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Math Help - Volume integral

  1. #1
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    Volume integral

    Is this correct?

    (2.secx / (2 + tanx))^2 = (2.tanx/(2+tanx) ???

    or

    (2.secx / (2 + tanx))^2= 2sec^2 x/ 4 + tan^2 x

    Thanks

    ANy help would be great
    Last edited by offahengaway and chips; July 14th 2009 at 10:04 AM.
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  2. #2
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    Quote Originally Posted by offahengaway and chips View Post
    Is this correct?

    (2.secx / (2 + tanx))^2 = (2.tanx/(2+tanx) ???

    or

    (2.secx / (2 + tanx))^2= 2sec^2 x/ 4 + tan^2 x
     \left(\frac{2\sec{x}}{2 + \tan{x}}\right)^2 = \frac{\left(2\sec{x}\right)^2}{\left(2 + \tan{x}\right)^2} =  \frac{4}{\left(2 \cos{x} + \sin{x} \right)^2}

    I don't think it simplifies further. Also, could you explain how you got the answers you did? Might be easier to spot where something may have gone wrong.
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  3. #3
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    Quote Originally Posted by pomp View Post
     \left(\frac{2\sec{x}}{2 + \tan{x}}\right)^2 = \frac{\left(2\sec{x}\right)^2}{\left(2 + \tan{x}\right)^2} =  \frac{4}{\left(2 \cos{x} + \sin{x} \right)^2}

    I don't think it simplifies further. Also, could you explain how you got the answers you did? Might be easier to spot where something may have gone wrong.
    im trying to integrate this

    \frac{\left(2\sec{x}\right)^2}{\left(2 + \tan{x}\right)^2}
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  4. #4
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    sorry ,

    im trying to integrate


    (2secx)^2/ (2+tanx)^2
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  5. #5
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    Oh OK. Then your first attempt was correct. You should be more clear what it is you're asking, as it was written it wasn't obvious you were integrating the expressions.

    For future reference, if you just want to check your answers, try: http://www.wolframalpha.com . Just type in integrate 'expression' and press enter.
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  6. #6
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    Why is it different from your attempt?
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  7. #7
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    Quote Originally Posted by pomp View Post
     \left(\frac{2\sec{x}}{2 + \tan{x}}\right)^2 = \frac{\left(2\sec{x}\right)^2}{\left(2 + \tan{x}\right)^2} =  \frac{4}{\left(2 \cos{x} + \sin{x} \right)^2}

    What I wrote is correct (note there are no integral signs)

    Whereas,

    \int \left(\frac{2\sec{x}}{2 + \tan{x}}\right)^2 dx = \frac{2 \tan{x}}{2+\tan{x}}  + C
    Last edited by pomp; July 14th 2009 at 10:36 AM. Reason: constant of integration ;)
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  8. #8
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    Did you use substitution to do this integration?
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  9. #9
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    Quote Originally Posted by offahengaway and chips View Post
    Did you use substitution to do this integration?

    Yep, set u = \tan{x} + 2 giving dx = \frac{du}{\sec^2{x}}

    Everything cancels nicely.
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  10. #10
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    still cant see it
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  11. #11
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    Sorry, but I originally assumed you had done the integral yourself in two diferent ways and arrived at different answers and wanted to check which was correct.

    Have you done integration by substitution before?

    Try working through setting u = \tan{x} + 2

    Do you know how to differentiate \tan{x} to get \frac{du}{dx} ? I already gave you the answer to this part in my previous post.

    Having done that simply substitute everything in.

    Post some working and let me know exactly where you get stuck.

    I'm going out now so won't be able to reply again, if you're still stuck use the link I gave you. type in 'integrate (2 sec(x) / (2 + tan(x)))^2' and the press enter. When it gives you the answer click on more steps.
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