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Math Help - Trig Integral

  1. #1
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    Trig Integral

    \int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})
    Then
    \int (1+TAN^2{5x})(SEC^2{5x})
    Then I multiply out
    \int (SEC^2{5x})+(SEC^2{5x}TAN^2{5x})

    I would assume U sub at this point
    U= SEC dU= SECTAN
    But I am not sure, what or how I should bring along with the SEC

    Any help?

    thanks
    Rich
    Last edited by Spoolx; January 1st 2011 at 10:56 AM.
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  2. #2
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    Quote Originally Posted by Spoolx View Post
    \int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})
    Then
    \int (1+TAN^2{5x})(SEC^2{5x})
    Then I multiply out
    \int (SEC^2{5x})(SEC^2{5x}TAN^2{5x})
    Fine so far, except you're missing a plus sign in the last line, which I assume is a typo. Now to substitute, which is really in order to work backwards through the chain rule.

    Just in case a picture helps...



    ... where (key in first spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    The general drift is...



    Spoiler:
    \int 1 + u^2\ du = u + \frac{1}{3} u^3

    and so we can explicate F...



    Hope this helps - if not, someone will probably show the sub with differentials. However, I say...
    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; January 1st 2011 at 11:52 AM. Reason: aberrant 'tan' in spoiler
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  3. #3
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    I edited the original to include the + sign, thanks for catching my mistake
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  4. #4
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    \displaystyle\int{sec^4(5x)}dx=\int{sec^2(5x)sec^2  (5x)}dx=\int{\left[1+tan^2(5x)\right]sec^2(5x)}dx


    Substitute

    u=tan(5x)

    \displaystyle\Rightarrow\frac{du}{dx}=5sec^2(5x)

    by the chain rule

    \displaystyle\Rightarrow\ sec^2(5x)dx=\frac{du}{5}

    The integral becomes

    \displaystyle\frac{1}{5}\int{\left[1+u^2\right]du
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  5. #5
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    Hello, Rich!

    A rehash of Archie's solution . . .


    \displaystyle \int \sec^4\!5x\,dx \:= \:\int (\sec^2\!5x)(\sec^2\!5x)\,dx

    Then: . \displaystyle \int (1+\tan^2\!5x)(\sec^2\!5x)\,dx

    Then I multiply out: . \displaystyle \int (\sec^2\!5x +\sec^2\!5x\tan^2\!5x)\,dx

    \displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)


    For the first integral, use: . \displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C


    For the second integral, let: . u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx

    . . and you have: . \displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C

    Got it?
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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, Rich!

    A rehash of Archie's solution . . .



    \displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)


    For the first integral, use: . \displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C


    For the second integral, let: . u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx

    . . and you have: . \displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C

    Got it?
    Thanks for the reply!
    So to refresh my algebra

    tan^2{5x} = (tan5x)^2 ?

    Also above, is your U supposed to be?
    u= tan5x ?
    if not where does the 5 go?
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  7. #7
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    Quote Originally Posted by Spoolx View Post
    Thanks for the reply!
    So to refresh my algebra

    tan^2{5x} = (tan5x)^2 ? Mr F says: Yes.

    Also above, is your U supposed to be?
    u= tan5x ? Mr F says: Yes. Soroban made a simple typo. You ought to be able to make the appropriate corrections in the calculations.
    if not where does the 5 go?
    ..
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  8. #8
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    Thanks, just wanted to verify. I can take it from here.. Appreciate the help
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