Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By skeeter

Math Help - Intergration by Substition Formula compared to Symbolic Algebra Software Answer

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    Australia
    Posts
    6

    Intergration by Substition Formula compared to Symbolic Algebra Software Answer

    For \int{\sqrt(9-25x^2)} dx

    I got the answer by substitution into a formula:

    \int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arctan(\frac{5x}{\sqrt(9-25x^2)} + C


    By using Maple (symbolic algebra software) I got:-

    \int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arcsin(\frac{5}{3}x) + C


    Can anyone help me explain the difference in the two answers?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,884
    Thanks
    673

    Re: Intergration by Substition Formula compared to Symbolic Algebra Software Answer

    if y = \arcsin\left(\frac{5x}{3} \right) ,

    \sin{y} = \frac{5x}{3} = \frac{opp}{hyp} \implies adj = \sqrt{9 - 25x^2}

    so,

    \tan{y} = \frac{opp}{adj} = \frac{5x}{\sqrt{9-25x^2}} \implies \arctan\left(\frac{5x}{\sqrt{9-25x^2}}\right) = y

    ... in other words, they are two expressions for the same angle.
    Thanks from Furyan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2012
    From
    Australia
    Posts
    6

    Re: Intergration by Substition Formula compared to Symbolic Algebra Software Answer

    Brilliant!! Thank you so much!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,672
    Thanks
    1490

    Re: Intergration by Substition Formula compared to Symbolic Algebra Software Answer

    Quote Originally Posted by tammmyl View Post
    For \int{\sqrt(9-25x^2)} dx

    I got the answer by substitution into a formula:

    \int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arctan(\frac{5x}{\sqrt(9-25x^2)} + C


    By using Maple (symbolic algebra software) I got:-

    \int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arcsin(\frac{5}{3}x) + C


    Can anyone help me explain the difference in the two answers?
    When you have an integral of the form \displaystyle \begin{align*} f\left(a^2 - b^2x^2\right) \end{align*}, it's often useful to substitute \displaystyle \begin{align*} x = \frac{a}{b}\sin{\theta} \end{align*} or \displaystyle \begin{align*} x = \frac{a}{b}\cos{\theta} \end{align*}, because it simplifies nicely with the Pythagorean Identity. In this case, \displaystyle \begin{align*} \int{\sqrt{9 - 25x^2}\,dx} \end{align*}, we substitute \displaystyle \begin{align*} x = \frac{3}{5}\sin{\theta} \implies dx = \frac{3}{5}\cos{\theta}\,d\theta \end{align*} and the integral becomes

    \displaystyle \begin{align*} \int{\sqrt{9 - 25x^2}\,dx} &= \int{\sqrt{9 - 25\left( \frac{3}{5}\sin{\theta} \right)^2}\,\frac{3}{5}\cos{\theta}\,d\theta} \\ &= \frac{3}{5}\int{\sqrt{9 - 9\sin^2{\theta}}\,\cos{\theta}\,d\theta} \\ &= \frac{3}{5}\int{ \sqrt{ 9\cos^2{\theta} } \, \cos{\theta} \, d\theta } \\ &= \frac{3}{5}\int{ 3\cos{\theta}\cos{\theta}\,d\theta } \\ &= \frac{9}{5}\int{\cos^2{\theta}\,d\theta} \\ &= \frac{9}{5}\int{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta} \\ &= \frac{9}{5}\left(\frac{1}{2}\theta + \frac{1}{4}\sin{2\theta}\right) + C \\ &= \frac{9}{10}\theta + \frac{9}{10}\sin{\theta}\cos{\theta} + C \\ &= \frac{9}{10}\theta + \frac{9}{10}\sin{\theta}\sqrt{1 - \sin^2{\theta}} +C  \\ &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{9}{10}\left(\frac{5}{3}x\right)\sqrt{1 - \left(\frac{5}{3}x\right)^2} + C \\ &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{3}{2}x\sqrt{1 - \frac{25x^2}{9}} + C \end{align*}

    \displaystyle \begin{align*}  &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{3}{2}x\sqrt{\frac{9 - 25x^2}{9}} \\ &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{3}{2}x\,\frac{\sqrt{9 - 25x^2}}{3}\\  &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{1}{2}x\sqrt{9 - 25x^2} + C\end{align*}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. learning algebra software
    Posted in the Math Forum
    Replies: 0
    Last Post: January 2nd 2012, 04:31 AM
  2. Help creating a formula for a software.
    Posted in the Math Software Forum
    Replies: 0
    Last Post: September 5th 2011, 02:11 PM
  3. Linear Algebra software/computers/calculators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 18th 2011, 12:30 AM
  4. Symbolic Algebra
    Posted in the Calculators Forum
    Replies: 1
    Last Post: December 6th 2009, 08:47 PM
  5. Replies: 2
    Last Post: August 27th 2005, 10:13 AM

Search Tags


/mathhelpforum @mathhelpforum