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

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Answer 1 · 2018-04-01T15:31:36

#int 3/sqrt(1-4x^2)dx = 3/2 arcsin(2x) + C#

Substitute:

#x =1/2 sint#

#dx = 1/2costdt#

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

Then:

#int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/sqrt(1-4(1/2sint)^2) #

#int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/sqrt(1-sin^2t) #

Now, for #t in (-pi/2,pi/2)#:

#sqrt(1-sin^2t) = cost#

so:

#int 3/sqrt(1-4x^2)dx = 3/2 int (costdt)/cost#

#int 3/sqrt(1-4x^2)dx = 3/2 intdt#

#int 3/sqrt(1-4x^2)dx = (3t)/2 + C#

and undoing the substitution:

#int 3/sqrt(1-4x^2)dx = 3/2 arcsin(2x) + C#