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Thread: diff. trig integral

  1. #1
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    diff. trig integral

    $\displaystyle \int tan^3(2t)sec^3(2t)dt = \int tan^2(2t)sec^2(2t)sec(2t)tan(2t)dt
    $

    $\displaystyle u = sec2t$
    $\displaystyle du = sec(2t)tan(2t)dt$

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

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

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

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

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

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

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

    Now do $\displaystyle u=\sec2t.$
    Last edited by Reckoner; Feb 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

    $\displaystyle
    \begin{aligned}
    u&=\sec 2t \\
    du&=2 \sec 2t \tan 2t\,dt.
    \end{aligned}
    $

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

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

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

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

    $\displaystyle u = \sec(2t)$

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

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

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

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

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

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

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

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

    Now do $\displaystyle u=\sec2t.$
    what would du =?
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  6. #6
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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.

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