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Math Help - Integration with square roots in the denominator

  1. #1
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    Integration with square roots in the denominator

    \int \frac {dx}{\sqrt {x^2+a^2}}

    In my solution book it says

    Let t=x+\sqrt {x^2+a^2}

    How do you figure what to use as substitution

    I'm too lazy to type this out, so here's a picture.





    In the circled part, how do they get from that trig expression to that?
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  2. #2
    MHF Contributor Amer's Avatar
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    \left( \frac{1+sinx}{cosx}\right)^2=\left(\frac{1}{cosx} +\frac{sinx}{cosx}\right)

    for your integration use the sub

    let atan(u)=x ..... asec^2(u) du=dx

    tanu=\frac{x}{a} as you can see in the attachment

    sec(u)=\frac{\sqrt{a^2+x^2}}{a} the rest for you

    Integration with square roots in the denominator-sesesese.jpg
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  3. #3
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    Go this website
    Type in the exact expression: integrate dx/sqrt[x^2 + a^2]
    Click the equal sign at the end. What do you get?
    Ckick the show steps button.
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  4. #4
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    I think you misunderstood my question.

    I asked how \frac {1}{\cos x} + \tan x = \sqrt {x^2+1} +x
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by chengbin View Post
    I think you misunderstood my question.

    I asked how \frac {1}{\cos x} + \tan x = \sqrt {x^2+1} +x
    At the start of the question it states let x = \tan{\theta}.

    Can you work out the \frac{1}{\cos} part?
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  6. #6
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    Thanks!
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  7. #7
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    OK, how about the first question?
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