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Homework Help: Trig substitution

  1. Jun 9, 2006 #1
    How would I go about solving this:

    [tex]\int cos^2x tan^3xdx[/tex]..all i did so far ..

    [tex]\int cos^2x tanx(tan^2x)[/tex]

    [tex][[ \frac{1}{cosx}^2 -1] = tan^2x[/tex]

    so...

    [tex] \int cos^2x[[ \frac{1}{cosx}]^2-1]tanx[/tex]

    is this right so far...now what?
     
    Last edited: Jun 9, 2006
  2. jcsd
  3. Jun 9, 2006 #2

    TD

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    Another way would be using tan(x) = sin(x)/cos(x), like:

    [tex]
    \cos ^2 x\tan ^3 x = \cos ^2 x\frac{{\sin ^3 x}}
    {{\cos ^3 x}} = \frac{{\sin ^3 x}}
    {{\cos x}}
    [/tex]

    Now if you use sin³x = sin²x.sinx and convert the sin²x to 1-cos²x, I smell a good substitution coming :smile:

    Your method was fine as well though, continuing could give:

    [tex]
    \cos ^2 x\tan ^3 x = \cos ^2 x\left( {\sec ^2 x - 1} \right)\tan x = \tan x - \cos ^2 x\tan x = \tan x - \sin x\cos x
    [/tex]
     
    Last edited: Jun 9, 2006
  4. Jun 9, 2006 #3
    ohh..ok so i got as far as [tex]\int \frac{sinx - sinxcos^2x}{cosx}[/tex]
     
  5. Jun 9, 2006 #4

    TD

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    Well, you can do many things with that. My choice would be rewriting as:

    [tex]
    \int {\frac{{\sin x - \sin x\cos ^2 x}}
    {{\cos x}}dx} = \int {\frac{{\sin x\left( {1 - \cos ^2 x} \right)}}
    {{\cos x}}dx}
    [/tex]

    Do you see a nice substitution now?
     
  6. Jun 9, 2006 #5
    yes I get [tex]ln|cosx| - \frac{1}{2}cos^2x + C[/tex]
     
  7. Jun 9, 2006 #6

    TD

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    You should check that again (signs?)
     
  8. Jun 9, 2006 #7
    im having trouble with another..ive worked it down..but i think its wrong..

    [tex] \int cot^5xsin^4xdx[/tex]

    [tex] \int \frac{cos^5x}{sinx}dx[/tex]...would i just simplify the top to cos^2x(cos^2x)(cosx)?
     
  9. Jun 9, 2006 #8

    TD

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    Sure, and switch each cos²x to 1-sin²x so you can substitute y = sin(x).

    Watch out though: your last one wasn't correct. I find:

    [tex]
    - \ln \left( {\cos x} \right) - \frac{{\sin ^2 x}}
    {2} + C
    [/tex]
     
  10. Jun 9, 2006 #9
    yea i messed up the signs thanks|
     
  11. Jun 9, 2006 #10

    TD

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    Ok, the other one should be fine now?

    [tex]
    \int {\frac{{\cos ^5 x}}
    {{\sin x}}dx} = \int {\frac{{\left( {1 - \sin ^2 x} \right)^2 }}
    {{\sin x}}\cos xdx}
    [/tex]

    Perfect for y = sin(x).
     
  12. Jun 9, 2006 #11
    how about [tex]\int sec^6tdt[/tex]

    where to start?

    sec^2t(sec^2t)(sec^2t)?
     
    Last edited: Jun 9, 2006
  13. Jun 9, 2006 #12
    The first thing you want to do is split up [tex]\int sec^6tdt[/tex] into components so some parts of it will cancel.

    You're on the right track, remember that sec^2t = 1 + tan^2t, and make a u substitution. If you go about it correctly, one of those sec^2ts will cancel and you will have the integral in terms of u. Once you get to that step, backsubstitute and tack on the C at the end.

    Best of luck.
     
  14. Jun 12, 2006 #13

    VietDao29

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    Uhm, sec6x = sec6x - sec4x + sec4x - sec2x + sec2x = sec4x (sec2x - 1) + sec2x (sec2x - 1) + sec2x.
    And you should also notice that:
    [tex]\frac{d}{dx} \tan x = \sec ^ 2 x[/tex]
    I'll give you an example.
    ---------------
    Example:
    [tex]\int \sec ^ 4 x dx = \int \left( \sec ^ 4 x - \sec ^ 2 x + \sec ^ 2 x \right) dx = \int \left( \sec ^ 2 x (\sec ^ 2 x - 1) + \sec ^ 2 x \right) dx[/tex]
    [tex]= \int ( \sec ^ 2 x \tan ^ 2 x ) dx + \int \sec ^ 2 x dx = \int \tan ^ 2 x d(\tan x) + \int \sec ^ 2 x dx = \frac{\tan ^ 3 x}{3} + \tan x + C[/tex]
    Can you go from here? :)
     
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