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Reduction formula

  1. Apr 20, 2007 #1
    Show that [tex] tan^n(x) = tan^{n-2}(x)(sec^2(x)-1) \\[/tex]. Hence if [tex] I_n = \int_0^{\frac{\pi}{4}}tan^n(x)dx \\[/tex]. Prove that [tex] I_n = \frac{1}{n-1} - I_{n-2} \\[/tex], and evaluate [tex]I_5{/tex]
    My effort:
    [tex] I_n = \int_0^{n-2}tan^{n-2}x(sec^2x-1)dx \\[/tex] [tex] du = (n-2)tan^{n-3}x sec^2x dx \\[/tex], and [tex] v = \int(sec^2x - 1) dx = tanx -x \\ [/tex]. Therefore:
    [tex] I_n = (tan^{n-2}x(tanx - x)) - (n-2)\int_0^{\frac{\pi}{4}}tan^{n-3}xsec^2x(tanx - x)dx \\ [/tex].
    Which implies, [tex] I_n = (tan^{n-2}x(tanx - x) - (n-2)\int_0^{\frac{\pi}{4}}tan^{n-2}sec^2x dx - (n-2)\int tan^{n-3}x(sec^2x -1) dx \\[/tex]. Therefore that implies :[tex]
    I_n = (tan^{n-1}x - tan^{n-2}(x) x) -(n-1)\int tan^{n-2}x(sec^2x -1)dx + (n-1)\int xtan^{n-3}x(sec^2x-1)dx \\[/tex]

    The end result I get is : [tex] I_n = 1 - \frac{\pi}{4} - (n-1)I_n + (n-1)\int_0^{\frac{\pi}{4}}tan^{n-2}(x) x dx [/tex]. I must have gone wrong some where to get this result. Thanks for the help.
     
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 20, 2007 #2

    AlephZero

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    The first identity is true because [tex]\sec^2 x = 1 + \tan^2 x[/tex]

    For the integral, use that identity:

    [tex]I_n = \int_0^{\frac \pi 4} tan^{n-2}x(sec^2x-1) dx[/tex]
    [tex]I_n = \int_0^{\frac \pi 4} tan^{n-2}xsec^2xdx - I_{n-2}[/tex]

    Evaluating the integral is straightforward - let u = tan x.
     
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