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How to get rid of the integral $\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dx}f(x)\right)^2}dx}$ when $f(x)=x^2$?

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What have you done so far? Have you calculated the integrand? What have you tried to solve the resulting integral? –  Johannes Kloos Mar 29 '12 at 8:33
    
@JohannesKloos I don't know where should I start. –  Garmen1778 Mar 29 '12 at 8:34
    
First step: What is $\sqrt{1+(\frac{d}{dx} x^2)^2}$? –  Johannes Kloos Mar 29 '12 at 8:34
    
A square root that can be expressed like $\sqrt{1+4x^2}$. And now what? –  Garmen1778 Mar 29 '12 at 8:37
2  
The formula describes an arc length of $f(x)$ on $[x_0,x]$. –  Salech Alhasov Mar 29 '12 at 8:46

4 Answers 4

up vote 4 down vote accepted

$$\int_{x_0}^x\!\!\!\sqrt{1+\left(\frac{d}{d\xi}f(\xi)\right)^2}\,d\xi\quad\text{with}\quad f(x)=x^2$$ $$ = \int_{x_0}^x\!\!\!\sqrt{1+4\xi^2}\,d\xi $$

Substituting $2\xi=\sinh(u) \Rightarrow d\xi = \frac{1}{2}\cosh(u)\,du$ results in $$ \int\!\frac{1}{2}\cosh(u)\sqrt{1+\sinh^2(u)}\,du $$ $$ = \frac{1}{2}\int \cosh^2(u)\, du $$

Solving this integral with partial integration gives $$ \int_a^b \cosh^2(u)\,du = \left[ \cosh(u)\sinh(u) \right]_{u=a}^{b} - \int_a^b \sinh^2(u)\, du $$ $$ = \left[ \cosh(u)\sinh(u) \right]_{u=a}^{b} - \int_a^b \left(\cosh^2(u)-1 \right)\,du $$ $$ \Rightarrow \int_a^b \cosh^2(u)\,du = \frac{1}{2}\left(\left[ \cosh(u)\sinh(u) \right]_{u=a}^{b} + (b-a)\right) $$

Pluging in this solution and subsequently undoing the substitution (so we can keep the old limits) gives $$ \frac{1}{4} \left(\left[ 2\xi\sqrt{1+4\xi^2} \right]_{\xi=x_0}^{x} + (\sinh^{-1}(2x)-\sinh^{-1}(2x_0))\right) $$

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In your final equation, aren't $\xi$ and $x$ the same? –  Garmen1778 Apr 25 '12 at 20:33
    
@Garmen1778 No, you still need to plug in the boundaries. I used $[f(\xi)]_{\xi=x_0}^x$ to denote $f(x)-f(x_0)$. –  example Apr 26 '12 at 5:24

As in other answers, let $2x=\tan(\theta)$ $$ \begin{align} \int\sqrt{1+4x^2}\mathrm{d}x &=\frac12\int\sec(\theta)\,\mathrm{d}\tan(\theta)\\ &=\frac12\int\sec^4(\theta)\,\mathrm{d}\sin(\theta)\\ &=\frac12\int\frac{\mathrm{d}\sin(\theta)}{(1-\sin^2(\theta))^2}\\ &=\frac18\int\left({\small\frac{1}{1-\sin(\theta)}+\frac{1}{(1-\sin(\theta))^2}+\frac{1}{1+\sin(\theta)}+\frac{1}{(1+\sin(\theta))^2}}\right)\mathrm{d}\sin(\theta)\\ &=\frac18\left(\log\left(\frac{1+\sin(\theta)}{1-\sin(\theta)}\right)+\frac{1}{1-\sin(\theta)}-\frac{1}{1+\sin(\theta)}\right)+C\\ &=\frac14\left(\log\left(\tan(\theta)+\sec(\theta)\right)+\tan(\theta)\sec(\theta)\right)+C\\ &=\frac14\log\left(2x+\sqrt{1+4x^2}\right)+\frac12x\sqrt{1+4x^2}+C\tag{1} \end{align} $$ Then take the difference of $(1)$ at the limits of integration.

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Summarising the comments, you'll get $$ \int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}f(t)\right)^2}dt} =\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}t^2\right)^2}dt} =\int\limits_{x_0}^{x}{\sqrt{1+4t^2}dt} $$ To solve the last one substitute $t=\tan(u)/2$ and $dt=\sec^2(u)/2du$. Then $\sqrt{1+4t^2}= \sqrt{\tan^2(u)+1}=\sec(u)$, so we get as antiderviative: $$ \begin{eqnarray} \frac{1}{2}\int \sec^3(u) du&=&\frac{1}{4}\tan(u)\sec(u)+\frac{1}{4}\int \sec(u)du+\text{const.}\\ &=&\frac{1}{4}\tan(u)\sec(u)+\frac{1}{4}\log(\tan(u)+\sec(u))+\text{const.} \\ &=& \frac{t}{2}\sqrt{1+4t^2}+\frac{1}{4}\log(2t+\sqrt{1+4t^2})+\text{const.}\\ &=& \frac{1}{4}\left(2t\sqrt{1+4t^2}+\sinh^{-1}(2t) \right)+\text{const.}. \end{eqnarray} $$ Put in your limits and your done: $$ \int\limits_{x_0}^{x}{\sqrt{1+4t^2}dx}=\left[\frac{1}{4}\left(2t\sqrt{1+4t^2}+\sinh^{-1}(2t) \right) \right]_{x_0}^x $$

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"putting in the limits" will not give you the "const.." term that should be well defined as the initial integral has both limits given. –  example Mar 29 '12 at 9:12
    
@example thanks for pointing that out –  draks ... Mar 29 '12 at 21:10

Go back to the basics! You need a derivative of the curve to begin with.

If the integrand has root of sum of squares in the denom Immediately draw a right triangle with appropriate sides and the rest is an algebraic walk.

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