Find solutions for $f'(\sin x) f(\cos x)=\sin x$ We are given $$f'(\sin x) f(\cos x)=\sin x$$ and we need to find out the function which respects this condition. The answer is $$f(x)=\sqrt{c}e^{\frac{1}{2c}(x^2-\frac{1}{2})}$$
I tried to expand the initial condition just as on the other problems, but I can't seem to be getting anywhere, and having trigonometrical functions doesn't make it better. The closest I've gotten to the answer is through an integration after which nothing else works.
 A: The given problem, find the function $t\mapsto f(t)$ satisfying
$$f'(\sin x)\>f(\cos x)=\sin x\qquad\forall\, x\ ,\tag{1}$$
does not belong to Calculus 101. What we have here is not an ODE, but a functional equation. An ODE involves an independent variable $t$ and an unknown function $f(t)$, whereby a relation between  variable $t$ and the values of $f$, $f'$ at this same point $t$ is stipulated:
$$\Phi\bigl(t, f(t),f'(t)\bigr)=0\qquad\forall\, t\ .$$ But in $(1)$ the values of $f$ and $f'$ at different points (related somehow) are involved. This is another story.
One has to be aware that a given functional equation need not have solutions that are expressible in terms of elementary functions. When an $f$ satisfies an interesting functional equation this is a miracle, and such a function appears in all mathematical catalogues. A simpler functional equation is $h(2x)-2h^2(x)+1\equiv0$. I leave it to you to find a solution.
As a rule, there is no standard method to attack something like $(1)$. One has to try various "Ansätze" in order to see whether something works. Of course $\sin^2 x+\cos^2 x\equiv1$ plays a rôle here. After many trials one might try
$$f(t):=e^{a t^2 +b}\tag{2}$$
for suitably chosen constants $a$ and $b$. Introducing this "Ansatz" into $(1)$ one first computes
$$f'(t)=2at \, e^{a t^2 +b}$$
and then obtains the condition
$$2a\sin x\exp\bigl(a\sin^2 x+b) \cdot\exp(a\cos^2 x+ b)\equiv \sin x\qquad\forall\, x\ .$$
This is fulfilled when
$$2a \exp(a+2b)=1\ ,\tag{3}$$
so that we may choose $a>0$ arbitrarily and then $b$ according to $(3)$. The resulting functions $(2)$ are the same as the functions listed in your source.
A: Setting $f(u) = e^{g(u)}$ makes the equation become $e^{g(\sin x)+g(\cos x)} g'(\sin x) = \sin x.$ Knowing how difficult it is to get rid of an exponential function suggest that $e^{g(\sin x)+g(\cos x)}$ should equal some constant $c>0$ and thus $g'(\sin x) = c^{-1}\sin x.$ The latter equation implies that we should have $g'(u) = c^{-1}u$ for $u\in [-1, 1].$ This is a differential equation with solutions $g(u) = \frac12 c^{-1}u^2 + b$ where $b$ is some constant. Inserting this into $e^{g(\sin x)+g(\cos x)}=c$ gives us $e^{\frac12 c^{-1} + 2b} = c,$ i.e. $e^b = \sqrt{c} e^{-\frac14 c^{-1}}.$ Thus our solutions are
$$
f(u) 
= e^{g(u)}
= e^{\frac12 c^{-1}u^2 + b}
= e^{\frac12 c^{-1}u^2} e^{b}
= e^{\frac12 c^{-1}u^2} \sqrt{c} e^{-\frac14 c^{-1}}
= \sqrt{c} e^{\frac14 c^{-1} (2u^2-1)}
.
$$
