Find all positive differentiable functions $f$ that satisfy $\int_0^x \sin(t) f(t) dt = [f(x)]^2.$ Can someone please help me out with this question:
Find all positive differentiable functions $f$ that satisfy $$\int_0^x \sin(t) f(t) dt = [f(x)]^2.$$
My homework also lists $$\int_0^x \sin(t) f(t) dt = [f(x)]^2-1$$ as the correct problem because the first problem has a subtle issue.
a) I know (well, think) that the antiderivative of $\sin (t)$ is $-\cos (t)$ but I don't understand the $f(t)$ part. Where do I continue to go?
b) What is the subtle issue? I was told to reread the question and what it's asking for but I don't see it.
Thank you in advance!
 A: If you differentiate both sides of your integral equation, you get
$$
f(x) \sin x = 2 f(x) f'(x)
$$
which implies that either $f(x) \equiv 0$ or $f'(x) = (\sin x)/2$.
Can you finish this?
A: As mentioned by  Henning Makholm, you probably wanted $f$ to be nonnegative instead of being positive.  Ignoring the nonnegativity requirement and the differentiability condition, there are strange solutions $f:\mathbb{R}\to\mathbb{R}$ to the integral equation
$$\int_0^x\,\sin(t)\,f(t)\,\text{d}t=\big(f(x)\big)^2\,\text{ for all }x\in\mathbb{R}\,.\tag{*}$$     I would like to also address that gt6989b's hint leads to $$\sin(x)\,f(x)=2\,f(x)\,f'(x)\text{ for all }x\in\mathbb{R}\,,$$
which only implies that, for all $x\in\mathbb{R}$, either
$$f(x)=0\text{ or }f'(x)=\frac12\,\sin(x)\,.$$  (Here, $f'$ is a weak derivative of $f$.)
Let $M$ be some (possibly empty and possibly infinite) set of real numbers with the following properties:


*

*$M$ is a discrete subset of $\mathbb{R}$;

*if $a\in M$ and $a\notin \pi\mathbb{Z}$, then there exists an integer $m\neq 0$ such that $(m-1)\pi< a< m\pi$;

*if $a\in M$ and $a\in\pi\mathbb{Z}$ , then there exists an integer $m\neq -1$ such that $a= m\pi$;

*if $a,b\in M$ with $a<b$ are adjacent elements and $a\notin \pi\mathbb{Z}$, then  $a+b\geq 2\left\lceil\dfrac{a}{\pi}\right\rceil\pi$;

*if $a,b\in M$ with $a<b$ are adjacent elements and $a\in\pi\mathbb{Z}$, then $b-a\geq 2\pi$.


Define
$$f_{M}(x):=\left\{
\begin{array}{ll}
\frac{1}{2}\,\big(\cos(a)-\cos(x)\big)\,,&\text{if }a\leq x\leq 2\left\lceil\frac{a}{\pi}\right\rceil\pi-a\text{ for some }a\in M\setminus(\pi\mathbb{Z})\,,\\
\frac{1}{2}\,\big((-1)^m-\cos(x)\big)\,,&\text{if }m\pi\leq x\leq (m+2)\pi\text{ for some }m\pi\in M\cap (\pi\mathbb{Z})\,,\\
0\,,&\text{else}\,.
\end{array}
\right.$$ 
Then, $f:=f_{M}$ satisfies (*).
In fact, any solution $f$ to (*) satisfies $f=f_{M}$ for some $M\subseteq \mathbb{R}$ satisfying the five listed conditions.  If $f$ is required to be nonnegative, then we conclude that $f=f_{M}$ with $M$ containing only real numbers $a$ such that $2m\pi\leq a<(2m+1)\pi$ for some $m\in\mathbb{Z}$.  However, even without the nonnegativity requirement, there are two kinds of differentiable solution to (*)---the zero function $f\equiv0$ (corresponding to $M:=\emptyset$) and the function $f(x)=\dfrac{1-\cos(x)}{2}=\sin^2\left(\dfrac{x}{2}\right)$ for all $x\in\mathbb{R}$ (corresponding to $M:=2\pi\mathbb{Z}$).

As for the integral equation
$$\int_0^x\,\sin(t)\,f(t)\,\text{d}t=\big(f(x)\big)^2-1\text{ for all }x\in\mathbb{R}\,,$$
it can be seen that there exists a unique solution $f:\mathbb{R}\to\mathbb{R}$ regardless of the nonnegativity assumption or the differentiability requirement on $f$.  See gt6989b's answer for a good hint.
A: $$\int_0^x\sin(t)f(t)dt=\int\sin(x)f(x)dx$$
So if:
$$\int\sin(x)f(x)dx=f^2(x)$$
Then differentiate both sides:
$$\sin(x)f(x)=2f(x)f’(x) \therefore \, \sin(x)=2f’(x) \therefore \, f(x)=\int\frac{\sin(x)}{2}dx=-\frac{\cos(x)}{2}+C$$
