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Math Help - integrate the function....

  1. #1
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    integrate the function....

    ok i have a problem and i dont remember how to integrate...


    1/(cos^2(u) + 3)du

    the function should read 1 over the function of cosine squared of u +3
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  2. #2
    A Plied Mathematician
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    WolframAlpha to the rescue. Click on Show Steps to see how they got it.
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  3. #3
    Senior Member DeMath's Avatar
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    The WolframAlpha shows not a very elegant result for this example

    \displaystyle{\begin{aligned}\int\dfrac{du}{\cos^2  u+3}&=\int\dfrac{du}{\cos^2u+3(\sin^2u+\cos^2u)}=\  int\dfrac{du}{4\cos^2u+3\sin^2u}\\&=\dfrac{1}{4}\i  nt\dfrac{1}{1+{\left(\dfrac{\sqrt3}{2}\tan{u}\righ  t)\!}^2}\dfrac{du}{\cos^2u}=\left\{\begin{gathered  }\dfrac{\sqrt3}{2}\tanu=t\hfill\\\dfrac{du}{\cos^2  u}=\dfrac{2t}{\sqrt3}\,dt\hfill\\\end{gathered}\ri  ght\}\\&=\dfrac{1}{2\sqrt3}\int\dfrac{dt}{1+t^2}=\  dfrac{\sqrt3}{6}\arctan{t}+C\\&=\dfrac{\sqrt3}{6}\  arctan\!\left(\dfrac{\sqrt3}{2}\tan{u}\right)+C\en  d{aligned}}
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DeMath View Post
    The WolframAlpha shows not a very elegant result for this example

    \displaystyle\left\{\begin{gathered}\dfrac{\sqrt3}  {2}\tanu=t\hfill\\\dfrac{du}{\cos^2u}=\dfrac{2t}{\  sqrt3}\,dt\hfill\\\end{gathered}\right\}
    You meant
    \displaystyle t = \frac{\sqrt{3}}{2} tan (u)

    I don't know why that didn't come out.

    -Dan
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  5. #5
    A Plied Mathematician
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    Quote Originally Posted by topsquark View Post
    You meant
    \displaystyle t = \frac{\sqrt{3}}{2} tan (u)

    I don't know why that didn't come out.

    -Dan
    No space between the \tan command and the u. Should have typed "\tan u" instead of "\tanu".
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  6. #6
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    ok so i need to get a pocket version of wolfram..lol thanks everybody i thought i was losing it..when i couldnt figure it out....
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