# Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ dx}$

I'm learning integral. Here is my homework:

$$\int_0^1 \sqrt{1+x^2}\;dx$$

I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)

thanks :)

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You could start by setting $x=\tan\theta$. Then $dx=\sec^2\theta\,d\theta$ and the square root simplifies nicely. –  David Mitra Mar 16 '12 at 16:12
Also, $\frac{d}{dx} \sinh^{-1} x = \dfrac{1}{\sqrt{x^2+1}}.$ –  user2468 Mar 16 '12 at 16:18
If the function under the integral sign has form $f(ax^2+bx+c)$ in which $ax^2+bx+c$ has no root then you should be setting $x=A\tan\theta +B$. –  Takasima Senko Mar 16 '12 at 16:19
This reminds me of an other integral –  AD. Mar 22 '12 at 17:02

Since the integrand is a function of $x$ and $\sqrt{ax^{2}+bx+c}$ another option is to use the Euler substitution $\sqrt{ax^{2}+bx+c}=\pm \sqrt{a}x\pm t$, with $a>0$. Choosing $\sqrt{1+x^{2}}=t-x$, squaring both sides and solving for $x$, we obtain $x=\frac{t^{2}-1}{2t}$ and $dx=\frac{ t^{2}+1}{2t^{2}}dt$. The integrand becomes an easily integrable rational fraction of $t$ $$\begin{equation*} \sqrt{1+\left( \frac{t^{2}-1}{2t}\right) ^{2}}\frac{t^{2}+1}{2t^{2}}=\frac{1 }{4}\frac{\left( t^{2}+1\right) ^{2}}{t^{3}}=\frac{1}{2t}+\frac{1}{4t^{3}}+\frac{1}{4}t. \end{equation*}$$ So $$\begin{eqnarray*} \int_{0}^{1}\sqrt{1+x^{2}}dx &=&\int_{1}^{\sqrt{2}+1}\left( \frac{1}{2t}+\frac{1}{4t^{3}}+\frac{t}{4}\right) dt \\ &=&\left. \frac{1}{2}\ln t-\frac{1}{8t^{2}}+\frac{1}{8}t^{2}\right\vert _{1}^{\sqrt{2}+1}. \end{eqnarray*}$$

Added: I've checked the final result:

$$\begin{eqnarray*} \left. \frac{1}{2}\ln t-\frac{1}{8t^{2}}+\frac{1}{8}t^{2}\right\vert _{1}^{\sqrt{2}+1} &=&\frac{\ln \left( \sqrt{2}+1\right) }{2}-\frac{1}{8\left( \sqrt{2} +1\right) ^{2}}+\frac{1}{8}\left( \sqrt{2}+1\right) ^{2}-0 \\ &=&\frac{\ln \left( \sqrt{2}+1\right) }{2}+\frac{\sqrt{2}}{2}. \end{eqnarray*}$$

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Edit in response to a downwote. –  Américo Tavares Apr 6 '13 at 13:39

Integrate by parts to reduce to table integral: $$\int_0^1 \sqrt{1+x^2} \mathrm{d} x = \left. x \sqrt{1+x^2} \right|_0^1 - \int_0^1 \frac{x^2 {\color\green{+1-1}}}{\sqrt{1+x^2}} \mathrm{d} x = \sqrt{2} - \int_0^1 \sqrt{1+x^2} \mathrm{d} x + \int_0^1 \frac{\mathrm{d} x}{\sqrt{1+x^2}}$$ Now solving the equation for $\int_0^1 \sqrt{1-x^2} \mathrm{d} x$, and using table anti-derivative $\int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \operatorname{arcsinh}(x)$: $$\int_0^1 \sqrt{1+x^2} \mathrm{d} x = \frac{1}{2} \left( \sqrt{2} + \operatorname{arcsinh}(1) \right) = \frac{1}{2} \left( \sqrt{2} + \log(1+\sqrt{2}) \right) \approx 1.1478$$

