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Math Help - trig substitution

  1. #1
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    trig substitution

    How would you do this
    integral of tan^3(x)dx
    I broke it down into integral of tan(x)(sec^2(x) - 1)dx
    u = tan(x)
    du = sec^2(x)
    so i got integral of u(du) - integral of (u)
    I am not so sure if this is correct so far, and I do not know where to go from here. Thanks for the help.
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  2. #2
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    Krizalid's Avatar
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    \int {\tan ^3 x~dx} = \int {\tan x(\sec ^2 x - 1)~dx} = \int {\tan x\sec ^2 x~dx} - \int {\tan x~dx}

    ... everything you want... you got it...
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    \int {\tan ^3 x~dx} = \int {\tan x(\sec ^2 x - 1)~dx} = \int {\tan x\sec ^2 x~dx} - \int {\tan x~dx}

    ... everything you want... you got it...
    I'm confused because the answer in the book is 1/2tan^2(x) + ln(cos(x)) + C
    I understand where the 1/2tan^2(x) but where does the ln(cos(x)) come from?
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  4. #4
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    Curious...

    (\tan^2x)'=(\sec^2x)'

    Well \int {\tan x~dx} = \int {\frac{{\sin x}}<br />
{{\cos x}}~dx}

    That's easy, don't ya?
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