# How do you integrate int x / sqrt(-x^2+8x) dx using trigonometric substitution?

Sep 25, 2016

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

#### Explanation:

Complete the square in the denominator:

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

Let $x - 4 = 4 \sin \theta$. Thus, $\mathrm{dx} = 4 \cos \theta d \theta$. Also note that $x = 4 \sin \theta + 4$. Substituting in:

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

Factoring out both of the $4$ terms and the $\sqrt{16}$ from the denominator:

$\frac{16}{\sqrt{16}} \int \frac{\left(\sin \theta + 1\right) \cos \theta}{\sqrt{1 - {\sin}^{2} \theta}} d \theta$

Note that ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$, so $\sqrt{1 - {\sin}^{2} \theta} = \cos \theta$:

$= 4 \int \left(\sin \theta + 1\right) d \theta$

$= - 4 \cos \theta + 4 \theta + C$

From $x - 4 = 4 \sin \theta$, we see that $\theta = \arcsin \left(\frac{x - 4}{4}\right)$. Furthermore, we see that $\sin \theta = \frac{x - 4}{4}$. Now, use $\cos \theta = \sqrt{1 - {\sin}^{2} \theta}$:

$\cos \theta = \sqrt{1 - {\left(\frac{x - 4}{4}\right)}^{2}} = \sqrt{\frac{16 - {\left(x - 4\right)}^{2}}{16}} = \frac{1}{4} \sqrt{- {x}^{2} - 8 x}$

Thus the integral equals:

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

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