# How do you find the antiderivative of int xsqrt(1-x^2) dx?

Feb 10, 2017

The answer is $= - \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

#### Explanation:

We perform this by substitution

Let $u = 1 - {x}^{2}$

$\mathrm{du} = - 2 x \mathrm{dx}$

$x \mathrm{dx} = - \frac{1}{2} \mathrm{du}$

Therefore,

$\int x \sqrt{1 - {x}^{2}} \mathrm{dx} = - \frac{1}{2} \int \sqrt{u} \mathrm{du}$

$= - \frac{1}{2} \cdot {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)$

$= - \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

Feb 10, 2017

$- \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

#### Explanation:

Here's an alternative answer using trig substitution.

Let $x = \sin \theta$ . Then $\mathrm{dx} = \cos \theta d \theta$.

$= \int \sin \theta \sqrt{1 - {\sin}^{2} \theta} \cdot \cos \theta d \theta$

$= \int \sin \theta \sqrt{{\cos}^{2} \theta} \cdot \cos \theta d \theta$

$= \int \sin \theta \cos \theta \cos \theta d \theta$

$= \int \sin \theta {\cos}^{2} \theta d \theta$

Now make a substitution. Let $u = \cos \theta$, then $\mathrm{du} = - \sin \theta d \theta \to d \theta = \frac{\mathrm{du}}{- \sin \theta}$.

$= \int \sin \theta {\cos}^{2} \theta \cdot \frac{\mathrm{du}}{- \sin \theta}$

$= - \int {u}^{2} \mathrm{du}$

$= - \frac{1}{3} {u}^{3} + C$

$= - \frac{1}{3} {\cos}^{3} \theta + C$

We know from our initial substitution that $\frac{x}{1} = \sin \theta$. This means that the side adjacent $\theta$ in our imaginary triangle measures $\sqrt{1 - {x}^{2}}$. Therefore, $\cos \theta = \sqrt{1 - {x}^{2}}$

$= - \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

Hopefully this helps!