Results 1 to 7 of 7

Math Help - Hyperbolic Integration by substitution

  1. #1
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1

    Hyperbolic Integration by substitution

    Hi, could someone please help me out on this question.

    Using the substitution x = \frac{3}{sinh \theta}, find the exact value of:

    \int_{4}^{3\sqrt{3}} \frac{1}{x\sqrt{x^2 + 9}} dx

    If x = \frac{3}{sinh \theta}, then \frac{dx}{d\theta} = \frac{-3cosh \theta}{sinh^2 \theta}.

    Putting in the substitution, I get \int_{4}^{3\sqrt{3}} \frac{1}{\frac{3}{sinh \theta}\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh^2 \theta} d\theta.

    From this I moved the sinh \theta to the top of the first fraction and cancelled with one of the sinh^2 \theta on the denominator of the second fraction to give me this.

    \int_{4}^{3\sqrt{3}} \frac{1}{3\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta

    I have no idea what to do next, would really appreciate some help?

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,346
    Thanks
    29
    Quote Originally Posted by craig View Post
    Hi, could someone please help me out on this question.

    Using the substitution x = \frac{3}{sinh \theta}, find the exact value of:

    \int_{4}^{3\sqrt{3}} \frac{1}{x\sqrt{x^2 + 9}} dx

    If x = \frac{3}{sinh \theta}, then \frac{dx}{d\theta} = \frac{-3cosh \theta}{sinh^2 \theta}.

    Putting in the substitution, I get \int_{4}^{3\sqrt{3}} \frac{1}{\frac{3}{sinh \theta}\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh^2 \theta} d\theta.

    From this I moved the sinh \theta to the top of the first fraction and cancelled with one of the sinh^2 \theta on the denominator of the second fraction to give me this.

    \int_{4}^{3\sqrt{3}} \frac{1}{3\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta

    I have no idea what to do next, would really appreciate some help?

    Thanks in advance
    Use the identity \cosh^2 \theta = 1 + \sinh ^2 \theta in your square root. Note: Be careful with your limits.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Danny View Post
    Be careful with your limits.
    Thanks I nearly forgot them, the new limits are \ln{\sqrt{3}} and \ln{2} respectively I think?

    Quote Originally Posted by Danny View Post
    Use the identity \cosh^2 \theta = 1 + \sinh ^2 \theta in your square root.
    This would give me:

    <br /> <br />
\int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\cosh^2 \theta - 1} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta<br />

    Sorry but not sure where to go from here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,346
    Thanks
    29
     <br />
\frac{1}{\sqrt{\frac{9}{\sinh ^2 \theta} + 9}} = \frac{1}{\sqrt{\frac{9 + 9 \sinh ^2 \theta}{\sinh ^2 \theta}}} = \frac{\sinh \theta}{3\sqrt{1 + \sinh ^2\theta}} = \frac{\sinh \theta}{3 \cosh \theta}<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2007
    Posts
    144
    Quote Originally Posted by craig View Post
    This would give me:

    <br /> <br />
\int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\cosh^2 \theta - 1} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta<br />

    Sorry but not sure where to go from here?
    Hi craig,

    We have \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\sinh^2\theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta

    = \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{9\sqrt{\frac{1+\sinh^2\theta}{\sinh^2\the  ta}}} \times \frac{-3cosh \theta}{sinh \theta} d\theta

     <br />
=\frac{-1}{3} \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{\sqrt{\coth^2\theta}}\times \frac{cosh \theta}{sinh \theta} d\theta<br />

    Can you continue?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Danny View Post
     <br />
\frac{1}{\sqrt{\frac{9}{\sinh ^2 \theta} + 9}} = \frac{1}{\sqrt{\frac{9 + 9 \sinh ^2 \theta}{\sinh ^2 \theta}}} = \frac{\sinh \theta}{3\sqrt{1 + \sinh ^2\theta}} = \frac{\sinh \theta}{3 \cosh \theta}<br />
    Thankyou!

    Simplifies down to \frac{-1}{3} \int_{\ln{2}}^{\ln{\sqrt{3}}} 1

    \left[\frac{-\theta}{3}\right]

    Putting in the limits you get

    \frac{1}{3}\left(\frac{1}{2}(\ln{4} - \ln{3}\right)

    Which gives you \frac{1}{6}\ln{\frac{4}{3}}

    Thanks again for the help!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Sean12345 View Post
    Hi craig,

    We have \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\sinh^2\theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta

    = \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{9\sqrt{\frac{1+\sinh^2\theta}{\sinh^2\the  ta}}} \times \frac{-3cosh \theta}{sinh \theta} d\theta

     <br />
=\frac{-1}{3} \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{\sqrt{\coth^2\theta}}\times \frac{cosh \theta}{sinh \theta} d\theta<br />

    Can you continue?
    Sorry just seen your reply now.

    Yep can manage the problem now, thanks for the reply
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. hyperbolic substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 30th 2011, 01:34 PM
  2. Integration by hyperbolic substitution
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 4th 2009, 11:30 AM
  3. Hyperbolic substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 1st 2008, 09:24 PM
  4. Replies: 3
    Last Post: April 17th 2008, 06:38 AM
  5. Replies: 6
    Last Post: April 13th 2008, 10:16 AM

Search Tags


/mathhelpforum @mathhelpforum