Mathematics - Definite Integration Question with Solution | TestHub

Given, 

$f\left(x\right)=x+{\int }_0^{\pi /2}\sin \left(x+y\right)f\left(y\right)dy$

$\Rightarrow f\left(x\right)=x+{\int }_0^{\pi /2}\left(\sin x\cos y+\cos x\sin y\right)f\left(y\right)dy$

$\Rightarrow f\left(x\right)=x+{\int }_0^{\pi /2}\left(\left(\cos yf\left(y\right)dy\right)\sin x+\left(\sin yf\left(y\right)dy\right)\cos x\right)$

And,$f\left(x\right)=x+\frac{a}{{\pi }^2-4}\sin x+\frac{b}{{\pi }^2-4}\cos x,x\in \mathrm{ℝ}$

On comparing both the equations of $f\left(x\right)$ we get,

$\Rightarrow \frac{a}{{\pi }^2-4}={\int }_0^{\pi /2}\cos yf\left(y\right)dy \dots \left(1\right)$

$\Rightarrow \frac{b}{{\pi }^2-4}={\int }_0^{\pi /2}\sin yf\left(y\right)dy \dots \left(2\right)$

Adding equation $\left(1\right)$ and equation $\left(2\right)$ we get,

$\frac{a+b}{{\pi }^2-4}={\int }_0^{\pi /2}\left(\sin y+\cos y\right)f\left(y\right)dy \dots \left(3\right)$

$\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$\therefore \frac{a+b}{{\pi }^2-4}={\int }_0^{\pi /2}\left(\sin y+\cos y\right)f\left(\frac{\pi }{2}-y\right)dy \dots \left(4\right)$

Add equation $\left(3\right)$ and equation $\left(4\right)$ we get,

$\frac{2\left(a+b\right)}{{\pi }^2-4}={\int }_0^{\pi /2}\left(\sin y+\cos y\right)\left(\frac{\pi }{2}+\frac{\left(a+b\right)}{{\pi }^2-4}\left(\sin y+\cos y\right)\right)dy$

$\Rightarrow \frac{2\left(a+b\right)}{{\pi }^2-4}=\pi +\frac{a+b}{{\pi }^2-4}\left(\frac{\pi }{2}+1\right)$

$\Rightarrow \left(a+b\right)=-2\pi \left(\pi +2\right)$