Let $$f(x) = x + {a \over {{\pi ^2} - 4}}\sin x + {b \over {{\pi ^2} - 4}}\cos x,x \in R$$ be a function which

satisfies $f(x) = x + \int\limits_0^{\pi /2} {\sin (x + y)f(y)dy}$. then $(a+b)$ is equal to

$f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$ <br/><br/> $f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)$ <br/><br/> On comparing with <br/><br/> $f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}$ then <br/><br/> $\Rightarrow \frac{a}{\pi^{2}-4}=\int_{0}^{\pi / 2} \cos y f(y) d y \quad....(2)$ <br/><br/> $\Rightarrow \frac{b}{\pi^{2}-4}=\int_{0}^{\pi / 2} \sin y f(y) d y \quad....(3)$ <br/><br/> <b>Add (2) and (3)</b> <br/><br/> $\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f(y) d y \quad....(4)$ <br/><br/> $\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y\quad....(5)$ <br/><br/> <b>Add (4) and (5)</b> <br/><br/> $\frac{2(a+b)}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y)\left(\frac{\pi}{2}+\frac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)\right) d y$ <br/><br/> $=\pi+\frac{a+b}{\pi^{2}-4}\left(\frac{\pi}{2}+1\right)$ <br/><br/> $(a+b)=-2 \pi(\pi+2)$