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

Question

Cameron Alexander · Accepted Answer

#= 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 #

Charles Arocha · Answer

undefined To integrate $ \sqrt{1 - x^2} $, you can use a trigonometric substitution. Let $ x = \sin(\theta) $. Then, $ dx = \cos(\theta) \, d\theta $.

$$ \int \sqrt{1 - x^2} \, dx = \int \sqrt{1 - \sin^2(\theta)} \cdot \cos(\theta) \, d\theta $$

Using the Pythagorean identity $ \cos^2(\theta) = 1 - \sin^2(\theta) $, we have $ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} $.

$$ \int \sqrt{1 - x^2} \, dx = \int \cos^2(\theta) \, d\theta $$

Now, recall that $ \cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta)) $.

$$ \int \sqrt{1 - x^2} \, dx = \frac{1}{2} \int (1 + \cos(2\theta)) \, d\theta $$

Integrating term by term, you get:

$$ \int \sqrt{1 - x^2} \, dx = \frac{1}{2} (\theta + \frac{1}{2} \sin(2\theta)) + C $$

Finally, substitute back $ x = \sin(\theta) $ and simplify the expression:

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