# How do you integrate int x^2sqrt(16-x^2) using trig substitutions?

Sep 5, 2016

$32 \arcsin \left(\frac{x}{4}\right) - 8 \sin \left(4 \arcsin \left(\frac{x}{4}\right)\right) + C$

#### Explanation:

Solving for:

$I = \int {x}^{2} \sqrt{16 - {x}^{2}} \mathrm{dx}$

Make the substitution $x = 4 \sin \theta$ where $\mathrm{dx} = 4 \cos \theta d \theta$:

$I = \int 16 {\sin}^{2} \theta \sqrt{16 - 16 {\sin}^{2} \theta} 4 \cos \theta d \theta$

Bringing all the multiplicative constants out, including the $\sqrt{16} = 4$:

$I = 256 \int {\sin}^{2} \theta \cos \theta \sqrt{1 - {\sin}^{2} \theta} d \theta$

Since ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$, we see that $\cos \theta = \sqrt{1 - {\sin}^{2} \theta}$:

$I = 256 \int {\sin}^{2} \theta {\cos}^{2} \theta d \theta$

Here, since $\sin 2 \theta = 2 \sin \theta \cos \theta$, we see that ${\sin}^{2} 2 \theta = 4 {\sin}^{2} \theta {\cos}^{2} \theta$:

$I = 64 \int 4 {\sin}^{2} \theta {\cos}^{2} \theta d \theta$

$I = 64 \int {\sin}^{2} 2 \theta d \theta$

Note that, according to the cosine double angle formula, $\cos 4 \theta = 1 - 2 {\sin}^{2} 2 \theta$, so ${\sin}^{2} 2 \theta = \frac{1 - \cos 4 \theta}{2}$:

$I = 32 \int \left(1 - \cos 4 \theta\right) d \theta$

$I = 32 \int d \theta - 32 \int \cos 4 \theta d \theta$

Integrating both (substitution can be used for the second integral, or just see it for the reverse chain rule of $\sin 4 \theta$ that it is):

$I = 32 \theta - 8 \int 4 \cos 4 \theta d \theta$

$I = 32 \theta - 8 \sin 4 \theta$

Since $x = 4 \sin \theta$, solving for $\theta$ gives $\theta = \arcsin \left(\frac{x}{4}\right)$ and $4 \theta = 4 \arcsin \left(\frac{x}{4}\right)$. Thus:

$I = 32 \arcsin \left(\frac{x}{4}\right) - 8 \sin \left(4 \arcsin \left(\frac{x}{4}\right)\right) + C$