Find the maximum and minimum value of $f(x)$

Question

$$f(x) = \sin x + \int_{-\frac \pi 2}^{\frac \pi 2} (\sin x + t\cos x)f(t)dt$$ 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.

Answer 1

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.

Answer 2

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

$$A = -\frac{1}{3}, B = -\frac{2}{3}$$