Q97 / 180 · COMEDK 2015 · Mathematics

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

  • A
    0
  • B
    $2 \pi$
  • C
    $\pi /4$
  • Correct answer
    D
    $\pi$

Solution

We have, $$ \begin{array}{ll} \quad \int_{0}^{\pi} x f(\sin x) & d x=A \int_{0}^{x / 2} f(\sin x) d x \\ \text { Let } \quad I & =\int_{0}^{\pi} x \cdot f(\sin x) d x \quad \text{...(i)} \\ \Rightarrow \quad I & =\int_{0}^{\pi}(\pi-x) f(\sin (\pi-x)) d x \\ \Rightarrow \quad I & =\int_{0}^{\pi}(\pi-x) f(\sin x) d x \quad \text{...(ii)} \end{array} $$ Adding Eqs. (i) and (ii), we get $$ \begin{aligned} &2 I=\pi \int_{0}^{\pi} f(\sin x) d x \Rightarrow I=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) d x \\ &\Rightarrow \quad I=\frac{\pi}{2} \times 2 \int_{0}^{\pi / 2} f(\sin x) d x \\ &\Rightarrow \quad I=\pi \int_{0}^{\pi / 2} f(\sin x) d x \end{aligned} $$ $\therefore$ On comparing, we get $A=\pi$

Hence, the correct answer is option: 'D'.

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