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Math Help - trigonometric integrals

  1. #1
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    trigonometric integrals

    Integral of sec^5x dx
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  2. #2
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    Trig Integral

    Hello homerb
    Quote Originally Posted by homerb View Post
    Integral of sec^5x dx
    This is very innocent looking, but is a bit of a brute. Here's my solution - there may be an easier one.

    Note first that \frac{d}{dx}(\tan x) = \sec^2 x, \frac{d}{dx}(\sec x) = \sec x\tan x and \sec^2x = 1 + \tan^2x

    Let I = \int \sec^5x\,dx

    = \int \sec^3x \sec^2x\,dx

    = \sec^3x\tan x - \int 3sec^2x\sec x\tan x\tan x\,dx, by Parts

    = sec^3x\tan x-3\int \sec^3x(\sec^2x -1)\,dx

    = \sec^3x\tan x-3I+3\int\sec^3x\,dx

    \Rightarrow 4I = \sec^3x\tan x + 3I', where

    I' = \int\sec^3x\,dx

    In a similar way you can show that

    2I' = \sec x \tan x + I'', where

    I'' = \int \sec x\,dx

    =\int\frac{\cos x}{\cos^2 x}\,dx

    = \int\frac{\cos x\,dx}{1-\sin^2 x}

    =\int\frac{ds}{1-s^2}, where s = \sin x

    = \frac{1}{2}\int\left(\frac{1}{1-s}+\frac{1}{1+s}\right)ds

    =\frac{1}{2}\text{ln}\left(\frac{1+s}{1-s}\right)

    =\frac{1}{2}\text{ln}\left(\frac{1+\sin x}{1-\sin x}\right)

    When you put all this back together again, you get

    I = \frac{1}{4}\sec^3x\tan x+\frac{3}{8}\sec x \tan x +\frac{3}{16}\text{ln}\left(\frac{1+\sin x}{1-\sin x}\right)

    Grandad
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