# How do you find the integral of int 5/sqrt(9-x^2)dx?

Let $x = 3 \cdot \sin t$ hence $\mathrm{dx} = 3 \cos t \cdot \mathrm{dt}$ then we have that

$\int \frac{5}{\sqrt{9 - {\left(3 \sin t\right)}^{2}}} \cdot 3 \cos t \cdot \mathrm{dt}$

$\int \frac{5}{\sqrt{9 - 9 \cdot {\sin}^{2} t}} \cdot 3 \cos t \cdot \mathrm{dt}$

$\int \frac{5}{\sqrt{9} \cdot \sqrt{1 - {\sin}^{2} t}} \cdot 3 \cdot \cos t \cdot \mathrm{dt}$

$\int \frac{5}{3 \cdot \sqrt{{\cos}^{2} t}} \cdot 3 \cdot \cos t \cdot \mathrm{dt}$

$\int \frac{5}{3 \cdot \cos t} \cdot 3 \cdot \cos t \cdot \mathrm{dt}$

$\int \frac{5}{\cancel{3} \cdot \cancel{\cos} t} \cdot \cancel{3} \cdot \cancel{\cos} t \cdot \mathrm{dt}$

$\int 5 \mathrm{dt} = 5 \cdot t + c$

But hence $\sin t = \frac{x}{3}$ or $t = \arcsin \left(\frac{x}{3}\right)$

We have that

$\int \frac{5}{\sqrt{9 - {x}^{2}}} \cdot \mathrm{dx} = 5 \cdot \arcsin \left(\frac{x}{3}\right) + c$

Nov 16, 2016

#### Explanation:

You can find the general form of the integral in any list of integrals:

$\int \frac{1}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx} = {\sin}^{-} 1 \left(\frac{x}{a}\right) + C$

However, I think you want to be shown how to do the trigonometric substitution that put this general form on the list. Let's begin:

Given: int5/sqrt(9 - x^2)dx = ?

Let $x = 3 \sin \left(\theta\right)$, then dx = 3cos(theta)d"theta

Substitute the above into the integral:

5int(3cos(theta))/sqrt(9 - 9sin^2(theta))d"theta =

Factor 9 from under the radical:

5int(3cos(theta))/(3sqrt(1 - sin^2(theta)))d"theta =

Use the identity cos(theta) = sqrt(1 - sin^2(theta):

5int(3cos(theta))/(3cos(theta))d"theta =

The numerator and denominator cancel to become 1:

5intd"theta = 5theta + C

To reverse the substitution, we need to describe $\theta$ in terms of x so we start with equation that we used to make the substution:

$x = 3 \sin \left(\theta\right)$

$\frac{x}{3} = \sin \left(\theta\right)$

theta = sin^-1(x/3); -3 <=x <= 3

int5/sqrt(9 - x^2)dx = 5sin^-1(x/3) + C; -3 <=x <= 3