How do you find the integral of int 3/sqrt(1-4x^2)dx∫3√1−4x2dx?

1 Answer
Andrea S.
Apr 1, 2018

int 3/sqrt(1-4x^2)dx = 3/2 arcsin(2x) + C∫3√1−4x2dx=32arcsin(2x)+C

Explanation:

Substitute:

x =1/2 sintx=12sint

dx = 1/2costdtdx=12costdt

with t in (-pi/2,pi/2)t∈(−π2,π2) as the integrand is defined only for abs (2x) < 1|2x|<1.

Then:

int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/sqrt(1-4(1/2sint)^2) ∫3√1−4x2dx=32∫costdt√1−4(12sint)2

int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/sqrt(1-sin^2t) ∫3√1−4x2dx=32∫costdt√1−sin2t

Now, for t in (-pi/2,pi/2)t∈(−π2,π2):

sqrt(1-sin^2t) = cost√1−sin2t=cost

so:

int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/cost∫3√1−4x2dx=32∫costdtcost

int 3/sqrt(1-4x^2)dx = 3/2 intdt∫3√1−4x2dx=32∫dt

int 3/sqrt(1-4x^2)dx = (3t)/2 + C∫3√1−4x2dx=3t2+C

and undoing the substitution:

int 3/sqrt(1-4x^2)dx = 3/2 arcsin(2x) + C∫3√1−4x2dx=32arcsin(2x)+C

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