The function f(x), that satisfies the condition $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy} $ is:

(A) $x + \frac{2}{3}(\pi  – 2)\sin x$
(B) $x + (\pi  + 2)\sin x$
(C) $x + \frac{\pi }{2}\sin x$
(D) $x + (\pi  – 2)\sin x$

Solution

We have, $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy}  = x + \sin x\int\limits_0^{\pi /2} {\cos yf(y)dy}  = x + \sin x.k$

Where, $k = \int\limits_0^{\pi /2} {\cos yf(y)dy}  = \int\limits_0^{\pi /2} {\cos y(y + \sin y.k)dy} $

$\therefore k = \int\limits_0^{\pi /2} {y\cos ydy}  + k\int\limits_0^{\pi /2} {\cos y\sin ydy} $

$ \Rightarrow k = \left. {y\sin y} \right|_0^{\pi /2} – \int\limits_0^{\pi /2} {\sin ydy}  + \frac{k}{2}\int\limits_0^{\pi /2} {sin2ydy}  = \frac{\pi }{2} + \left. {\cos y} \right|_0^{\pi /2} – \frac{k}{2}.\frac{1}{2}\left. {\cos 2y} \right|_0^{\pi /2}$

$ \Rightarrow k = \frac{\pi }{2} – 1 – \frac{k}{4}( – 1 – 1) = \frac{\pi }{2} – 1 + \frac{k}{2}$

$ \Rightarrow \frac{k}{2} = \frac{\pi }{2} – 1 = \frac{{\pi  – 2}}{2}$

$\therefore k = \pi  – 2$

Now, $f(x) = x + \sin x.k = x + (\pi  – 2)\sin x$

Answer: (D)