# How do you integrate int sqrt(9-x^2) using trig substitutions?

Aug 1, 2016

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

#### Explanation:

Let $x = 3 \sin \left(u\right) \implies \mathrm{dx} = 3 \cos \left(u\right) \mathrm{du}$

$x = 3 \sin \left(u\right) \implies u = {\sin}^{- 1} \left(\frac{x}{3}\right)$

$\int \sqrt{9 - {x}^{2}} \mathrm{dx} = 3 \int \sqrt{9 - 9 {\sin}^{2} \left(u\right)} \cos \left(u\right) \mathrm{du}$

$= 9 \int \sqrt{1 - {\sin}^{2} \left(u\right)} \cos \left(u\right) \mathrm{du} = 3 \int \sqrt{{\cos}^{2} \left(u\right)} \cos \left(u\right) \mathrm{du}$

$= 9 \int {\cos}^{2} \left(u\right) \mathrm{du}$

From double angle formula

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1 \implies {\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$

$= \frac{9}{2} \int \left(1 + \cos \left(2 u\right)\right) \mathrm{du}$

$= \frac{9}{2} \left[u + \frac{1}{2} \sin \left(2 u\right)\right] + C$

Apply double angle formula:

$= \frac{9}{2} \left[u + \sin \left(u\right) \cos \left(u\right)\right] + C$

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

$= \frac{9}{2} \left[u + \sin \left(u\right) \sqrt{1 - {\sin}^{2} \left(u\right)}\right] + C$

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

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

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

Aug 1, 2016

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

#### Explanation:

Let us subst. $x = 3 \sin t$, so, $\mathrm{dx} = 3 \cos t \mathrm{dt}$

Also, $\sqrt{9 - {x}^{2}} = \sqrt{9 - 9 {\sin}^{2} t} = \sqrt{9 {\cos}^{2} t} = 3 \cos t$.

Hence, $I = \int \sqrt{9 - {x}^{2}} \mathrm{dx}$

$= \int 3 \cos t \cdot 3 \cos t \mathrm{dt} = 9 \int {\cos}^{2} t \mathrm{dt} = \frac{9}{2} \int \left(1 + \cos 2 t\right) \mathrm{dt}$

$= \frac{9}{2} \left\{t + \sin \frac{2 t}{2}\right\} = \frac{9}{2} \left\{t + \sin t \cos t\right\}$

Here, $x = 3 \sin t \Rightarrow t = \arcsin \left(\frac{x}{3}\right)$.

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

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