Results 1 to 4 of 4

Math Help - Trgnometric integrals

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    67

    Trgnometric integrals

    How would you evaluate this indefinite integral?
    tan^4x dx

    I began with Seperating it into two tans. Both to the 2nd power.
    Then I got lost. How do I get to my answer?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,519
    Thanks
    1404
    Quote Originally Posted by radioheadfan View Post
    How would you evaluate this indefinite integral?
    tan^4x dx

    I began with Seperating it into two tans. Both to the 2nd power.
    Then I got lost. How do I get to my answer?
    Use the chain rule.

    y = \tan^4{x} = (\tan{x})^4.


    Let u = \tan{x} so that y = u^4.


    \frac{du}{dx} = \sec^2{x}


    \frac{dy}{du} = 4u^3

     = 4\tan^3{x}.


    Therefore

    \frac{dy}{dx} = 4\sec^2{x}\tan^3{x}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,628
    Thanks
    430
    Quote Originally Posted by radioheadfan View Post
    How would you evaluate this indefinite integral?
    tan^4x dx

    I began with Seperating it into two tans. Both to the 2nd power.
    Then I got lost. How do I get to my answer?
    \int \tan^4{x} \, dx

    \int \tan^2{x}(\sec^2{x} - 1) \, dx

    \int \tan^2{x}\sec^2{x} - \tan^2{x} \, dx

    \int \tan^2{x}\sec^2{x} \, dx - \int \tan^2{x} \, dx

    \int \tan^2{x}\sec^2{x} \, dx - \int \sec^2{x} - 1 \, dx

    first integral ... u =tan{x} ... du = \sec^2{x} \, dx

    \frac{\tan^3{x}}{3} - \tan{x} + x + C
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,519
    Thanks
    1404
    Quote Originally Posted by skeeter View Post
    \int \tan^4{x} \, dx

    \int \tan^2{x}(\sec^2{x} - 1) \, dx

    \int \tan^2{x}\sec^2{x} - \tan^2{x} \, dx

    \int \tan^2{x}\sec^2{x} \, dx - \int \tan^2{x} \, dx

    \int \tan^2{x}\sec^2{x} \, dx - \int \sec^2{x} - 1 \, dx

    first integral ... u =tan{x} ... du = \sec^2{x} \, dx

    \frac{\tan^3{x}}{3} - \tan{x} + x + C

    Oops, I read that it had to be differentiated. LOL.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 09:23 PM
  2. integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2010, 01:54 PM
  3. Replies: 1
    Last Post: December 6th 2009, 07:43 PM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 04:52 PM
  5. Some integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 20th 2008, 01:41 AM

Search Tags


/mathhelpforum @mathhelpforum