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Trig substitution help

  1. Feb 2, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex] ((sin(x))^3/(cos(x)) )*dx



    3. The attempt at a solution

    alright i have been trying to use

    u= cosx
    -du = sinx

    but it dosnt make sense bause there is still a sinx^2 to account for

    so i know i need to make a trig substitution but i cant figure out the appropriate
    substitution for sinx^2 even though i think the 1/2 ( 1- cos2x) is hte only one


    question... when u use a u substituion for u = cosx -du = sindx would u write

    1/2 [tex]\int[/tex] ((sinx*( 1 - cos2x)) / cosx)*dx


    - 1/2 [tex]\int[/tex]( (1 - 2u) / u ) *du



    if u = cosx and u have to change a cos2x... what would you write 2u... that wouldn't make sensebecause it would come out 2*cosx which does not equatl cos2x

    idk where im lost... i cant logically follow through this problem for some reason, any advice would be amazing, thanks for your time.
     
    Last edited: Feb 2, 2008
  2. jcsd
  3. Feb 2, 2008 #2
    Assuming your work is right, just write (1 - 2u)/u = 1/u - 2
     
  4. Feb 2, 2008 #3
    according the calculator ti 89 the answer is


    -ln abs(cosx) - sinx^2 / 2


    accordign to what i ended up w/ i got

    -1/2 ( ln abs(cosx) - 2*cosx )

    which dosnt make sense my work is wrong it has to be lol

    if u have a few mins to try to work through it on paper let me kno what u see
     
  5. Feb 2, 2008 #4
    Oops, I misread. Just a second.

    Edit: Your work is wrong. cos(2x) =/= 2u if u = cos(x).

    I'd suggest instead of using [itex]sin^{2}(x) = \frac{1 - cos(2x)}{2}[/itex] that you use a more basic identity.
     
    Last edited: Feb 2, 2008
  6. Feb 2, 2008 #5
    i know how to evaluate those easy integrals but the challenge is turning a complex integral into a easy integral
     
  7. Feb 2, 2008 #6
    1/u du is

    ln(u)

    2 =2u

    u = cosx
     
  8. Feb 2, 2008 #7
    yea i have been trying to figure out a simpler substitution and it just isnt working
     
  9. Feb 2, 2008 #8
    What's the very first trig identity you ever learned?
     
  10. Feb 2, 2008 #9
    what about using [tex]sin(x)^{2}=1-cos(x)^{2}[/tex] ?

    you get 2 simple integrals i guess...
     
  11. Feb 2, 2008 #10
    sinx^2 + cosx^2 = 1
     
  12. Feb 2, 2008 #11
    omg i made that way to hard >.< lol thanks
    ill reevaluate it and see what happens
     
  13. Feb 3, 2008 #12
    [tex]\int\frac{\sin^3 x}{\cos x}dx=\int\frac{\sin x\frac{1-\cos(2x)}{2}}{\cos x}dx=\frac{1}{2}\int\frac{\sin x(1-\cos(2x))}{\cos x}\frac{d(\cos x)}{-\sin x}=-\frac{1}{2}\int\frac{1-\cos(2x)}{\cos x}d(\cos x)=[/tex]
    [tex]=-\frac{1}{2}\int\frac{1-(2\cos^2 x-1)}{\cos x}d(\cos x)=-\frac{1}{2}\int\frac{2}{\cos x}-2\cos x d(\cos x)=-\ln\cos x+\frac{\cos^2 x}{2}+C[/tex]

    where [tex]d(\cos x)=-\sin x dx[/tex]
    [tex]dx=\frac{d(\cos x)}{-\sin x}[/tex]
    [tex]\cos(2x)=2\cos^2 x-1[/tex]

    Answer should be [tex]-\ln\cos x+\frac{1}{4}\cos (2x)+C[/tex] according to http://integrals.wolfram.com/index.jsp


    [tex]\int\frac{\sin^3 x}{\cos x}dx=\int\frac{\sin x(1-\cos^2 x)}{\cos x}dx=\int\frac{\sin x(1-\cos^2 x)}{\cos x}\frac{d(\cos x)}{-\sin x}=-\int\frac{1-\cos^2 x}{\cos x}d(\cos x)=[/tex]
    [tex]=-\int\frac{1}{\cos x}-\cos x d(\cos x)=-(\ln\cos x-\frac{\cos^2 x}{2})+C[/tex]
    where [tex]d(\cos x)=-\sin x dx[/tex]
    [tex]dx=\frac{d(\cos x)}{-\sin x}[/tex]

    Does this http://integrals.wolfram.com/index.jsp integrator giving wrong answer? Becouse [tex]\frac{\cos^2 x}{2}[/tex] not equal to [tex]\frac{1}{4}\cos (2x)[/tex]
     
    Last edited: Feb 3, 2008
  14. Feb 3, 2008 #13
    [tex]\frac{\cos^2(x)}{2}=\frac{\cos(2x)}{4}+C_1[/tex]
     
  15. Feb 3, 2008 #14
    Interesting
    [tex]\frac{\cos^2 0.5}{2}\approx 0.385075576[/tex]
    [tex]\frac{\cos 1}{4}\approx 0.135075576[/tex]
    [tex]\frac{\cos^2 0.5}{2}-\frac{\cos 1}{4}=0.25[/tex]
    [tex]\frac{\cos^2 0.3}{2}-\frac{\cos 0.6}{4}=0.25[/tex]
    ...
    But why?
     
  16. Feb 3, 2008 #15
    cos^2(x) = 1/2(1 + cos(2x))
     
  17. Feb 3, 2008 #16
    [tex]\int[/tex] 1 / (sinx - 1) dx

    wth i cant figure this out, thinkin about it to hard again

    i broke a more complex integral down into this final part, diffrent problem


    i know it equals

    cosx / (sinx - 1)
     
  18. Feb 3, 2008 #17
    Try conjugation.
     
  19. Feb 3, 2008 #18
    reduces back into itself
     
  20. Feb 3, 2008 #19
    What about integrating f(x)=1/sin(x)?
     
  21. Feb 3, 2008 #20
    1/sinx does not equal 1 / (sinx -1 )

    in that case u suggest f(x) = ln ( sinx / cosx + 1 )
     
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