Proof for differentiable functions Find all positive differentiable functions $f$ that satisfy $$\int_0^x \sin(t) f(t) dt = [f(x)]^2-1$$for all real numbers $x$.
So, it appears that this question has been asked in the past (at this link Find all positive differentiable functions $f$ that satisfy $\int_0^x \sin(t) f(t) dt = [f(x)]^2.$). However, given the hint that this user provided in the previous post, I believe I am still unsure how to go proceed. Also, as the previous poster said, this is the correct problem and I believed people answered about the incorrect problem. My reasoning is that if you differentiate both sides, you can get a function f such that it will only be positive. However, I’m not sure that this I’m fact works. Can someone please help? Thanks. (I’m also sorry about my previous posts - please don’t downvote anymore) 
 A: hint
Differentiate both sides to get
$$f(x)\sin(x)=2f(x)f'(x)$$
If $f(x)\ne 0$ then
$$f'(x)=\frac{\sin(x)}{2}$$ at a certain intervall.
$$f(x)=\frac{-\cos(x)}{2}+C$$
plugg it to get $C$.
A: @Maam Should be  $$\frac{\cos^2(x)}{4}-C\cos x-\frac{1}{4}+C=\frac{(\cos x)^2}{4}-C\cos x +C^2-1.$$ 
Did you use $u=-\cos(t) $ and integrate $ (\frac{u}{2} + C ) du $ ?
$C^2 -C -\frac{3}{4} $ so the only solution that works to make $f(x)$ positive is $C= \frac {3}{2}$ 
Note: $f(x) =0 $ is not a solution as its not a positive function.
A: (I use Fatima's solution)
Usually, we continue as follows:
$f(x)=-\frac{\cos x}{2} + C$, plug it in the given equation
$$\int_0^x \sin t \left(-\frac{\cos t}{2} + C\right) dt=\left(-\frac{\cos x}{2} + C\right)^2-1.$$
The integral is easy to compute, but there is a problem with $C$ which is assumed to be a constant. If I am not wrong, we get $$\frac{\cos(2x)}{8}-C\cos x-\frac{1}{8}+C=\frac{(\cos x)^2}{4}-C\cos x +C^2-1.$$ 
It seems to me that the equation does not have other solution than $f(x)=0.$
