<s>[INST] Inverse Trig Functions 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
 [/INST] 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:
$f(x) = 4\arcsin(x) +  1 - 4(\pi/6)$
</s>