MCQ
Let $f (x) = \frac{{\sin x}}{x}$ , then $\int\limits_0^{\frac{\pi }{2}} {f(x)\,\,f\left( {\frac{\pi }{2} - x} \right)\,dx} =$
  • $\frac{2}{\pi }\int\limits_0^\pi  {f(x)\,dx} $
  • B
    $\int\limits_0^\pi  {f(x)\,dx} $
  • C
    $\pi \,\int\limits_0^\pi  {f(x)\,dx} $
  • D
    $\frac{1}{\pi }\,\int\limits_0^\pi  {f(x)\,dx} $

Answer

Correct option: A.
$\frac{2}{\pi }\int\limits_0^\pi  {f(x)\,dx} $
a
$I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x\cos x}}{{x\left( {\frac{\pi }{2} - x} \right)}}\,dx} $ $= \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin 2x}}{{x(\pi  - 2x)}}\,dx} $; put $2x = t$
$I = \int\limits_0^\pi  {\frac{{\sin t}}{{t(\pi  - t)}}dt} $ $= \frac{1}{\pi }\,\int\limits_0^\pi  {\left( {\frac{{\sin t}}{t} + \frac{{\sin t}}{{(\pi  - t)}}} \right)\,dt} $ $=\frac{2}{\pi }\,\int\limits_0^\pi  {\frac{{\sin t}}{t}\,dt} $ 

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