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Math Help - what is integral of sqrt(1+4x^2)?

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    what is integral of sqrt(1+4x^2)?

    what is integral of sqrt(1+4x^2)?

    I tried looking at the trig rules but it was only if that part was in the denominator ... how would I find the integral of that?
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    Quote Originally Posted by circuscircus View Post
    what is integral of sqrt(1+4x^2)?

    I tried looking at the trig rules but it was only if that part was in the denominator ... how would I find the integral of that?
    \sqrt{1+4x^2} = \sqrt{1+(2x)^2}
    Use substitutiton t=2x to bring integral of \sqrt{1+t^2}. Now use arc-tangent substitution.
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    Quote Originally Posted by circuscircus View Post
    what is integral of sqrt(1+4x^2)?

    I tried looking at the trig rules but it was only if that part was in the denominator ... how would I find the integral of that?
    we will use trigonometric substitution:

    \int \sqrt {1 + 4x^2}~dx

    We proceed by trig substitution:

    Let 2x = \tan \theta

    \Rightarrow 2~dx = \sec^2 \theta ~d \theta

    \Rightarrow dx = \frac {1}{2} \sec^2 \theta ~d \theta

    So our integral becomes

    \frac {1}{2} \int \sec^2 \theta \cdot \sec \theta~d \theta

    Now continue, remember to back substitute when finished to get the answer in terms of x
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    Quote Originally Posted by ThePerfectHacker View Post
    \sqrt{1+4x^2} = \sqrt{1+(2x)^2}
    Use substitutiton t=2x to bring integral of \sqrt{1+t^2}. Now use arc-tangent substitution.
    I got up to that but afterwards the only formulas I found were for 1/sqrt(a^2+u^2)

    The problem I had was the sqrt part was in the demoniator so none of the trig formulas worked for me
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by circuscircus View Post
    I got up to that but afterwards the only formulas I found were for 1/sqrt(a^2+u^2)
    The problem I had was the sqrt part was in the demoniator so none of the trig formulas worked for me
    i gave you the integral you have to calculate, forget the formula and do it

    you need to find \frac {1}{2} \int \sec^2 \theta \cdot \sec \theta~d \theta
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    Quote Originally Posted by Jhevon View Post
    i gave you the integral you have to calculate, forget the formula and do it

    you need to find \frac {1}{2} \int \sec^2 \theta \cdot \sec \theta~d \theta
    I didn't understand what happened to the sqrt and the 1+ inside the sqrt... sry I'm a bit slow...
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  7. #7
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    \int\sqrt{1+4x^2}\,dx=\frac12\int\sqrt{1+4\left(\f  rac{\tan^2\theta}4\right)}\sec^2\theta\,d\theta=\f  rac12\int\sec\theta\sec^2\theta\,d\theta

    Where 1+\tan^2\theta=\sec^2\theta

    Is it clearer now?

    You can do the remaining integral using integration by parts.
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