Results 1 to 5 of 5

Math Help - Integration Problem

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    8

    Integration Problem

    Find the integral of (tanx/sec^2x)dx

    I tried using u substitution, and I get:

    u=tanx
    du=(sec^2)x

    At first it seems like it was going to be simple.

    But in the problem there is a sec^-2
    And in my substitution there is a sec^2.

    Any help would be appreciated.
    Last edited by ninjaku; April 4th 2010 at 07:00 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by ninjaku View Post
    Find the integral of (tanx/(sec^2)x)dx.

    I tried using u substitution, and I get:

    u=tanx
    du=(sec^2)x.

    At first it seems like it was going to be simple.

    But in the problem there is a sec^-2.
    And in my substitution there is a sec^2.

    Any help would be appreciated.
    \frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}

    now integrate
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    8
    Quote Originally Posted by skeeter View Post
    \frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}

    now integrate
    Ok after your suggestion. I did the problem two ways.

    The first way I did using:

    u=sinx
    du=cosx

    and got the answer to be  \frac{sin^2x}{2} + c.

    However when I checked it in my calculator, it said the answer was instead, -\frac{cos^2x}{2} +c.

    Which I found to be the answer when I set :

    u=cosx
    du=-sinx

    I would just like to know if both of those are correct, or only one.
    And if only one of them is correct, how would I know which term I should pick to use for the substitution?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by ninjaku View Post
    Ok after your suggestion. I did the problem two ways.

    The first way I did using:

    u=sinx
    du=cosx

    and got the answer to be  \frac{sin^2x}{2} + c.

    However when I checked it in my calculator, it said the answer was instead, -\frac{cos^2x}{2} +c.

    Which I found to be the answer when I set :

    u=cosx
    du=-sinx

    I would just like to know if both of those are correct, or only one.
    And if only one of them is correct, how would I know which term I should pick to use for the substitution?
    \frac{sin^2x}{2} + c = \frac{1-cos^2x}{2} + c= \frac{1}{2}-\frac{cos^2x}{2} +c = -\frac{cos^2x}{2} +c'

    Note:  c' = c + \frac{1}{2}

    these two terms  \frac{sin^2x}{2} + c and -\frac{cos^2x}{2} +c' differ only by a constant. Either answer works in this case.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2010
    Posts
    8
    That makes a lot of sense, thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 21st 2010, 12:33 PM
  2. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  3. Integration Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 22nd 2009, 06:34 PM
  4. Integration problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 1st 2008, 04:47 AM
  5. Just another integration problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 21st 2006, 03:52 AM

Search Tags


/mathhelpforum @mathhelpforum