Results 1 to 9 of 9

Math Help - integral of \sqrt(\tan(x))dx

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    integral of \sqrt(\tan(x))dx

    this one has a long answer:
    \int\sqrt{\tan{x}}dx

    What do i start with first?
    need advices or hints
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2007
    Posts
    237
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    this one has a long answer:
    \int\sqrt{\tan{x}}dx

    What do i start with first?
    need advices or hints
    You may try this substitution:u=sqrt(tanx).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Look at here.

    Don't be afraid about the language, just follow the steps.

    In the third one, I applied partial fractions.

    I hope it helps.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    thanks krizalid,
    question:
    how is \int \frac{2u^2}{u^4+1}du being factored? coz i used quickmath and gave no results in partial fractions.
    once again thanks so much for all the help
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    u^4+1=\left(u^2+\sqrt2u+1\right)\left(u^2-\sqrt2u+1\right)

    Then set \frac{2u^2}{u^4+1}=\frac{ax+b}{u^2+\sqrt2u+1}+\fra  c{cx+d}{u^2-\sqrt2u+1}

    This is a little hard.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2007
    Posts
    237
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    how is \int \frac{2u^2}{u^4+1}du being factored?
    At this step you may use the reciprocal subsititution:u=1/t. This may make the integral easier.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2007
    Posts
    237
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    thanks krizalid,
    question:
    \int \frac{2u^2}{u^4+1}du
    Maple gives the following result:
    Attached Thumbnails Attached Thumbnails integral of \sqrt(\tan(x))dx-aug04.gif  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    That's the same result as mine
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    After the brute force applied on the problem, there's another way to solve this (elegant by the way).

    Suppose it remains to compute

    \int\frac{x^2}{x^4+1}\,dx

    Let's use (x^2+1)+(x^2-1)=2x^2

    Next, we split the integral in two pieces and in each one of them we divide by x^2

    After some simple calculations we happily get that

    \int\frac{x^2}{x^4+1}\,dx=\frac12\left(\frac1{\sqr  t2}\arctan\frac{x-x^{-1}}{\sqrt2}+\frac1{2\sqrt2}\ln\left|\frac{x+x^{-1}-\sqrt2}{x+x^{-1}+\sqrt2}\right|\right)+k

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integral 2/sqrt(x^2-4)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 28th 2010, 09:13 PM
  2. Integral of sin^3 (sqrt(x)) / (sqrt(x))
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 21st 2010, 04:48 PM
  3. Integral of sqrt(9-x^2)
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 18th 2009, 09:34 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 11
    Last Post: January 6th 2008, 09:33 AM

Search Tags


/mathhelpforum @mathhelpforum