Results 1 to 3 of 3

Math Help - How do I simplify this trigonometric integral even further?

  1. #1
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597

    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!
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by s3a View Post
    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!
    Dear s3a,

    Your final expression is,

    \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

    Hope this will help you.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    202

    quoting the wrong reply unintentionally

    Quote Originally Posted by Sudharaka View Post
    Dear s3a,

    Your final expression is,

    \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

    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.
    Last edited by Pulock2009; February 27th 2010 at 11:25 PM. Reason: quoting the wrong reply
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplify the trigonometric expression
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: December 15th 2010, 03:16 AM
  2. Simplify Trigonometric Fraction
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 22nd 2009, 04:37 PM
  3. how to start/ simplify these trigonometric functions
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: January 22nd 2009, 08:46 AM
  4. simplify trigonometric
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 13th 2008, 09:05 PM
  5. Simplify This Trigonometric Expression
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 8th 2007, 08:21 PM

Search Tags


/mathhelpforum @mathhelpforum