
On adding Eqs. (i) and (ii), we get
$ 2I=\int_{0}^{\pi }{\pi }f(\sin x)dx $
$ \Rightarrow $ $ I=\frac{\pi }{2}\int_{0}^{\pi }{f(\sin x)dx=\pi }\int_{0}^{\pi /2}{f(\sin x)}dx $
$ \Rightarrow $ $ A\int_{0}^{\pi /2}{f(\sin x)dx=\pi }\int_{0}^{\pi /2}{f(\sin x)}dx $
$ \Rightarrow $ $ A=\pi $