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Please Help with Trig Substitution Integration

  1. Feb 3, 2006 #1
    I am not too good with trig identities. I cant seem to figure out how to simplify these trig intergrals. I know I can use a triangle to turn the second problem into a trig integral, but once I have the trig integral, I am lost. Any help would be greatly appriciated.:redface:

    [tex]\int\tan(x)\sec^3(x)dx[/tex]
    [tex]\int\frac{1}{x^2&\sqrt{16-x^2}}dx[/tex]
     
    Last edited: Feb 3, 2006
  2. jcsd
  3. Feb 3, 2006 #2
    Have you tried a u-substitution on the first one yet? Here is a hint what is the derivative of sec?
     
  4. Feb 3, 2006 #3
    see, thats what I thought, but there seems to be an extra secant in there. The derivative of sec is sectan. So, u=sec(x), du=sec(x)tan(x)dx. That gives tan(x)sec^2(x)... right?
     
  5. Feb 3, 2006 #4

    Tide

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    HINT: Let

    [tex]u = \cos x[/tex]

    It will save you a lot of work.
     
  6. Feb 3, 2006 #5
    [tex] u=sec(x) [/tex] [tex]du=sec(x) tan(x)dx [/tex]

    [tex] \int u^2du [/tex]
     
    Last edited: Feb 3, 2006
  7. Feb 3, 2006 #6
    brilliant! Thanks Tide. I should have seen that.
    [tex]\int\frac{sin(x)}{cos^4(x)}dx = \frac{1}{3cos^3(x)}[/tex]
     
  8. Feb 3, 2006 #7
    right, yes I can see that too Valhalla. Thanks so much. Looks like that one was much easier than I made it.
     
  9. Feb 3, 2006 #8
    Any ideas on the second problem? Using a triangle, I have changed it to:

    [tex]\int\frac{1}{(4sin\Theta)^2(4cos\Theta)}d\Theta[/tex]
     
  10. Feb 3, 2006 #9
    wait, I think got it, [tex]u=sin\Theta[/tex]
     
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