# Math Help - Integration help: 7*sqrt(x^2+9)

1. ## Integration help: 7*sqrt(x^2+9)

I need to integrate the following:

7*sqrt(x^2+9)

Here is my work so far:

x=3tan(theta)
d-x=3sec^2(theta)*d-theta

Then:
7*int[sqrt{9tan^2(theta)+9}*3sec^2(theta)*d-theta

Then:
63*int[sec^3(theta)*d-theta]
And I don't know what to do from there. Thank you!

2. I'm pretty sure you can solve the integral of sec^3 by doing integration by parts

Take sec^2 and sec as your two parts and try that

3. stupid!!! Ah I feel dumb. I need sleep :P

Thanks man!

4. Originally Posted by dillonmhudson
I need to integrate the following:

7*sqrt(x^2+9)

Here is my work so far:

x=3tan(theta)
d-x=3sec^2(theta)*d-theta

Then:
7*int[sqrt{9tan^2(theta)+9}*3sec^2(theta)*d-theta

Then:
63*int[sec^3(theta)*d-theta]
And I don't know what to do from there. Thank you!
$\int{7\sqrt{x^2 + 9}\,dx}$

Use a hyperbolic substitution instead - it's easier...

Substitute $x = 3\sinh{t}$ so that $dx = 3\cosh{t}\,dt$.

Note that $t = \sinh^{-1}{\frac{x}{3}}$.

The integral becomes

$\int{7\sqrt{(3\sinh{t})^2 + 9}\cdot 3\cosh{t}\,dt}$

$= \int{21\cosh{t}\sqrt{9\sinh^2{t} + 9}\,dt}$

$= \int{21\cosh{t}\sqrt{9(\sinh^2{t} + 1)}\,dt}$

$= \int{63\cosh{t}\sqrt{\cosh^2{t}}\,dt}$

$= 63\int{\cosh^2{t}\,dt}$

$= \frac{63}{2}\int{\cosh{2t} + 1\,dt}$

$= \frac{63}{2}\left[\frac{1}{2}\sinh{2t} + t\right] + C$

$= \frac{63}{2}\left[\sinh{t}\cosh{t} + t\right] + C$

$= \frac{63}{2}\left[\sinh{t}\sqrt{1 + \sinh^2{t}} + t\right] + C$

$= \frac{63}{2}\left[\frac{x}{3}\sqrt{1 + \left(\frac{x}{3}\right)^2} + \sinh^{-1}{\frac{x}{3}}\right] + C$

$= \frac{7x\sqrt{9 + x^2}}{2} + \frac{63\sinh^{-1}{\frac{x}{3}}}{2} + C$

5. how about a solution without hyperbolics?