Find the maximum and minimum value of $f(x)$ 
$$f(x) = \sin x + \int_{-\frac \pi 2}^{\frac \pi 2} (\sin x + t\cos x)f(t)dt$$
  Find the minimum and maximum value of $f(x)$.

My attempt:
Rewrite the functional equation as
$$f(x) = \sin x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) + \cos x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
Then differentiate both sides
$$f'(x) = \cos x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) - \sin x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
for maxima/minima, $f'(x)$ = 0
$$\cos x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) = \sin x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
I got stuck at this point.
 A: I found a solution. Since 
$$f(x) = \sin x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) + \cos x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
We can rewrite it as $$f(x) = A\sin x + B\cos x$$
This gives us the equations
$$\begin{gather}
A = 1 + \int_{-\frac \pi 2}^{\frac \pi 2}f(t)dt \tag{1} \\
B = \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt \tag{2} 
\end{gather}$$
Using some integration properties, it is easy to see that $\int_{-\frac \pi 2}^{\frac \pi 2}f(t) = \int_{-\frac \pi 2}^{\frac \pi 2}B\cos t dt$ and $\int_{-\frac \pi 2}^{\frac \pi 2}tf(t)dt = \int_{-\frac \pi 2}^{\frac \pi 2}At\sin t dt$. Evaluating these integrals and substituting in the equations, they simplify to
$$\begin{gather}
A = 1 + 2B \tag{1} \\
B = 2A \tag{2}
\end{gather}$$
Solving this system gives $A = -\frac 13$ and $B = -\frac 23$. Thus
$$f(x) = -\frac 13 \sin x - \frac 23 \cos x$$
The maximum and minimum values of $f(x)$ are $\frac{\sqrt{5}}{3}$ and $-\frac{\sqrt{5}}{3}$ respectively.
A: The solution is a bit convoluted, but here goes
Step 1: Differentiate the given equation by $x$
$$f'(x) = \cos x + \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} (\cos x - t\sin x)f(t)dt$$
Step 2: Consider the following sum
$$f'(x)\cos x + f(x) \sin x = 1 + \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(t)dt = f(\frac{\pi}{2})$$
Now, if you solve this differential equation in $f(x)$, you would get the general solution to be 
$$f(x) = A\sin x + B\cos x$$
Now, to satisfy the given functional equation, the constants need to satisfy the following
$$f(0) = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}tf(t)dt$$
$$f(\frac{\pi}{2}) = 1 + \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(t)dt$$
This should give you two equations in $A, B$, and you can solve to get 
$$A = -\frac{1}{3}, B = -\frac{2}{3}$$
