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

Feb 25, 2016

$\sqrt{16 + {x}^{2}} + \frac{16}{\sqrt{16 + {x}^{2}}}$

#### Explanation:

Remember that
${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

For the case is convenient that

$x = 4 \tan y$
$\mathrm{dx} = 4 {\sec}^{2} y \cdot \mathrm{dy}$

Then

$\int {x}^{3} / {\left(\sqrt{16 + {x}^{2}}\right)}^{3} \mathrm{dx} = \int \frac{64 {\tan}^{3} y}{\sqrt{16 + 16 {\tan}^{2} y}} ^ 3 \cdot 4 {\sec}^{2} y \cdot \mathrm{dy}$
$= \int \frac{256 {\tan}^{3} y \cdot {\sec}^{2} y}{4 \sec y} ^ 3 \mathrm{dy} = \int \frac{4 {\tan}^{3} y}{\sec} y \mathrm{dy}$
$= 4 \cdot \int {\sin}^{3} \frac{y}{\cos} ^ 3 y \cdot \cos y \cdot \mathrm{dy} = 4 \int \sin \frac{y}{\cos} ^ 2 y \cdot \left(1 - {\cos}^{2} y\right) \mathrm{dy}$
$= 4 \int \sin \frac{y}{\cos} ^ 2 y \cdot \mathrm{dy} - 4 \int \sin y \cdot \mathrm{dy}$ [A]

$\to \int \sin \frac{y}{\cos} ^ 2 y \cdot \mathrm{dy} =$
Making $\cos y = z$ and $- \sin y \cdot \mathrm{dy} = \mathrm{dz}$
$= - \int \frac{\mathrm{dz}}{z} ^ 2 = \frac{1}{z} = \frac{1}{\cos} y$

Inserting the result above in expression [A]:
$= \frac{4}{\cos} y + 4 \cos y + c o n s t .$ [B]

But

$4 \tan y = x$ => $\sin y = \frac{x}{4} \cdot \cos y$
$\to {\cos}^{2} y + {\sin}^{2} y = 1$ => ${\cos}^{2} y + {x}^{2} / 16 \cdot {\cos}^{2} y = 1$ => $\left(1 + {x}^{2} / 16\right) {\cos}^{2} y = 1$ => $\cos y = \sqrt{\frac{16}{16 + {x}^{2}}} = \frac{4}{\sqrt{16 + {x}^{2}}}$

Inserting the result above in expression [B]:

$= \frac{\cancel{4}}{\frac{\cancel{4}}{\sqrt{16 + {x}^{2}}}} + 4 \cdot \frac{4}{\sqrt{16 + {x}^{2}}} + c o n s t .$
$= \sqrt{16 + {x}^{2}} + \frac{16}{\sqrt{16 + {x}^{2}}} + c o n s t .$