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) $x + (\pi + 2)\sin x$
(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)