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Math Help - help with integrating fractions with e in them

  1. #1
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    help with integrating fractions with e in them

    Hey everyone, first post ever in here. This is my 2nd time taking Calculus 2 and I feel like I am worse off than the first time.

    I have a quick set of 4 problems that Im not sure I understand how to do. Apparently 1, 2 and 3 can be solved with u-substitution and a formula, while 4 just needs a u-substitution.

    Here are the problems, sorry about the format:

    1: \int\frac{1}{\sqrt{9-e^{2x}}}\,dx

    2: \int\frac{e^x}{\sqrt{9-e^{2x}}}\,dx

    3: \int\frac{1}{e^x\sqrt{9-e^{2x}}}\,dx

    4: \int\frac{e^2x}{\sqrt{9-e^{2x}}}\,dx

    I feel like Ive lost a lot of my fundemental calc skills.

    Thanks in advance!!

    -WhiteX
    Last edited by whitex; January 21st 2009 at 07:53 PM.
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  2. #2
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    Hello, whitex!

    Welcome aboard!
    I'll run through the first one . . . it takes quite a bit of work.


    1)\;\;\int \frac{dx}{\sqrt{9-e^{2x}}}
    We use a Trig Substitution . . .

    Let: e^x \:=\:3\sin\theta \quad\Rightarrow\quad x \:=\:\ln(3\sin\theta) \quad\Rightarrow\quad dx \:=\:\frac{3\cos\theta}{3\sin\theta}d\theta \:=\:\cot\theta\,d\theta

    . . and: . \sqrt{9-e^{2x}} \:=\:\sqrt{9-9\sin^2\!\theta} \:=\:\sqrt{9(1-\sin^2\!\theta)} \:=\:\sqrt{9\cos^2\!\theta} \:=\:3\cos\theta


    Substitute: . \int\frac{\cot\theta\,d\theta}{3\cos\theta} \;=\;\frac{1}{3}\int\frac{\frac{\cos\theta}{\sin\t  heta}}{\cos\theta}\,d\theta \;=\;\frac{1}{3}\int\frac{d\theta}{\sin\theta} \;=\;\frac{1}{3}\int\csc\theta\,d\theta

    Integrate: . \frac{1}{3}\ln|\csc\theta - \cot\theta| + C .[1]


    Back-substitute: . 3\sin\theta \:=\:e^x\quad\Rightarrow\quad \sin\theta \:=\:\frac{e^x}{3} \:=\:\frac{opp}{hyp}

    We see that \theta is in a right triangle with: . opp = e^x,\;hyp = 3

    . . Using Pythagorus, we have: . adj = \sqrt{9-e^{2x}}

    Hence: . \csc\theta \:=\:\frac{3}{e^x},\;\cot\theta \:=\:\frac{\sqrt{9-e^{2x}}}{e^x}


    Substitute into [1]: . \frac{1}{3}\ln\left|\frac{3}{x} - \frac{\sqrt{9-e^{2x}}}{e^x}\right| + C \;\;=\;\;\frac{1}{3}\ln\left|\frac{3-\sqrt{9-e^{2x}}}{e^x}\right| + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    This answer can be simplified further . . .

    . . \frac{1}{3}\bigg[\ln\left|3 - \sqrt{9-e^{2x}}\right| - \ln\left(e^x\right)\bigg] + C \;\;=\; \;\frac{1}{3}\bigg[\ln\left|3-\sqrt{9-e^{2x}}\right| - x\bigg] + C

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  3. #3
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    Wow thanks I appreciate it.

    This site is great. Any way you want to assist me in the other 3? If not, thats ok!
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  4. #4
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    Hello, whitex!

    #3 and #4 are similar to the first one.
    #2 is the easiest of them.



    2)\;\;\int\frac{e^x\,dx}{\sqrt{9-e^{2x}}}

    Let: u \:=\:e^x\quad\Rightarrow\quad du \:=\:e^x\,dx

    Substitute: . \int\frac{ du}{\sqrt{9-u^2}} \;=\;\sin^{-1}\left(\tfrac{u}{3}\right) + C

    Back-substitute: . \sin^{-1}\left(\frac{e^x}{3}\right) + C




    3)\;\;\int\frac{dx}{e^x\sqrt{9-e^{2x}}}
    This one uses the same substitution as #1 . . .

    Let: e^x \:=\:3\sin\theta \quad\Rightarrow\quad x \:=\:\ln(3\sin\theta) \quad\Rightarrow\quad dx \:=\:\cot\theta\,d\theta
    . . and: . \sqrt{9-e^{2x}} \:=\:3\cos\theta


    Substitute: . \int\frac{\cot\theta\,d\theta}{3\sin\theta\cdot3\c  os\theta} \;=\;\frac{1}{9}\int\frac{d\theta}{\sin^2\!\theta} \;=\;   \frac{1}{9}\int\csc^2\!\theta\,d\theta \;=\;  -\frac{1}{9}\cot\theta + C


    Back-substitute: . e^x \:=\:3\sin\theta \quad\Rightarrow\quad\sin\theta \:=\:\frac{e^x}{3} \:=\:\frac{opp}{hyp}
    . . Then: . adj \:=\:\sqrt{9-e^{2x}}


    And we have: . -\frac{1}{9}\,\frac{\sqrt{9-e^{2x}}}{e^x} + C




    4)\;\;\int\frac{e^{2x}\,dx}{\sqrt{9-e^{2x}}}
    Same subsitution as #3 . . .


    Substitute: . \int\frac{(9\sin^2\!\theta)(\cot\theta\,d\theta)}{  3\cos\theta} \;=\;3\int\sin\theta\,d\theta \;=\;-3\cos\theta + C


    Back-substitute: . -\frac{1}{3}\cdot\frac{\sqrt{9-e^{2x}}}{3} + C \;=\;-\sqrt{9-e^{2x}} + C

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