Evaluate the definite integral: $y(x) = \int_{0}^{\pi} \sin(x+y(x)) dx$

Question

We were recently asked to evaluate this - $y(x) = \int_{0}^{\pi} \sin(x+y(x)) dx$

I think we can start by breaking up the integral as $y(x) = \int_{0}^{\pi} \sin(x)\cos(y(x)) dx + \int_{0}^{\pi} \cos(x)\sin(y(x)) dx$ and then assuming the form that $y$ would take as $y(x) = A\sin(x) + B\cos(x) + D$.

Answer 1

As @joriki suggested, the main point in solving this is in realizing that y is not a function of x but, contrary to what seems in the question, a constant because RHS is a definite integral! So let's assume that $y = K$(say). So the equation can be rewritten as-

$K = \int_{0}^{\pi} \sin(x+K) dx$ and we're left with simply finding the value of $K$. So, $K = -(\cos(\pi+K) - \cos(0 + K)) = 2\cos(K)$ and finally, y is the solution of the equation: $K/2 = \cos(K)$ for which $y\approx1.02987$ as calculated here

Answer 2

As written, the equation is syntactically ambiguous... the integration variable can't have the same name as a free variable, or else you can't tell which $x$ is meant by each occurrence within the integrand. There are four possible disambiguations:

1. $y(x) = \int_{0}^{\pi} \sin(t+y(t)) dt.$
2. $y(x) = \int_{0}^{\pi} \sin(t+y(x)) dt = 2 \cos(y(x))$.
3. $y(x) = \int_{0}^{\pi} \sin(x+y(t)) dt = \sin(x)\int_{0}^{\pi}\cos(y(t))dt + \cos(x)\int_{0}^{\pi}\sin(y(t))dt.$
4. $y(x) = \int_{0}^{\pi} \sin(x+y(x)) dt = \pi\sin(x+y(x))$.

In the first case, $y(x)$ is a constant $K$ satisfying $K=\int_{0}^{\pi}\sin(t+K)dt=2\cos K$; the unique solution is $K=1.0298665...$.

In the second case, $y(x)$ must satisfy $y=2\cos(y)$ at each point independently; since this has a unique solution, the result is the same as the first case.

In the third case, $y(x)=A\cos(x)+B\sin(x)$, where $A$ and $B$ satisfy a coupled integral equation.

In the fourth case, $y(x)$ satisfies an implicit equation that depends on $x$ at each point independently.