Proof of an integral property $$\int_0^1  f(x)g'(x)dx=\pi$$ If $f(1)g(1)=f(0)g(0)$ then $$\int_0^1 f'(x)g(x)dx= -\pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from  $f(1)g(1)$ to  $f(0)g(0)$

Hint:
$$\dfrac{d(f(x)\cdot g(x))}{dx}=?$$
Integrate both sides with respect to $x$ between $[0,1]$
