# How do you integrate int sqrt(5-4x-x^2) dx using trigonometric substitution?

Sep 6, 2016

$\frac{\left(x + 2\right) \sqrt{5 - 4 x - {x}^{2}} + 9 \arcsin \left(\frac{x + 2}{3}\right)}{2} + C$

#### Explanation:

We have:

$I = \int \sqrt{5 - 4 x - {x}^{2}} \mathrm{dx} = \int \sqrt{- {\left(x + 2\right)}^{2} + 9} \mathrm{dx}$

From here, let $3 \sin \theta = x + 2$. Thus, $3 \cos \theta d \theta = \mathrm{dx}$. Plugging these in:

$I = 3 \int \cos \theta \sqrt{- 9 {\sin}^{2} \theta + 9} d \theta$

Factoring a $\sqrt{9} = 3$ from the square root:

$I = 9 \int \cos \theta \sqrt{- {\sin}^{2} \theta + 1} d \theta$

Recall that since ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$, we know that $\sqrt{- {\sin}^{2} \theta + 1} = \cos \theta$.

$I = 9 \int {\cos}^{2} \theta d \theta$

Recall that $\cos 2 \theta = 2 {\cos}^{2} \theta - 1$. Thus, ${\cos}^{2} \theta = \frac{1}{2} \left(\cos 2 \theta + 1\right)$.

$I = \frac{9}{2} \int \left(\cos 2 \theta + 1\right) d \theta = \frac{9}{2} \int \cos 2 \theta d \theta + \frac{9}{2} \int d \theta$

The first integral can be approached with substitution, or just the reverse chain rule. The second is simple:

$I = \frac{9}{4} \int 2 \cos 2 \theta d \theta + \frac{9}{2} \theta = \frac{9}{4} \sin 2 \theta + \frac{9}{2} \theta$

Using $\sin 2 \theta = 2 \sin \theta \cos \theta$:

$I = \frac{9}{2} \sin \theta \cos \theta + \frac{9}{2} \theta$

Writing all in terms of sine:

$I = \frac{9}{2} \sin \theta \sqrt{1 - {\sin}^{2} \theta} + \frac{9}{2} \theta$

Recall that $\sin \theta = \frac{x + 2}{3}$, so we also see that $\theta = \arcsin \left(\frac{x + 2}{3}\right)$:

$I = \frac{9}{2} \left(\frac{x + 2}{3}\right) \sqrt{1 - {\left(\frac{x + 2}{3}\right)}^{2}} + \frac{9}{2} \arcsin \left(\frac{x + 2}{3}\right)$

$I = \frac{3 \left(x + 2\right)}{2} \sqrt{\frac{9 - {\left(x + 2\right)}^{2}}{9}} + \frac{9}{2} \arcsin \left(\frac{x + 2}{3}\right)$

The $\sqrt{\frac{1}{9}} = \frac{1}{3}$ will come out of the square root and cancel with the $3$:

$I = \frac{x + 2}{2} \sqrt{9 - {\left(x + 2\right)}^{2}} + \frac{9}{2} \arcsin \left(\frac{x + 2}{3}\right)$

$I = \frac{\left(x + 2\right) \sqrt{5 - 4 x - {x}^{2}} + 9 \arcsin \left(\frac{x + 2}{3}\right)}{2} + C$