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Integral 3x (sinx/cos^4x) dx

  1. Oct 15, 2006 #1

    413

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    can anybody help me with this problem
    Evaluate :
    integral 3x (sinx/cos^4x) dx
     
    413, Oct 15, 2006
  2. jcsd
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  3. Oct 15, 2006 #2

    courtrigrad

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    Is it [tex] \int \frac{3x\sin x}{\cos^{4} x} [/tex]?

    Rewrite it as [tex] 3\int x\tan x\ sec^{3} x [/tex] and use integration by parts
     
    Last edited: Oct 15, 2006
    courtrigrad, Oct 15, 2006
  4. Oct 16, 2006 #3

    413

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    i know the equation for intergration by parts is
    intergral u dv = uv -intergral v du

    can u tell me which variable is which?...u, du, v, dv=?...there seems to have 3 different variable.
     
    413, Oct 16, 2006
  5. Oct 17, 2006 #4

    courtrigrad

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    Let [tex] u = x [/tex] and [tex] dv = \ tan x \sec^{3} x [/tex]. You will then need to use integration by parts on [tex] dv [/tex] to get [tex] v [/tex].
     
    Last edited: Oct 17, 2006
    courtrigrad, Oct 17, 2006
  6. Oct 17, 2006 #5

    HallsofIvy

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    I'm sure courtrigrad meant [itex]u= x[/itex] and [itex]dv= tan x sec^3 xdx[/itex] (with out the "x" in dv).
     
    HallsofIvy, Oct 17, 2006
  7. Oct 17, 2006 #6

    courtrigrad

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    yep, thanks catching that :smile:
     
    courtrigrad, Oct 17, 2006
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