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If we choose to take a (purely) trigonometric route, we might start like this: \eqalign{ x&=\tan\theta\\ dx&=\sec^2\theta\,d\theta\\ I&=\int_0^1\sqrt{1+x^2}\,dx\\ &=\int_0^\frac{\pi}{4}\sec^3\theta\,d\theta\\ } However at this stage, we actually need the substitution $t=\sec\theta+\tan\theta$, believe it or not: \eqalign{ t&=\sec\theta+\tan\theta\\ dt&=\left(\sec\theta\tan\theta+\sec^2\theta\right)\,d\theta =t\sec\theta\,d\theta\\ \frac{dt}{t}&=\sec\theta\,d\theta } And then we need some trigonometry inspirations: \eqalign{ t & = \sec\theta+\tan\theta = \frac{1+\sin\theta}{\cos\theta} = \frac{\cos\theta}{1-\sin\theta} \\ t-\frac1t & = \frac{\cos\theta}{1-\sin\theta} - \frac{\cos\theta}{1+\sin\theta} = 2\tan\theta \\ \tan\theta & = \frac12 \left( t-\frac1t \right) = \frac12 \left( t-t^{-1} \right) \\ \tan^2\theta & = \frac{t^2+t^{-2}}{4} - \frac12 \\ \sec^2\theta & = \frac{t^2+t^{-2}}{4} + \frac12 } We can can then proceed as follows: \eqalign{ I & = \int_0^\frac{\pi}{4}\sec^3\theta\,d\theta = \int_1^{1+\sqrt{2}} \left( \frac{t^2+t^{-2}}{4}+\frac12 \right)\,\frac{dt}{t} \\& = \frac14 \int_1^{1+\sqrt{2}} \left(t+t^{-3}\right)\,dt + \frac12 \int_1^{1+\sqrt{2}} t^{-1}\,dt \\& = \frac18\left[t^2-t^{-2}\right]_1^{1+\sqrt2} + \frac12\left[\ln t \right]_1^{1+\sqrt2} \\& = \frac{\sqrt2+\ln{\left(1+\sqrt2\right)}}{2} } where it is helpful to notice that $$\left(1+\sqrt2\right)^{-1} = \frac{1}{1+\sqrt2} \cdot \frac{1-\sqrt2}{1-\sqrt2} = \frac{1-\sqrt2}{-1}$$ so that $$\left(1+\sqrt2\right)^{-2} = \left(1-\sqrt2\right)^2.$$

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Since we are adding multiple answers, here is one using the Differentiation under the integral sign technique.

If $$F(a) = \int_{0}^{a} f(a,x) \text{ dx}$$ then, under suitable hypotheses,

$$F'(a) = f(a,a) + \int_{0}^{a} \frac{\partial f(a,x)}{\partial a} \text{ dx}$$

(A more general version is here: Wiki page on Differentiating under integral sign)

Now let

$$F(a) = \int_{0}^{a} \sqrt{a^2 + x^2} \text{ dx} \tag{1}$$

A substitution $x = at$ shows us that

$$F(a) = a^2 \int_{0}^{1} \sqrt{1 + t^2} \text{ dt} = Ka^2$$

(we are trying to find the value of $K$).

Thus we must have that $$F'(a) = 2Ka \tag{2}$$

Now go back to (1) and use the techinque of differentiating under the integral sign.

We get

$$F'(a) = \sqrt{2} a + \int_{0}^{a} \frac{a}{\sqrt{a^2 + x^2}} \text{ dx} = \sqrt{2}a + a \sinh^{-1}(x/a)|_0^a = a (\sqrt{2} + \sinh^{-1}(1) - \sinh^{-1}(0))$$

Compare this with (2) and we get the value of $K$.

A similar approach was used here: Definite integral: $\displaystyle\int^{4}_0 (16-x^2)^{\frac{3}{2}} dx$

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The way I've seen this done before starts as bgins did with the substitution

$$x=\tan\theta,\quad dx=\sec^2\theta \; d\theta$$

$$\int \sqrt{1+x^2} \; dx=\int\sec^3\theta \; d\theta$$

From here, we use integration by parts.

$$u=\sec\theta,\quad du=\sec\theta\tan\theta \; d\theta$$

$$dv=\sec^2\theta dx,\quad v=\tan\theta$$

$$\int\sec^3\theta d\theta=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta \; d\theta$$

$$\int\sec\theta\tan^2\theta d\theta=\int\sec\theta(\sec^2\theta-1) \; d\theta=\int\sec^3\theta d\theta-\int\sec\theta d\theta$$

Combining, we get

$$\int\sec^3\theta d\theta=\sec\theta\tan\theta-\int\sec^3\theta \; d\theta+\int\sec\theta d\theta$$

$$2\int\sec^3\theta d\theta=\sec\theta\tan\theta+\int\sec\theta \; d\theta$$

$$\int\sec^3\theta \; d\theta=\frac12\sec\theta\tan\theta+\frac12\ln|\sec\theta+\tan\theta|+C$$

At which point you could substitute back for $x$ or, since your case is a definite integral, evaluate for the corresponding vaules of $\theta$.

Let's try another method just for kicks. We'll try to get rid of that square root with the substitution

$$x=\frac{e^y}2-\frac{e^{-y}}2,\quad dx=\frac{e^y}2+\frac{e^{-y}}2 \; dy$$

$$\int\sqrt{1+x^2}dx=\int\sqrt{1+\left(\frac{e^y}2-\frac{e^{-y}}2\right)^2}\left(\frac{e^y}2+\frac{e^{-y}}2\right) \; dy$$

$$=\int \left(\frac{e^y}2+\frac{e^{-y}}2\right)\sqrt{1+\frac{e^{2y}}4+\frac{e^{-2y}}4-\frac12} \; dy=$$

$$\int \left(\frac{e^y}2+\frac{e^{-y}}2\right)\sqrt{\frac{e^{2y}}4+\frac{e^{-2y}}4+\frac12} \; dy$$

