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

1. ## 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?

2. Originally Posted by circuscircus
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?
$\displaystyle \sqrt{1+4x^2} = \sqrt{1+(2x)^2}$
Use substitutiton $\displaystyle t=2x$ to bring integral of $\displaystyle \sqrt{1+t^2}$. Now use arc-tangent substitution.

3. Originally Posted by circuscircus
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:

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

We proceed by trig substitution:

Let $\displaystyle 2x = \tan \theta$

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

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

So our integral becomes

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

Now continue, remember to back substitute when finished to get the answer in terms of $\displaystyle x$

4. Originally Posted by ThePerfectHacker
$\displaystyle \sqrt{1+4x^2} = \sqrt{1+(2x)^2}$
Use substitutiton $\displaystyle t=2x$ to bring integral of $\displaystyle \sqrt{1+t^2}$. 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

5. Originally Posted by circuscircus
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 $\displaystyle \frac {1}{2} \int \sec^2 \theta \cdot \sec \theta~d \theta$

6. Originally Posted by Jhevon
i gave you the integral you have to calculate, forget the formula and do it

you need to find $\displaystyle \frac {1}{2} \int \sec^2 \theta \cdot \sec \theta~d \theta$
I didn't understand what happened to the sqrt and the 1+ inside the sqrt... sry I'm a bit slow...

7. $\displaystyle \int\sqrt{1+4x^2}\,dx=\frac12\int\sqrt{1+4\left(\f rac{\tan^2\theta}4\right)}\sec^2\theta\,d\theta=\f rac12\int\sec\theta\sec^2\theta\,d\theta$

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

Is it clearer now?

You can do the remaining integral using integration by parts.

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# 루트 4t^2 1 적분

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