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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!

Thanks in advance!

quoting the wrong reply unintentionally

Quote:

Originally Posted by

**Sudharaka** Dear s3a,

Your final expression is,

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

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

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

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

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

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

Hope this will help you.

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.