But the term outside the parentheses is the square root of the term under the radical. So we have

$$\int \left(\frac{e^{2y}}4+\frac{e^{-2y}}4+\frac12\right) \; dy = \frac{e^{2y}}8-\frac{e^{-2y}}8+\frac y2+C$$

Only problem now is back-substitution is painful.

$$\frac{e^{2y}}8-\frac{e^{-2y}}8+\frac y2+C=\frac12\left(\frac{e^y}2+\frac{e^{-y}}2\right)\left(\frac{e^y}2-\frac{e^{-y}}2\right)+\frac y2+C$$

$$=\frac{x\sqrt{1+x^2}}2+\frac12\ln(x+\sqrt{1+x^2})+C$$

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Before we proceed let us recall some basics on the hyperbolic functions $$\sinh x=\frac{e^x-e^{-x}}{2}\qquad \text{and }\qquad \cosh x=\frac{e^x+e^{-x}}{2}$$ Below are some relations that are easy to prove $$\cosh^2 x - \sinh^2x = 1\\ \cosh2x=\cosh^2 x + \sinh^2 x\qquad \sinh2x=2\cosh x\sinh x\\ (\cosh x)' =\sinh x\qquad\qquad (\sinh x)' =\cosh x$$

Next, we turn to the integral of the OP.

Let us make the substitution $x=\sinh t$, then $$\int_0^1\sqrt{1+x^2}dx=\int_0^{\rm{arcsinh 1}}\sqrt{1+\sinh^2t}\cosh t\,dt\\ \qquad =\int_0^{\rm{arcsinh 1 }}\cosh^2 t\,dt =\frac12 \int_0^{\rm{arcsinh 1 }}1+\cosh 2 t\,dt\\ =\frac12\left({\rm{arcsinh}}1 + \frac{\sinh2({\rm{arcsinh}} 1)}{2}\right)$$ Now, a calculation based on the definition shows that ${\rm{arcsinh}}x=\log(\sqrt{1+x^2}+x)$, and hence, after using the definition of $\sinh$ once more we arrive at $$\int_0^1\sqrt{1+x^2}dx=\frac{\log(\sqrt{2}+1)}{2} + \frac{\sinh(2\log(\sqrt{2}+1))}{4}\\ = \frac{\log(\sqrt{2}+1)}{2} + \frac{(\sqrt{2}+1)^2 -\frac{1}{(\sqrt{2}+1)^2}}{8} = \frac{\log(\sqrt{2}+1)}{2} + \frac{\sqrt{2}}{2}\\$$

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The Wikipedia article titled Integral of secant cubed explains how that integral comes from $$\int \sqrt{a^2+x^2} \; dx$$ and explains that that arises in rectification of the parabola, rectification of the Archimedean spiral, and quadrature of the helicoid. Then it explains how to find the integral by two different methods.

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As a different aproach. You can go usual trigonemetric integral way via transform $x=iu$

$$\int \sqrt{1+x^2}\;dx=i\int \sqrt{1-u^2}\;du$$

and then $u=\sin(p)$

$$i\int \sqrt{1-u^2}\;du=i\int \sqrt{1-\sin(p)^2} \cos(p) \; dp = i\int ]\cos^2(p) \; dp=i\int \frac{\cos(2p)+1}{2}\;dp=i\frac{\sin(2p)}{4}+i\frac{p}{2}+c=i\frac{\sin(p)\cos(p)}{2}+i\frac{p}{2}+c=i\frac{u \sqrt{1-u^2}}{2}+i\frac{\arcsin(u)}{2}+c$$

$$\int \sqrt{1+x^2}\;dx=i\int \sqrt{1-u^2}\;du=i\frac{u \sqrt{1-u^2}}{2}+i\frac{\arcsin(u)}{2}+c$$$$=\frac{x \sqrt{1+x^2}}{2}+i\frac{\arcsin(-ix)}{2}+c$$

$$\int _0^1 \sqrt{1+x^2}\;dx=\frac{1 \sqrt{2}}{2}+i\frac{\arcsin(-i)}{2}+c - \left(i\frac{\arcsin(0)}{2}+c\right)$$

we need to find $\arcsin(-i)$

$\arcsin(-i)=k$

$\sin(k)=-i$

$\sin(k)= \dfrac{e^{ik}-e^{-ik}}{2i}=-i$

$e^{ik}-e^{-ik}=2$

$e^{ik}=z$

$z^2-2z-1=0$

$z=1+\sqrt{2}$ /// we will need $\ln(z)$ , thus ignore negative root $1-\sqrt{2}$

$ik=\ln(z)=\ln(1+\sqrt{2})$

$k=\dfrac{\ln(1+\sqrt{2})}{i}$

$$\int _0^1 \sqrt{1+x^2}\;dx=\frac{\sqrt{2}}{2}+\frac{\ln(1+\sqrt{2})}{2}$$

Note: I don't claim that it is easy way to find the solution. I just wanted show different approach to the problem. Sometimes we can use complex numbers in such kind problems too.

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