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

Feb 27, 2017

$\frac{1}{8} {\sin}^{-} 1 \left(x\right) + \frac{1}{8} x \left(2 {x}^{2} - 1\right) \sqrt{1 - {x}^{2}} + C$

#### Explanation:

$I = \int {x}^{2} \sqrt{1 - {x}^{2}} \mathrm{dx}$

Let $x = \sin \left(\theta\right)$. This implies that $\mathrm{dx} = \cos \left(\theta\right) d \theta$. Then:

$I = \int {\sin}^{2} \left(\theta\right) \sqrt{1 - {\sin}^{2} \left(\theta\right)} \left(\cos \left(\theta\right) d \theta\right)$

$I = \int {\sin}^{2} \left(\theta\right) {\cos}^{2} \left(\theta\right) d \theta$

Now let ${\cos}^{2} \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)$:

$I = \int {\sin}^{2} \left(\theta\right) \left(1 - {\sin}^{2} \left(\theta\right)\right) d \theta$

$I = \int {\sin}^{2} \left(\theta\right) d \theta - \int {\sin}^{4} \left(\theta\right) d \theta$

Now use the identity ${\sin}^{2} \left(\alpha\right) = \frac{1}{2} \left(1 - \cos \left(2 \alpha\right)\right)$:

$I = \frac{1}{2} \int \left(1 - \cos \left(2 \theta\right)\right) d \theta - \int {\left(\frac{1}{2} \left(1 - \cos \left(2 \theta\right)\right)\right)}^{2} d \theta$

$I = \frac{1}{2} \int d \theta - \frac{1}{2} \int \cos \left(2 \theta\right) d \theta - \frac{1}{4} \int \left(1 - 2 \cos \left(2 \theta\right) + {\cos}^{2} \left(2 \theta\right)\right) d \theta$

Use the identity ${\cos}^{2} \left(\alpha\right) = \frac{1}{2} \left(1 + \cos \left(2 \alpha\right)\right)$:

$I = \frac{1}{2} \int d \theta - \frac{1}{2} \int \cos \left(2 \theta\right) d \theta - \frac{1}{4} \int d \theta + \frac{1}{2} \int \cos \left(2 \theta\right) d \theta - \frac{1}{8} \int \left(1 + \cos \left(4 \theta\right)\right) d \theta$

$I = \frac{1}{4} \int d \theta - \frac{1}{8} \int d \theta - \frac{1}{8} \int \cos \left(4 \theta\right) d \theta$

Solving through a substitution:

$I = \frac{1}{8} \theta - \frac{1}{32} \sin \left(4 \theta\right)$

We now have to return to $x$ from $\theta$. Recall that $x = \sin \left(\theta\right)$. Thus $\theta = {\sin}^{-} 1 \left(x\right)$. Also, use the identity $\sin \left(2 \alpha\right) = 2 \sin \left(\alpha\right) \cos \left(\alpha\right)$ to break down $\sin \left(4 \theta\right)$:

$I = \frac{1}{8} {\sin}^{-} 1 \left(x\right) - \frac{1}{16} \sin \left(2 \theta\right) \cos \left(2 \theta\right)$

Reusing the sine double-angle identity and also applying $\cos \left(2 \alpha\right) = 1 - 2 {\sin}^{2} \left(\alpha\right)$:

$I = \frac{1}{8} {\sin}^{-} 1 \left(x\right) - \frac{1}{8} \sin \left(\theta\right) \cos \left(\theta\right) \left(1 - 2 {\sin}^{2} \left(\theta\right)\right)$

Flipping the order of the terms in the last parenthesis with the preceding negative sign and using $\cos \left(x\right) = \sqrt{1 - {\sin}^{2} \left(x\right)}$:

$I = \frac{1}{8} {\sin}^{-} 1 \left(x\right) + \frac{1}{8} \sin \left(\theta\right) \sqrt{1 - {\sin}^{2} \left(\theta\right)} \left(2 {\sin}^{2} \left(\theta\right) - 1\right)$

Finally using $x = \sin \left(\theta\right)$:

$I = \frac{1}{8} {\sin}^{-} 1 \left(x\right) + \frac{1}{8} x \left(2 {x}^{2} - 1\right) \sqrt{1 - {x}^{2}} + C$