# \intx^3\sqrt(1-x^2)dx?

## I tried to get the answer found on Symbolab, but instead I got: $- \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + \frac{1}{5} {\left(1 - {x}^{2}\right)}^{\frac{5}{2}} + C$

Apr 13, 2018

#### Explanation:

You're going to have to use trig substitution for this question.
Let $x = \sin \theta$. Then $\mathrm{dx} = \cos \theta d \theta$.

$I = \int {\sin}^{3} \theta \sqrt{1 - {\sin}^{2} \theta} \cos \theta d \theta$

$I = \int {\sin}^{3} \theta \sqrt{{\cos}^{2} \theta} \cos \theta d \theta$

$I = \int {\sin}^{3} \theta {\cos}^{2} \theta d \theta$

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

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

$I = \int \sin \theta {\cos}^{2} \theta - \sin \theta {\cos}^{4} \theta d \theta$

$I = \int \sin \theta {\cos}^{2} \theta d \theta - \int \sin \theta {\cos}^{4} \theta d \theta$

Let $u = \cos \theta$. You should be left with $\mathrm{du} = - \sin \theta d \theta$ and $d \theta = \frac{\mathrm{du}}{- \sin \theta}$.

$I = - \int {u}^{2} \mathrm{du} + \int {u}^{4} \mathrm{du}$

$I = - \frac{1}{3} {u}^{3} + \frac{1}{5} {u}^{5} + C$

$I = - \frac{1}{3} {\cos}^{3} \theta + \frac{1}{5} {\cos}^{5} \theta + C$

Recall from our initial substitution that $x = \sin \theta$, and since ${\cos}^{2} x + {\sin}^{2} x = 1$, we can see that $\cos \theta = \sqrt{1 - {x}^{2}}$.

$I = \frac{1}{5} {\left(1 - {x}^{2}\right)}^{\frac{5}{2}} - \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

I checked and our answer is the same as the one shown on symbolab, except ours is simplified a little further.

Hopefully this helps!