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Math Help - tough trig integral

  1. #1
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    tough trig integral

    okay so, obviously this is pretty urgent (due tomorrow) and i've been literally working on it all night, to no avail...

    it's the integral of (sec(x))^5 dx

    i have to do it using the integration by parts method

    so far i have..

    integral (sec(x))^5 dx= integral (sec(x))^3*(sec(x))^2 dx

    then i tried to say that u=(sec(x))^3 and thus du= 3(sec(x))^2*(secxtanx) and dv= (sec(x))^2 dx so v= tan x

    it gets really hairy from here..

    i said tanxsec^3(x) - integral tanx*(3sec^3(x)tanx)dx

    after that i got horribly confused.. this is due at 8:30 tomorrow morning, i'm so sorry to come off as someone putting this off until the last minute but i really have been trying to work on it

    someone please help?
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  2. #2
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    Hello, buttonbear!

    Your start is fine . . .


    I \;=\;\int\sec^5\!x\,dx

    i have to do it using the integration by parts method

    So far: . I \;=\;\int\sec^5\!x\,dx \;=\;\int \sec^3\!x\cdot\sec^2\!x\,dx

    Then: . \begin{array}{ccccccc}u&=&\sec^3\!x & & dv &=&\sec^2\!x\,dx \\ du&=& 3\sec^3\!x\tan x\,dx & & v&=& \tan x\end{array}

    I said: . I \;=\;\sec^3\!x\tan x - 3\!\!\int\sec^3\!x\tan^2\!x\,dx . . . . Right!

    Then we have: . I \;=\;\sec^3\!x\tan x - 3\!\!\int\sec^3\!x(\sec^2\!x-1)\,dx

    . . . . . . . . . . . I \;=\;\sec^3\!x\tan x - 3\!\!\int\left(\sec^5\!x - \sec^3\!x\right)\,dx

    . . . . . . . . . . . I \;=\;\sec^3\!x\tan x - 3\underbrace{\int\sec^5\!x\,dx}_{\text{This is }I} +\; 3\!\!\int\sec^3\!x\,dx

    So we have: . I \;=\;\sec^3\!x\tan x - 3I + 3\!\!\int\sec^3\!x\,dx

    . . . . . . . . . 4I \;=\;\sec^3\!x\tan x + 3\!\!\int\sec^3\!x\,dx


    Therefore: . I \;=\;\tfrac{1}{4}\sec^3\!x\tan x + \tfrac{3}{4}\int\sec^3\!x\,dx + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Now, the new integral, \int\sec^3\!x\,dx, can be done by parts.

    Or if you're lucky, you've already memorized its formula:
    . . . \int\sec^3\!x\,dx \;=\;\tfrac{1}{2}\bigg[\sec x\tan x + \ln|\sec x + \tan x|\bigg] + C

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  3. #3
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    oh thank you so much!!!
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