Inverse Trig Functions

Question

Find $f(x)$ if $f'(x)=4/\sqrt{1-x^2}$ and $f(1/2)=1$

So far I have integrated $f'(x)$ and have found: $$f(x) =y = 4\arcsin(x), x=4\sin(y)$$

$$1/2=4\sin(1)$$ $$1/2=4(\pi/2)$$ $$1/2=2\pi$$

So is $f(x)=1/2$ or $2\pi$?

Thanks

Answer 1

If $f'(x) = \frac{4}{\sqrt{1-x^2}}$ then $f(x) = 4\arcsin(x) + c$, now if $f(1/2) = 1$ we have: $4\arcsin(1/2) + c = 1 \Rightarrow c = 1 - 4(\pi/6)$, therefore:

$f(x) = 4\arcsin(x) + 1 - 4(\pi/6)$

Answer 2

$f\left(x\right)=4\arcsin\left(x\right)+C$ where $c$ is a constant to be determinated. Taking $x=\frac{1}{2}$ , one obtains $$1=f\left(\frac{1}{2}\right)=4\arcsin\left(\frac{1}{2}\right)+C=4\frac{\pi}{6}+C$$ whence $C=1-\frac{2\pi}{3}$ .