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Homework Help: Some-what fancy trig inegral

  1. May 14, 2007 #1
    find the integral of (cosx)^2(tanx)^3

    so first of all i rewrote tan to be (sinx)^3/(cosx)^3 so that i could cancle out the (cosx)^2 on top

    and now i have the integral of (1/cosx)(sinx)^3

    i rewrote that to be (tanx)(sinx)^2...and now i am stuck

    i don't know of any IDs that could make the integral simplier...i am open to suggestions
  2. jcsd
  3. May 14, 2007 #2
    [tex]\frac{\sin^3{x}}{\cos{x}} = \frac{\sin^2{x}sin{x}}{\cos{x}} [/tex]
  4. May 14, 2007 #3
    right...so sinx/cosx would be tanx and that would it it (sinx)^2(tanx).....right?
  5. May 14, 2007 #4
    No. sin^2 would be 1-cos^2
  6. May 14, 2007 #5
    ok let me try that
  7. May 14, 2007 #6
    now i am getting the integral of tanx-sinxcosx

    i know how to integrate sinxcosx but not tanx
  8. May 14, 2007 #7
    That's strange...

    tanx is just sinxcosx, except that the cosx is placed in the denominator. :wink:
  9. May 14, 2007 #8
    oohhh ok...let me try it
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