find f, if f ' (x) = 4/sqrt(1-x^2)?

Question

find f,
f ' (x) = 4/(1-x^2)^(1/2)
and
f(1/2) = 1

I think it is 4*arcsin(x) + C,
but I am cofused about how to figure C.

Answer 1

The way you find the constant of integration is by plugging in the given x-value, setting it equal to the given value of the function, and solving for C. Thus

f(1/2) = 4 arcsin (1/2) + C = 1
4π/6 + C = 1
C=1-2π/3

Your antiderivative is perfect, BTW.

Answer 2

f ' (x) = 4/(1-x^2)^(1/2)

x = sin z
dx = cos z dz
dz/dx = 1/cos z ---(1)

df/dx = 4/(1-x^2)^(1/2)
df/dz . dz/dx = 4/(1-sin^2 z)^(1/2)
from (1)
df/dz . (1/cos z) = 4/(1-sin^2 z)^(1/2)
df/dz . (1/cos z) = 4/cos z
df/dz = 4
df = 4 dz
f(z) = 4z + A where A is constant

so,
f(x) = 4*arcsin(x) + C where C is constant

f(1/2)=1
4*arcsin(1/2) + C = 1
4*0.5236 + C =1
2.0944 + C = 1
C = -1.0944

f(x) = 4*arcsin(x) -1.0944