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Thread: Need Guidance...

  1. #1
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    Exclamation Need Guidance...

    Help with calculus problems needed? - Yahoo! Answers


    I could use some advice im just looking for hints as to how to attack these problems... It's the beginning of Calc 2 and i didnt get the background i should have due to taking 1 in highschool with a not so great teacher.

    Thank you
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  2. #2
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    ok I figured out the x+1 over sqrt(16-x^2) one...


    down to only looking for help on the other two
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Juk3boxh3ro View Post
    ok I figured out the x+1 over sqrt(16-x^2) one...


    down to only looking for help on the other two
    $\displaystyle \int\frac{e^x\,dx}{e^{2x}-4}$

    Make the subsitution $\displaystyle u=e^x$ to get $\displaystyle \int\frac{\,du}{u^2-4}$.

    Can you take it from here?
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  4. #4
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    is the answer 1/2 arc tan e^x / 2 + C??
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Juk3boxh3ro View Post
    is the answer 1/2 arc tan e^x / 2 + C??
    Not!!

    $\displaystyle \int {\frac{{{e^x}}}
    {{{e^{2x}} - 4}}dx} = \left\{ \begin{gathered}
    {e^x} = u, \hfill \\
    {e^x}dx = du \hfill \\
    \end{gathered} \right\} = \int {\frac{{du}}
    {{{u^2} - 4}}} = \int {\frac{{du}}
    {{\left( {u - 2} \right)\left( {u + 2} \right)}}} .$

    $\displaystyle \frac{1}{{\left( {u - 2} \right)\left( {u + 2} \right)}} = \frac{A}{{u - 2}} + \frac{B}{{u + 2}} \Leftrightarrow 1 = \left( {u + 2} \right)A + \left( {u - 2} \right)B.$

    $\displaystyle {\text{If }}u = 2{\text{ then }}1 = 4A \Leftrightarrow A = \frac{1}
    {4}.$

    $\displaystyle {\text{If }}u = - 2{\text{ then }}1 = - 4B \Leftrightarrow B = - \frac{1}{4}.$

    $\displaystyle \int {\frac{{du}}
    {{{u^2} - 4}}} = \frac{1}
    {4}\int {\left( {\frac{1}
    {{u - 2}} - \frac{1}
    {{u + 2}}} \right)du} = \frac{1}
    {4}\left( {\ln \left| {u - 2} \right| - \ln \left| {u + 2} \right|} \right) + C =$

    $\displaystyle = \frac{1}
    {4}\ln \left| {\frac{{u - 2}}
    {{u + 2}}} \right| + C = \frac{1}
    {4}\ln \left| {\frac{{{e^x} - 2}}
    {{{e^x} + 2}}} \right| + C.$
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  6. #6
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    oh i see... sorry i used the wrong rule my notes are squeezed together and i lined them up wrong

    thanks alot tho...

    how about the algebraic form of tan (arc cos x/2)

    i drew the right triangle x is on the adjacent side 2 is on the hypotenuse then you take the arc cos rule of u prime over square root of u^2 - 1

    so...

    the opposite side is 1/4 x^2 - 1??

    idk sorry im so clueless...
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