How do you integrate ${(1 - x^{2})}^{.5}$ $(1-x^2)^.5$ ?

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Answer 1 · 2016-07-04T18:26:14

$= \frac{1}{2} x \sqrt{1 - x^{2}} + \frac{arcsin x}{2} + C$ $= 1/2 x sqrt(1-x^2) + arcsin x/2 + C$

$int dx qquad sqrt(1-x^2)$

trig sub $x = sin phi, dx = cos phi dphi$

$implies int dphi qquad cos phi sqrt(1-sin^2 phi)$

$= int dphi qquad cos^2 phi$

double angle $cos 2A = 2 cos^2 A - 1$

$= int dphi qquad (cos 2 phi + 1)/2$

$= 1/4 sin 2 phi + phi/2 + C$

double angle $sin 2A = 2 sin A cos A$

$= 1/2 x sqrt(1-x^2) + arcsin x/2 + C$

How do you integrate (1-x^2)^.5(1−x2).5 (1-x^2)^.5?