How do you integrate #(1-x^2)^.5#?

1 Answer
George C.
Oct 12, 2016

#int (1-x^2)^0.5 dx = 1/2 x sqrt(1-x^2) + 1/2 sin^(-1)x + C#

Explanation:

Let #x = sin theta#

Then:

#int (1-x^2)^0.5 dx = int (1-sin^2 theta)^0.5 ((d sin theta)/(d theta)) d theta#

#color(white)(int (1-x^2)^0.5 dx) = int (cos^2 theta)^0.5 (cos theta) d theta#

#color(white)(int (1-x^2)^0.5 dx) = int cos^2 theta d theta#

#color(white)(int (1-x^2)^0.5 dx) = int 1/2(cos 2 theta + 1) d theta#

#color(white)(int (1-x^2)^0.5 dx) = 1/4 sin 2 theta + theta/2 + C#

#color(white)(int (1-x^2)^0.5 dx) = 1/2 sin theta cos theta + theta/2 + C#

#color(white)(int (1-x^2)^0.5 dx) = 1/2 x sqrt(1-x^2) + 1/2 sin^(-1)x + C#

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