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Math Help - diff. trig integral

  1. #1
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    diff. trig integral

    \int tan^3(2t)sec^3(2t)dt = \int tan^2(2t)sec^2(2t)sec(2t)tan(2t)dt<br />

    u = sec2t
    du = sec(2t)tan(2t)dt

    \int (tan^2(2t)u^2)du

    Am I going in the right direction?
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  2. #2
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    Quote Originally Posted by saiyanmx89 View Post
    \int tan^3(2t)sec^3(2t)dt = \int tan^2(2t)sec^2(2t)sec(2t)tan(2t)dt<br />
    Try this:

    \int\tan^32t\sec^32t\,dt

    =\int\tan^22t\tan2t\sec^32t\,dt

    =\int\left(\sec^22t-1\right)\tan2t\sec^32t\,dt

    =\int\tan2t\sec^52t\,dt-\int\tan2t\sec^32t\,dt

    =\int(\sec2t\tan2t)\sec^42t\,dt-\int(\sec2t\tan2t)\sec^22t\,dt

    Now do u=\sec2t.
    Last edited by Reckoner; February 24th 2009 at 05:19 PM. Reason: Careless error.
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  3. #3
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    You are going in the right direction, only

    <br />
\begin{aligned}<br />
u&=\sec 2t \\<br />
du&=2 \sec 2t \tan 2t\,dt.<br />
\end{aligned}<br />

    Now you can use the fact that \tan^2 t+1=\sec^2 t.
    Last edited by Scott H; February 24th 2009 at 05:51 PM.
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  4. #4
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    \int \tan^3(2t)\sec^3(2t) \, dt

    \int \tan^2(2t)\sec^2(2t)\sec(2t)\tan(2t) \, dt

    \int [\sec^2(2t) - 1]\sec^2(2t)\sec(2t)\tan(2t) \, dt

    \int [\sec^4(2t) - \sec^2(2t)]\sec(2t)\tan(2t) \, dt

    u = \sec(2t)

    du = 2\sec(2t)\tan(2t) \, dt

    \frac{1}{2} \int [\sec^4(2t) - \sec^2(2t)] \cdot 2\sec(2t)\tan(2t) \, dt

    \frac{1}{2} \int u^4 - u^2 \, du
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  5. #5
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    Quote Originally Posted by Reckoner View Post
    Try this:

    \int\tan^32t\sec^32t\,dt

    =\int\tan^22t\tan2t\sec^32t\,dt

    =\int\left(\sec^22t-1\right)\tan2t\sec^32t\,dt

    =\int\tan2t\sec^52t\,dt-\int\tan2t\sec^32t\,dt

    =\int(\sec2t\tan2t)\sec^42t\,dt-\int(\sec2t\tan2t)\sec^22t\,dt

    Now do u=\sec2t.
    what would du =?
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  6. #6
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by saiyanmx89 View Post
    what would du =?
    You should learn and know the derivatives of all six of the basic trigonometric functions.

    \frac d{dx}[\sec x]=\sec x\tan x
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