How do you integrate #int 1/sqrt(1-4x^2)# by trigonometric substitution?

1 Answer
George C.
Oct 12, 2016

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

Explanation:

Let #x = 1/2 sin theta#

Then:

#int 1/(sqrt(1-4x^2)) dx = int 1/sqrt(1 - sin^2 theta) (d 1/2 sin theta)/(d theta) d theta#

#color(white)(int 1/(sqrt(1-4x^2)) dx) = int (1/2 cos theta)/cos theta d theta#

#color(white)(int 1/(sqrt(1-4x^2)) dx) = int 1/2 d theta#

#color(white)(int 1/(sqrt(1-4x^2)) dx) = 1/2 theta + C#

#color(white)(int 1/(sqrt(1-4x^2)) dx) = 1/2 sin^(-1) (2x) + C#

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