# How do you integrate int x^3/sqrt(16-x^2) by trigonometric substitution?

Feb 4, 2017

The integral is ${\left(16 - {x}^{2}\right)}^{\frac{3}{2}} / 3 - 16 \sqrt{16 - {x}^{2}} + C$

#### Explanation:

Apply the substitution $x = 4 \sin \theta$. Then $\mathrm{dx} = 4 \cos \theta d \theta$.

$\implies 4 \int {\left(4 \sin \theta\right)}^{3} / \sqrt{16 - {\left(4 \sin \theta\right)}^{2}} \cdot \cos \theta d \theta$

$\implies 4 \int \frac{64 {\sin}^{3} \theta}{\sqrt{16 - 16 {\sin}^{2} \theta}} \cdot \cos \theta d \theta$

$\implies 4 \int \frac{64 {\sin}^{3} \theta}{\sqrt{16 \left(1 - {\sin}^{2} \theta\right)}} \cdot \cos \theta d \theta$

$\implies 4 \int \frac{64 {\sin}^{3} \theta}{\sqrt{16 {\cos}^{2} \theta}} \cdot \cos \theta d \theta$

$\implies 4 \int \frac{64 {\sin}^{3} \theta}{4 \cos \theta} \cdot \cos \theta d \theta$

$\implies 4 \int \frac{64 {\sin}^{3} \theta}{4} d \theta$

$\implies 4 \int 16 {\sin}^{3} \theta d \theta$

$\implies 64 \int \sin \theta \left({\sin}^{2} \theta\right) d \theta$

$\implies 64 \int \sin \theta \left(1 - {\cos}^{2} \theta\right) d \theta$

Let $u = \cos \theta$. Then $\mathrm{du} = - \sin \theta d \theta$ and $d \theta = - \frac{\mathrm{du}}{\sin} \theta$.

$\implies - 64 \int \sin \theta \left(1 - {u}^{2}\right) \cdot \frac{\mathrm{du}}{\sin} \theta$

$\implies - 64 \int 1 - {u}^{2} \mathrm{du}$

$\implies - 64 \left(u - \frac{1}{3} {u}^{3}\right) + C$

$\implies - 64 u + \frac{64}{3} {u}^{3} + C$

Reverse the substitution.

$\implies - 64 \cos \theta + \frac{64}{3} {\cos}^{3} \theta + C$

Since the initial substitution was $\sin \theta = \frac{x}{4}$, then $\cos \theta = \frac{\sqrt{16 - {x}^{2}}}{4}$.

$\implies - \frac{64 \sqrt{16 - {x}^{2}}}{4} + \frac{64}{3} {\left(\frac{\sqrt{16 - {x}^{2}}}{4}\right)}^{3} + C$

$\implies - 16 \sqrt{16 - {x}^{2}} + {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} / 3 + C$

Hopefully this helps!

Feb 4, 2017

$\int {x}^{3} / \left(\sqrt{16 - {x}^{2}}\right) \mathrm{dx} =$
$\frac{1}{3} {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} - 16 \sqrt{16 - {x}^{2}} + \text{C}$

#### Explanation:

$\int {x}^{3} / \left(\sqrt{16 - {x}^{2}}\right) \mathrm{dx}$

If we look at the denominator, it looks like a rearrangement of Pythagoras' theorem, ${a}^{2} + {b}^{2} = {c}^{2} \rightarrow {c}^{2} - {b}^{2} = {a}^{2}$

So the length of the hypotenuse of our triangle is $4$ and the length of the opposite is $x$. This means that the length of the adjacent is $\sqrt{16 - {x}^{2}}$.

If we rewrite $x$ using trigonometric ratios, we get $\sin \theta = \frac{x}{4} \rightarrow x = 4 \sin \theta$

We can also rewrite the adjacent in the same way:

$\cos \theta = \frac{\sqrt{16 - {x}^{2}}}{4}$

$\sqrt{16 - {x}^{2}} = 4 \cos \theta$

If we differentiate $x$, we get:

$\frac{\mathrm{dx}}{d \theta} = 4 \cos \theta$

$\mathrm{dx} = 4 \cos \theta d \theta$

$\int {x}^{3} / \left(\sqrt{16 - {x}^{2}}\right) \mathrm{dx} = \int {x}^{3} \frac{4 \cos \theta}{4 \cos \theta} d \theta = \int {x}^{3} d \theta$

Let's rewrite ${x}^{3}$ using our $x$ function:

$x = 4 \sin \theta$

${x}^{3} = {\left(4 \sin \theta\right)}^{2} = 64 {\sin}^{3} \theta$

Now our integral is $\int 64 {\sin}^{3} \theta d \theta = 64 \int {\sin}^{3} \theta d \theta$.

$= 64 \int {\sin}^{2} \theta \sin \theta d \theta = 64 \int \sin \theta \left(1 - {\cos}^{2} \theta\right) d \theta$

$u = \cos \theta$

$\frac{\mathrm{du}}{d \theta} = - \sin \theta$

$d \theta = - \left(\frac{1}{\sin} \theta\right) \mathrm{du}$

$64 \int \sin \theta \left(1 - {\cos}^{2} \theta\right) d \theta = - 64 \int \cancel{\sin} \theta \left(1 - {u}^{2}\right) \cancel{\frac{1}{\sin} \theta} \mathrm{du}$

$= - 64 \int \left(1 - {u}^{2}\right) \mathrm{du} = - 64 \left(u - \frac{1}{3} {u}^{3}\right) + \text{C}$

Substitute $u$ for $\cos \theta$ and $\cos \theta$ for $\frac{\sqrt{16 - {x}^{2}}}{4}$

$- 64 \left(u - \frac{1}{3} {u}^{3}\right) + \text{C"=-64(sqrt(16-x^2)/4-(1/3)(16-x^2)^(3/2)/64)+"C}$

$= \frac{1}{3} {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} - 16 \sqrt{16 - {x}^{2}} + \text{C}$

Feb 9, 2017

Contd.

#### Explanation:

We solve the Problem is required to be solved using trigo. substn.,