Results 1 to 4 of 4

Thread: integration

  1. #1
    Junior Member
    Joined
    Feb 2006
    Posts
    29

    integration

    I'm having problems integrating

    $\displaystyle
    \int\frac {sin(x)}{cos^2(x)}dx
    $

    the answer in the book is

    $\displaystyle
    \frac{1}{cos(x)}+c
    $

    I can't see how this answer is obtained.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Let u=cos(x) and du=-sin(x)

    This becomes $\displaystyle \int \frac{1}{u^2}(-1du)$

    When you integrate this, the negatives cancel out and you get $\displaystyle \frac{1}{u}+C$ or $\displaystyle \frac{1}{\cos(x)}+C$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2006
    Posts
    29
    Tremendous

    Thanks Jameson
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, macca101!

    Another method . . .


    $\displaystyle \int\frac{\sin(x)}{\cos^2(x)}\,dx$

    The answer in the book is: .$\displaystyle \frac{1}{\cos(x)}+C
    $
    I'm surprised that they left the answer like that . . .

    We have: .$\displaystyle \int\frac{\sin x}{\cos^2x}\,dx\;=\;\int\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\,dx\;=$ $\displaystyle \int\sec x\tan x\,dx \;= \;\sec x + C$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: Nov 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum