Regarding integrals of type $f(x)=\int_{a}^{b}g{(x+f(x))}dx$

If my integral was defined recursively like $f(x)= \int_{0}^{\pi}\sin{(x+f(x))} \,dx$, how would I calculate the value of the integral? Or is such an integral definition even possible? I was thinking about going via Integration by parts.

$$f(x)=\int_{0}^{\pi}\sin{(x+f(x))}dx$$ $$\implies f(x)=\sin{(x+f(x))}\int_{0}^{\pi}dx - \int_{0}^{\pi}(\frac{d}{dx}(\sin{(x+f(x))})(\int_{0}^{\pi}dx)dx$$ I am stuck here, how to proceed from here?

Solution 1:

Since, as you write, the integral is a constant, it might be more helpful to write your expression as

$$k = \int_0^\pi \sin(x + k) \; dx. $$

Evaluating the integral yields $k = \cos(k) - \cos(\pi + k) = 2\cos(k)$. In this particular case, we'd need to solve numerically; according to Wolfram Alpha, $k \approx 1.02987$.