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.

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.
