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Integral of tan^n (x), reduction fromula?

  1. Jul 31, 2016 #1
    1. The problem statement, all variables and given/known data

    reduction.jpg
    The reduction formula for tannx is a confusing matter for me,
    First how do you derive it? Lets use tan3 as an example
    Here in an integral calculator when it gets to tan4(x) it uses the reduction formula

    reduction.jpg

    BUT

    When I ask it to integrate Tan3(x) the answer resembles nothing like the reduction formula.

    reduction.jpg

    So why do you use the reduction formula for tan^4 but not tan^3?
    how do you derive the reduction formula using integration by parts?

    2. Relevant equations

    integration by parts ∫udv=uv-∫vdu
    reduction formula ?

    3. The attempt at a solution

    ∫tan^3 (x) dx

    u= tan^3(x) dv=1
    du= 3tan2(x)sec2(x)dx v = x

    = xtan3 - ∫x3tan2(x)sec2(x)dx

    again, ∫x3tan2(x)sec2(x)dx

    u = x dv= tan3sec2
    du = dx v =(tan ^4 )/¼

    This is going no where to what the reduction formula is supposed to look like...
     
  2. jcsd
  3. Jul 31, 2016 #2

    vela

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    If you apply the reduction formula to ##\tan^3 x##, you get
    $$\int \tan^3 x\,dx = \frac{\tan^2 x}{2} - \int \tan x\,dx.$$ You just have to recognize that since ##\sec^2 x = \tan^2 + 1##, the two results only differ by a constant.

    To derive the reduction formula, you don't want to use integration by parts. Instead, write ##\tan^n x = \tan^{n-2} x \ \tan^2 x = (\tan^{n-2} x)(\sec^2 x-1)## and go from there.
     
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