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

Sep 10, 2016

$- \sqrt{16 - {x}^{2}} + C$

#### Explanation:

Although this is well set up for a shorter, non-trigonometric substitution (see: u=16-x^2), we can make the trigonometric substitution $x = 4 \sin \theta$. Note that this means that $\mathrm{dx} = 4 \cos \theta d \theta$.

Thus:

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

Note that $\sqrt{16 - 16 {\sin}^{2} \theta} = 4 \sqrt{1 - {\sin}^{2} \theta} = 4 \cos \theta$.

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

Integrating sine gives:

$= - 4 \cos \theta + C$

Write this in terms of sine:

$= - 4 \sqrt{1 - {\sin}^{2} \theta} + C$

Since $\sin \theta = \frac{x}{4}$:

$= - 4 \sqrt{1 - {x}^{2} / 16} + C = - 4 \sqrt{\frac{16 - {x}^{2}}{16}} + C$

Bringing the $\frac{1}{16}$ from the square root as a $\frac{1}{4}$:

$= - \sqrt{16 - {x}^{2}} + C$