# How do you integrate int (x+5)/sqrt (9-(x-3)^2) dx?

Aug 4, 2016

$8 {\sin}^{- 1} \left(\frac{x - 3}{3}\right) - \sqrt{9 - {\left(x - 3\right)}^{2}} + C$

#### Explanation:

I hope you like substitutions, because we're about to do a lot of them. To start, let $u = x - 3 \implies \mathrm{du} = \mathrm{dx}$

Integral becomes:

$\int \frac{u + 8}{\sqrt{9 - {u}^{2}}} \mathrm{du}$

Now we want to deal with the denominator. This type of function will often involve a trig substitution - if you learn/can derive the pythagorean identities they are very helpful in figuring out what will cancel out etc. Whatever we choose for $u$ in this case, it should have a coefficient of 3 so that we can take the 9 outside the square root.

Let's go with $u = 3 \sin \theta \implies \mathrm{du} = 3 \cos \theta d \theta \mathmr{and} \theta = {\sin}^{- 1} \left(\frac{u}{3}\right)$

Integral becomes:

$3 \int \frac{3 \sin \theta + 8}{3 \sqrt{1 - {\sin}^{2} \theta}} \cos \theta d \theta$

$1 - {\sin}^{2} \theta = {\cos}^{2} \theta$ so:

$3 \int \frac{3 \sin \theta + 8}{3 \sqrt{{\cos}^{2} \theta}} \cos \theta d \theta$

$= \int \left(3 \sin \theta + 8\right) d \theta$

Can split this up into two seperate integrals:

$= 3 \int \sin \theta d \theta + 8 \int d \theta$

$= - 3 \cos \theta + 8 \theta + C$

Substitute back in that $\theta = {\sin}^{- 1} \left(\frac{u}{3}\right)$

$= 8 {\sin}^{- 1} \left(\frac{u}{3}\right) - 3 \cos \left({\sin}^{- 1} \left(\frac{u}{3}\right)\right) + C$

We need to figure out what the general form of $\cos \left({\sin}^{- 1} \left(\phi\right)\right)$ is:

Consider $y = {\sin}^{- 1} \left(\phi\right)$

$\implies \phi = \sin \left(y\right)$

${\phi}^{2} = {\sin}^{2} \left(y\right)$

${\phi}^{2} = 1 - {\cos}^{2} \left(y\right)$

$\cos \left(y\right) = \cos \left({\sin}^{- 1} \left(\phi\right)\right) = \sqrt{1 - {\phi}^{2}}$

Hence $\cos \left({\sin}^{- 1} \left(\frac{u}{3}\right)\right) = \sqrt{1 - {\left(\frac{u}{3}\right)}^{2}} = \frac{1}{3} \sqrt{9 - {u}^{2}}$

Solution becomes:

$8 {\sin}^{- 1} \left(\frac{u}{3}\right) - \sqrt{9 - {u}^{2}} + C$

Now back substitute $u = x - 3$

Solution is:

$8 {\sin}^{- 1} \left(\frac{x - 3}{3}\right) - \sqrt{9 - {\left(x - 3\right)}^{2}} + C$