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

Mar 5, 2016

The answer is: $I = - \frac{1}{3} \sqrt{16 - {x}^{2}} \cdot \left({x}^{2} + 32\right) + c$.

#### Explanation:

If we make this substitution:

$x = 4 \sin t \Rightarrow \mathrm{dx} = 4 \cos t \cdot \mathrm{dt}$,

then:

$\int \frac{64 {\sin}^{3} t}{\sqrt{16 - 16 {\sin}^{2} t}} \cdot 4 \cos t \cdot \mathrm{dt} =$

$= \int \frac{64 {\sin}^{3} t}{4 \sqrt{1 - {\sin}^{2} t}} \cdot 4 \cos t \cdot \mathrm{dt} =$

$= \int \frac{64 {\sin}^{3} t}{4 \cos t} \cdot 4 \cos t \cdot \mathrm{dt} = 64 \int {\sin}^{3} t \cdot \mathrm{dt} =$

$64 \int \sin t {\sin}^{2} t \cdot \mathrm{dt} = 64 \int \sin t \left(1 - {\cos}^{2} t\right) \mathrm{dt} =$

$64 \int \left(\sin t - \sin t {\cos}^{2} t\right) \mathrm{dt} = 64 \left(- \cos t + {\cos}^{3} \frac{t}{3}\right) + c = \left(1\right)$.

Now, remembering our substitution:

$\sin t = \frac{x}{4} \Rightarrow \cos t = \sqrt{1 - {\sin}^{2}} = \sqrt{1 - {x}^{2} / 16} = \frac{1}{4} \sqrt{16 - {x}^{2}}$.

So:

$\left(1\right) = 64 \left(- \frac{1}{4} \sqrt{16 - {x}^{2}} + \frac{1}{64} \cdot \frac{1}{3} \cdot \sqrt{16 - {x}^{2}} \cdot \left(16 - {x}^{2}\right)\right) + c =$

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

$= \frac{1}{3} \sqrt{16 - {x}^{2}} \left(- 48 + 16 - {x}^{2}\right) + c =$

$= - \frac{1}{3} \sqrt{16 - {x}^{2}} \cdot \left({x}^{2} + 32\right) + c$.