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

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 $y(x) = A\sin(x) + B\cos(x) + D$.

-
You have $\mathrm dy$ in the title and $\mathrm dx$ in the body. If you meant $\mathrm dx$, then you have $x$ as a dummy variable on the right and a free variable on the left. The result is that $y(x)$ is constant, which is probably not what you wanted. – joriki Jan 7 '12 at 11:40
You've just given me the solution! I'll answer it later though... – tipycalFlow Jan 7 '12 at 11:52

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

-

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.

-
+1, nice answer...I think this was a trick question, hence the ambiguity seems to have been put in deliberately. It wasn't intended to be too difficult so I'm assuming it's the first case. The third case takes it up one(rather, multiple) notch. The fourth is too bland. I'll try to solve the third case and post it too... – tipycalFlow Jan 7 '12 at 15:09