The function f ( x ) , that satisfies the condition f ( x ) = x + ∫...

Given the function $f\left(x\right) = x + {∫}_{0}^{\dfrac{\pi}{2}}\sin x \cdot cosyf$\left(y\right)dy$</p><p>$f\left(x\right) = x + ksinx . . . \left(i\right)$</p><p>$f\left(y\right) = y + ksiny . . . \left(ii\right)

k = {∫}_{0}^{\dfrac{\pi}{2}}$cosyf \left(y\right)$dy$</p><p>From equations \left(i\right)\text{\text{ and } }$\left(ii\right)

$f\left(x\right)$ = x + {∫}_{0}^${\dfrac{\pi}{2}}$\sin x \cos y \left(y + ksiny\right)$dy$

$f\left(x\right)$ = x + \sin x{∫}_{0}^${\dfrac{\pi}{2}}$ycosydy + $\dfrac{k}{2}$\sin x{∫}_{0}^${\dfrac{\pi}{2}}$\sin 2ydy$</p><p>$f\left(x\right)= x + \sin x \left[y{∫}_{0}^{\dfrac{\pi}{2}}cosydy - {∫}_{0}^{\dfrac{\pi}{2}}\left\{\dfrac{d}{dy}y{∫}_{0}^{\dfrac{\pi}{2}}cosydy\right\}dy\right] +\dfrac{k}{2}\sin x{∫}_{0}^{\dfrac{\pi}{2}}$\sin 2ydy$

$f\left(x\right)$ = x + \sin x \left[\dfrac{\pi}{2} -1\right]$+$\dfrac{k}{2}$\sin x \times \left(1\right)$

$f\left(x\right) = x + \dfrac{( \pi - 2 ) \sin x}{2} + \dfrac{ksinx}{2} . . . \left(iii\right)</p><p>from$( i ) and ( iii )$</p><p>$k = $\dfrac{\pi}{2} -$ 1 + $\dfrac{k}{2}</p><p>$k = \pi - 2$</p><p>Hence,$f ( x ) = x + ( \pi - 2 ) \sin x$</p>$