Results 1 to 5 of 5

Math Help - Very "Messy" Integrals

  1. #1
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Very "Messy" Integrals

    Math Challenge
    Evaluate the following Integrals.

    1) \int \sqrt{x+\sqrt{a^2+x^2}}dx

    2) \int \sqrt{x+a\sqrt{ax-a^2}}dx

    Moderator approved by CaptainBlack
    Last edited by CaptainBlack; December 18th 2011 at 01:06 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: Very "Messy" Integrals

    Quote Originally Posted by sbhatnagar View Post
    Math Challenge
    Evaluate the following Integrals.

    1) \int \sqrt{x+\sqrt{a^2+x^2}}dx

    2) \int \sqrt{x+a\sqrt{ax-a^2}}dx
    1)


    \begin{array}{l}x + \sqrt {{x^2} + {a^2}}  = u\\\\\left( {1 + \frac{1}{{2 \cdot \sqrt {{x^2} + {a^2}} }} \cdot 2x} \right)dx = du\\\\\frac{{x + \sqrt {{x^2} + {a^2}} }}{{\sqrt {{x^2} + {a^2}} }}dx = du\\\\dx = \sqrt {{x^2} + {a^2}}  \cdot \frac{{du}}{u}\end{array}




    \begin{array}{l}\sqrt {{x^2} + {a^2}}  = u - x\\\\{x^2} + {a^2} = {u^2} - 2ux + {x^2}\\\\2ux = {u^2} - {a^2}\\\\x = \frac{{{u^2} - {a^2}}}{{2u}}\end{array}




    \begin{array}{l}dx = \sqrt {\frac{{{{\left( {{u^2} - {a^2}} \right)}^2}}}{{4{u^2}}} + {a^2}}  \cdot \frac{{du}}{u}\\\\dx = \sqrt {\frac{{{u^4} -2{a^2}{u^2} + {a^4} + 4{a^2}{u^2}}}{{4{u^2}}}}  \cdot \frac{{du}}{u}\\\\dx = \frac{{{u^2} + {a^2}}}{{2{u^2}}}du\end{array}




    \begin{array}{l}\int {\sqrt u  \cdot \frac{{{u^2} + {a^2}}}{{2{u^2}}}du = \int {\frac{{{u^2} + {a^2}}}{{2{u^{\frac{3}{2}}}}}du} } \\\\ = \frac{1}{2}\int {{u^{\frac{1}{2}}}} du + \frac{{{a^2}}}{2}\int {{u^{\frac{{ - 3}}{2}}}} du = \\\\\frac{1}{3}{u^{\frac{3}{2}}} - {a^2} \cdot {u^{\frac{{ - 1}}{2}}} + C = \\\\= \frac{1}{3}\sqrt {{{\left( {x + \sqrt {{x^2} + {a^2}} } \right)}^3}}  - {a^2}\frac{1}{{\sqrt {x + \sqrt {{x^2} + {a^2}} } }} + C\end{array}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: Very "Messy" Integrals

    Quote Originally Posted by sbhatnagar View Post
    Math Challenge
    Evaluate the following Integrals.

    1) \int \sqrt{x+\sqrt{a^2+x^2}}dx
    Substitute x = a\sinh u. Then

    \begin{aligned}\int \sqrt{x+\sqrt{a^2+x^2}}dx &= \int \sqrt{a(\sinh u + \cosh u)}\, a\cosh u\,du \\ &= \tfrac12a^{3/2}\int e^{u/2}(e^u + e^{-u})\,du \\ &= \tfrac12a^{3/2}\int(e^{3u/2} - e^{-u/2})\,du \\ &= \tfrac12a^{3/2}\bigl(\tfrac23e^{3u/2} - 2e^{-u/2}\bigr) + C \\ &= \sqrt{ae^u}\bigl(\tfrac13ae^u - ae^{-u}\bigr) + C \\ &= \sqrt{ae^u}\bigl(\tfrac43a\sinh u - \tfrac23a\cosh u\bigr) + C \\ &=  \sqrt{x+\sqrt{a^2+x^2}}\,\bigl(\tfrac43x - \tfrac23\sqrt{a^2+x^2}\bigr) + C. \end{aligned}

    I haven't checked to see whether this answer agrees with princeps's. They may well be equivalent.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Very "Messy" Integrals

    Your answers are correct are correct!

    Hint for Q2)

    Spoiler:

    Try any of the following substitutions

    u=x+a\sqrt{ax-a^2}

    u=\sqrt{ax-a^2}
    Last edited by sbhatnagar; December 19th 2011 at 05:14 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2010
    Posts
    57

    Re: Very "Messy" Integrals

    in reply to Princeps' signature, it's its own cure!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 17th 2011, 03:50 PM
  2. Replies: 2
    Last Post: June 4th 2011, 01:11 PM
  3. Replies: 2
    Last Post: April 24th 2011, 08:01 AM
  4. Replies: 1
    Last Post: October 25th 2010, 05:45 AM
  5. Replies: 1
    Last Post: June 4th 2010, 11:26 PM

Search Tags


/mathhelpforum @mathhelpforum