# Thread: How do I simplify this trigonometric integral even further?

1. ## How do I simplify this trigonometric integral even further?

I get the correct final answer as Wolfram Alpha confirms: http://www.wolframalpha.com/input/?i=is+(sqrt(x^2+%2B+9))^3+%2F3+-+9*sqrt(x^2+%2B+9)+%3D+1%2F3+*+(x^2+-+18)+*+sqrt(x^2+%2B+9)%3F

My work is attached and I would just like to know how to get from where I am to the answer the book provides in the back (I wrote the answer the book provides in red on the attached file).

(The problem is #39 of the attached pdf)

Any help would be greatly appreciated!

2. Originally Posted by s3a
I get the correct final answer as Wolfram Alpha confirms: http://www.wolframalpha.com/input/?i=is+(sqrt(x^2+%2B+9))^3+%2F3+-+9*sqrt(x^2+%2B+9)+%3D+1%2F3+*+(x^2+-+18)+*+sqrt(x^2+%2B+9)%3F

My work is attached and I would just like to know how to get from where I am to the answer the book provides in the back (I wrote the answer the book provides in red on the attached file).

(The problem is #39 of the attached pdf)

Any help would be greatly appreciated!
Dear s3a,

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

Talking $\sqrt{x^2+9}$ out will give you,

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

$\frac{\sqrt{x^{2}+9}\left[x^{2}+9-27\right]}{3}+C$

$\frac{\sqrt{x^{2}+9}\left[x^{2}-18\right]}{3}+C$

$\frac{1}{3}(x^2-18)\sqrt{x^2+9}+C$

3. ## quoting the wrong reply unintentionally

Originally Posted by Sudharaka
Dear s3a,

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

Talking $\sqrt{x^2+9}$ out will give you,

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

$\frac{\sqrt{x^{2}+9}\left[x^{2}+9-27\right]}{3}+C$

$\frac{\sqrt{x^{2}+9}\left[x^{2}-18\right]}{3}+C$

$\frac{1}{3}(x^2-18)\sqrt{x^2+9}+C$