# How do you integrate int sqrt(-x^2-10x)/xdx using trigonometric substitution?

Mar 20, 2018

Use the substitution $x + 5 = 5 \sin \theta$.

#### Explanation:

Let

$I = \int \frac{\sqrt{- {x}^{2} - 10 x}}{x} \mathrm{dx}$

Complete the square in the square root:

$I = \int \frac{\sqrt{25 - {\left(x + 5\right)}^{2}}}{x} \mathrm{dx}$

Apply the substitution $x + 5 = 5 \sin \theta$:

$I = \int \frac{5 \cos \theta}{5 \sin \theta - 5} \left(5 \cos \theta d \theta\right)$

Simplify:

$I = 5 \int {\cos}^{2} \frac{\theta}{\sin \theta - 1} d \theta$

Apply the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$:

$I = 5 \int \frac{1 - {\sin}^{2} \theta}{\sin \theta - 1} d \theta$

Apply the difference of squares ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$:

$I = - 5 \int \left(1 + \sin \theta\right) d \theta$

Integrate directly:

$I = - 5 \left(\theta - \cos \theta\right) + C$

Reverse the substitution:

$I = \sqrt{25 - {\left(x + 5\right)}^{2}} - 5 {\sin}^{- 1} \left(\frac{x + 5}{5}\right) + C$

Mar 20, 2018

The answer is $= - 5 \arcsin \left(\frac{1}{5} \left(x + 5\right)\right) + 5 \sqrt{1 - {\left(\frac{x + 5}{5}\right)}^{2}} + C$

#### Explanation:

Complete the square :

$- {x}^{2} - 10 x = 25 - {\left(x + 5\right)}^{2}$

Therefore, the integral is

$I = \int \frac{\sqrt{- {x}^{2} - 10 x} \mathrm{dx}}{x} = \int \frac{\sqrt{25 - {\left(x + 5\right)}^{2}} \mathrm{dx}}{x}$

Let $u = x + 5$, $\implies$, $\mathrm{du} = \mathrm{dx}$

$I = \int \frac{\sqrt{25 - {u}^{2}} \mathrm{du}}{u - 5}$

Let $u = 5 \sin v$, $\implies$, $\mathrm{du} = 5 \cos v \mathrm{dv}$

Therefore,

$I = \int \frac{\left(5 \cos v\right) \cdot 5 \cos v \mathrm{dv}}{5 \left(\sin v - 1\right)}$

$= 5 \int \frac{{\cos}^{2} v \mathrm{dv}}{- \left(1 - \sin v\right)}$

$= - 5 \int \left(1 + \sin v\right) \mathrm{dv}$

$= - 5 v + 5 \cos v$

$= - 5 \arcsin \left(\frac{u}{5}\right) + 5 \sqrt{1 - {\left(\frac{u}{5}\right)}^{2}}$

$= - 5 \arcsin \left(\frac{1}{5} \left(x + 5\right)\right) + 5 \sqrt{1 - {\left(\frac{x + 5}{5}\right)}^{2}} + C$