How do I simplify this trigonometric integral even further?

• Feb 27th 2010, 07:35 PM
s3a
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!
• Feb 27th 2010, 07:47 PM
Sudharaka
Quote:

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,

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

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

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

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

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

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

• Feb 27th 2010, 10:22 PM
Pulock2009
Quote:

Originally Posted by Sudharaka
Dear s3a,

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

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

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

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

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

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

after getting the expression in tan and sec convert it to sin and cos and then solve converting the expression having an odd pwer to the differentiation of the expression having even power. u should get
integration sin^3(theta)/cos^4(theta).d(theta)
make this integration sin^2(theta).sin(theta)/cos^4(theta)
then (1-cos^(theta)).sin(theta)/cos^4(theta).
hope u are able to folow. its easy!

my mistake. i should have quoted s3a post but i quoted sudharkas unintentionally.