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Find the integral Please HELP

  1. Mar 20, 2008 #1
    Find the integral...Please HELP!!!!

    1. The problem statement, all variables and given/known data

    Find the integral involving sin, cos, sec, and tan:
    A. sin^3x * sqrt(cosx)dx
    B. sec^3(2x) * tan(2x)dx

    2. Relevant equations

    3. The attempt at a solution
    I will start with part A first...
    A. sin^3x * sqrt(cosx) dx
    sin^2*x * sqrt(cosx) * sinx dx
    (1-cos^2*x) * sqrt(cosx) * sinx dx
    sinx * sqrt(cosx) - sinx * cosx dx

    but not I don't know what to do??
  2. jcsd
  3. Mar 20, 2008 #2

    Gib Z

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    You did well, but just stop on the second last line, make a substitution u= sin x, then you have a simple integral in powers of u no?
  4. Mar 20, 2008 #3
    So I stop here....

    (1-cos^2*x) * sqrt(cosx) * sinx dx


    I am not understanding how to use u-substitution with this? u = sinx, du = cosx but thats not it?
  5. Mar 20, 2008 #4

    Gib Z

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    SOrry sorry! I meant u=cos x, du = -sin x dx.
  6. Mar 20, 2008 #5
    Umm but I am still not understanding lol im sorry...

    (1-cos^2*x) * sqrt(cosx) * sinx dx

    u=cos x, du = -sin x dx

    So if I do u^1/2 the integral is 2/3 * u^3/2

    but that only gets me....... sqrt(cosx) * -sinx dx

  7. Mar 20, 2008 #6

    Gib Z

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    When you make the substitution,

    [tex]\int (1- \cos^2 x) \sqrt{\cos x} \sin x dx = - \int (1-u^2) \sqrt{u} du = -\int ( u^{1/2} - u^{5/2}) du [/tex]

    You should be able to finish it off with the power rule, and replace back in u=cos x at the end.
    Last edited: Mar 20, 2008
  8. Mar 20, 2008 #7
    so if I leave it at...

    1 - u^2 * sqrt(u) du

    I could write it as

    x - cosx^2 * sqrt(cosx)
  9. Mar 20, 2008 #8

    Gib Z

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    Did you follow my last post :( ? Do you know how to integrate those terms I had in the last post?
  10. Mar 20, 2008 #9
    woops that wasn't done right...

    (x - 1/3*cosx^3) * 2/3*sqrt(cosx)^3/2

    can I write it like this...I know its not completely reduced yet but I will do that.
  11. Mar 20, 2008 #10

    Gib Z

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    Well when I evaluate the integral I posted, I get [tex]2\cos^{3/2} x \left( \frac{1}{3} - \frac{\cos^2 x}{7}} \right) [/tex]
  12. Mar 20, 2008 #11
    Why am I not understanding this....

    (1-u^2) * 2/3 u^3/2

    so I plug u=cosx back in and integrate...

    (x - 1/3*cosx^3) * 2/3*(cosx)^3/2

    can you show me how you got your answer?
  13. Mar 20, 2008 #12

    Gib Z

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    Do you follow post 6? I get my answer from post 6 by directly using the power rule.
  14. Mar 20, 2008 #13
    I thought it would be....

    u^3/2 - u^5/2
  15. Mar 20, 2008 #14
    [tex]\int\sin^3 x\sqrt{\cos x}dx[/tex]

    [tex]\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx[/tex]

    [tex]u=\cos x[/tex]
    [tex]du=-\sin xdx[/tex]

    [tex]-\int(1-u^2)\sqrt udu[/tex]

    [tex]-\int(u^{\frac 1 2}-u^{\frac 5 2})du[/tex]
  16. Mar 20, 2008 #15

    Gib Z

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    Check the power rule again.

    I believe that was already well established in this thread.
  17. Mar 21, 2008 #16
    Okay I don't know why I am not seeing what you are doing but when you have...
    (1-u^2) * sqrt u

    don't you multiply u^1/2 by (1-u^2) ???

    and get u^1/2 - u^3/2

    so I don't understand why the second term is ^5/2?
  18. Mar 21, 2008 #17
    BuBbLeS ... what is 1.5 and 2.5 in fractions?
  19. Mar 21, 2008 #18
    [tex]x^a\cdot x^b = x^a^+^b[/tex]

    Now add 1/2 and 2.
  20. Mar 22, 2008 #19
    OMG I feel like such an IDIOT right now...thanks!
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