Results 1 to 3 of 3

Math Help - Intergration With Trigonometry

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    15

    Intergration With Trigonometry

    \int \frac{\sec^5 \theta}{\tan^2\theta}d\theta

    I'm having troubles starting this. I've tried substituting for tangent,splitting up the secants, and a couple other methods but I get stuck after a couple steps. How should i start out?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,695
    Thanks
    1511
    Quote Originally Posted by twoteenine View Post
    \int \frac{\sec^5 \theta}{\tan^2\theta}d\theta

    I'm having troubles starting this. I've tried substituting for tangent,splitting up the secants, and a couple other methods but I get stuck after a couple steps. How should i start out?
    integral[((Sec[x])&# x5e;5)/((Tan[x]&#x29 ;^2)] - Wolfram|Alpha

    Click Show Steps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    715
    \frac{ \sec^5(x) }{ \tan^2(x) }

     = \frac{ \sec^4(x)  }{ \tan^2(x) } \cdot \sec(x)

     = \frac{ ( \tan^2(x) + 1)^2 }{ \tan^2(x) }\cdot \sec(x)

     = \frac{ \tan^4(x) + 2\tan^2(x) + 1}{\tan^2(x) } \cdot \sec(x)

     = [ \tan^2(x) + 2 + \frac{1}{\tan^2(x)} ] \cdot \sec(x)

     = \tan^2(x)\sec(x) + 2\sec(x) + \frac{1}{ \tan(x) \cdot \tan(x) \cdot \cos(x) }

     = \tan^2(x)\sec(x) + 2\sec(x) + \csc(x)\cot(x)

    its integral  = \int  \tan^2(x)\sec(x)   ~dx +  2\int \sec(x)~dx + \int  \csc(x)\cot(x) ~dx

     =  \int  \tan^2(x)\sec(x)  ~dx  +    2\int \sec(x)~dx  - \csc(x) + C

    Consider  I = \int  \tan^2(x)\sec(x)  ~dx

     I = \int [ \sec^3(x) - \sec(x) ]~dx  (1)

    but integration by parts gives

     I = \sec(x) \cdot \tan(x) - \int \sec^3 (x)~dx

    therefore ,

     \int \sec^3(x) ~dx = \sec(x) \tan(x) - I  (2)

    By substituting  (2) into  (1)

    we have

     I = \sec(x) \tan(x) - I  - \int \sec(x)~dx    ~~~ \implies

     I = \frac{1}{2} \sec(x) \tan(x) - \frac{1}{2} \int \sec(x)~dx

    The integral




     \int \frac{ \sec^5(x) }{ \tan^2(x) }  ~dx =

       \frac{1}{2} \sec(x) \tan(x)  + \frac{3}{2}  \int \sec(x)~dx   - \csc(x) + C

     = \frac{1}{2} \sec(x) \tan(x)  + \frac{3}{2} \cosh^{-1}[\sec(x)]   - \csc(x) + C
    Last edited by simplependulum; March 17th 2010 at 11:45 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry to Memorize, and Trigonometry to Derive
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: August 21st 2013, 12:03 PM
  2. Intergration Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 7th 2009, 01:49 AM
  3. Intergration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 16th 2008, 10:33 PM
  4. need help! intergration
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 18th 2008, 06:24 PM
  5. intergration help!!!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 23rd 2007, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum