# How do you integrate int x^2/sqrt(16-x^2) by trigonometric substitution?

Jun 12, 2018

$8 {\sin}^{-} 1 \left(\frac{x}{4}\right) - \frac{x \sqrt{16 - {x}^{2}}}{2} + C$

#### Explanation:

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

Let's use the substitution $x = 4 \sin \theta$. This substitution implies that couple important things:

• ${x}^{2} = 16 {\sin}^{2} \theta$
• $\sqrt{16 - {x}^{2}} = \sqrt{16 - 16 {\sin}^{2} \theta} = 4 \sqrt{1 - {\sin}^{2} \theta} = 4 \cos \theta$
• $\mathrm{dx} = 4 \cos \theta d \theta$

Now, let's substitute these back into our integral:

$I = \int \frac{16 {\sin}^{2} \theta}{4 \cos \theta} \left(4 \cos \theta d \theta\right) = 16 \int {\sin}^{2} \theta d \theta$

To solve this, I like to use the identity ${\sin}^{2} \theta = \frac{1}{2} \left(1 - \cos 2 \theta\right)$. If you've never seen this before, remember that $\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta$, which implies that $\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$, from which you can solve for ${\sin}^{2} \theta$ to get the identity from before. Using the identity in the integral:

$I = 16 \int \frac{1}{2} \left(1 - \cos 2 \theta\right) d \theta = 8 \left(\theta - \frac{1}{2} \sin 2 \theta\right) + C$

We need to get our final answer back in terms of $x$ instead of $\theta$. Use $\sin 2 \theta = 2 \sin \theta \cos \theta$:

$I = 8 \theta - 8 \sin \theta \cos \theta + C$

Recall we started with $x = 4 \sin \theta$. Thus:

• $\sin \theta = \frac{x}{4}$
• $\cos \theta = \sqrt{1 - {\sin}^{2} \theta} = \sqrt{1 - {x}^{2} / 16} = \frac{\sqrt{16 - {x}^{2}}}{4}$
• $\theta = {\sin}^{-} 1 \left(\frac{x}{4}\right)$

And finally:

$I = 8 {\sin}^{-} 1 \left(\frac{x}{4}\right) - 8 \left(\frac{x}{4}\right) \left(\frac{\sqrt{16 - {x}^{2}}}{4}\right) + C$

$I = 8 {\sin}^{-} 1 \left(\frac{x}{4}\right) - \frac{x \sqrt{16 - {x}^{2}}}{2} + C$

Jun 12, 2018

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 8 \arcsin \left(\frac{x}{4}\right) - \frac{x \sqrt{16 - {x}^{2}}}{2} + C$

#### Explanation:

Let $x = 4 \sin t$ with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, $\mathrm{dx} = 4 \cos t \mathrm{dt}$, then:

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = \int \frac{{\left(4 \sin t\right)}^{2} \left(4 \cos t \mathrm{dt}\right)}{\sqrt{16 - 16 {\sin}^{2} t}}$

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 16 \int \frac{{\sin}^{2} t \cos t \mathrm{dt}}{\sqrt{1 - {\sin}^{2} t}}$

For $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the cosine is positive, so:

$\sqrt{1 - {\sin}^{2} t} = \sqrt{{\cos}^{2} t} = \cos t$

then:

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 16 \int \frac{{\sin}^{2} t \cos t \mathrm{dt}}{\cos} t = 16 \int {\sin}^{2} t \mathrm{dt}$

Use now the trigonometric identity:

$2 {\sin}^{2} \alpha = 1 - \cos \left(2 \alpha\right)$

to have:

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 8 \int \left(1 - \cos 2 t\right) \mathrm{dt}$

and based on the linearity of the integral:

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 8 \int \mathrm{dt} - 4 \int \cos 2 t d \left(2 t\right)$

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 8 t - 4 \sin 2 t + C$

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 8 t - 8 \sin t \cos t + C$

and undoing the substitution:

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 8 \arcsin \left(\frac{x}{4}\right) - 2 x \sqrt{1 - {\left(\frac{x}{4}\right)}^{2}} + C$

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx} = 8 \arcsin \left(\frac{x}{4}\right) - \frac{x \sqrt{16 - {x}^{2}}}{2} + C$