I picked this example for practice and got stuck with it. Someone moderate me if I am in the right path.

I need to calculate the length of arc s, on the section of the curve $y=x^2$ with $0≤x≤1$

My Workings: The formula is $s=\int^b_a\sqrt{1+[f'(x)]^2}dx$

I work out everything under the $\sqrt{}$

my $f(x) =x^2$. I need to get $f'(x)$, which =$2x$

So our function becomes $s=\int^1_0\sqrt{1+[2x]^2}dx$

Then I work out $1+[f'(x)]^2 = 1+4x^2 $

$s=\int^1_0\sqrt{1+[f'(x)]^2}dx =\int^1_0\sqrt{(1+4x^2)}dx$

$s=\int^1_0\sqrt{(1+4x^2)}dx =\int^1_0(1+4x^2)^\frac{1}{2}dx$

I reckon I have to integrate $\int^1_0(1+4x^2)^\frac{1}{2}dx$

The integration part:
I used u-substitution:

let $ u = 1+4x^2$ then $du = 8x dx$

hence $dx = \frac{du}{8x}$

But, now there is trouble with the $8x$ because if I plug in $dx$, I get:

$\int^1_0(u)^\frac{1}{2}\frac{du}{8x} = \int^1_0\dfrac{(u)^\frac{1}{2}}{8x}du$

$= \dfrac{1}{8}\int^1_0\dfrac{1}{x}.{(u)^\frac{1}{2}}du$

Doesn't this give me $(\dfrac{1}{8}.ln|x|.\dfrac{2}{3}.u^\frac{3}{2})]^1_0$

...but... my answer doesn't look right......shouldnt' $x >0$?

How do I end this?

  • 2
    $\begingroup$ When you integrate, you can't anti-differentiate the 1/x part when you don't have a dx. $\endgroup$ – MCT Mar 12 '15 at 21:14
  • 1
    $\begingroup$ $\;x=\frac12\sqrt{u-1}\;$ so you can't consider it a constant wrt $\;u\;$ . In fact, it seems like your substitution doesn't really make things easier. Try $\;2x=\sinh u\;$ and be prepared to go hyperbolic. $\endgroup$ – user177692 Mar 12 '15 at 21:14
  • $\begingroup$ Are you familiar with hyperbolic trigonometric functions? $\endgroup$ – MCT Mar 12 '15 at 21:15

Whenever you see $\sqrt{a^2 \pm b^2x^2}$ in an integral, whether in the numerator or denominator, that is often an indicator that it can be solved through trig sub (or hyperbolic trig, which is pretty useful if you know it.) If we imagine a right triangle with angle $\theta$, opposite leg of length $2x$ and adjacent leg of length $1$, we can solve for the length of the hypotenuse of this triangle with the Pythagorean theorem to get $\sqrt{1+4x^2}$. We can also work out the following relations from the triangle:

$$\tan(\theta) = 2x \qquad \frac{1}{2}\sec^2(\theta)d\theta = dx \qquad \sec(\theta) = \sqrt{1+4x^2}$$ Hence $$\begin{align}\int_0^1 \sqrt{1+4x^2}\space dx = \int_0^{\arctan(2)}\sec(\theta)\frac{1}{2}\sec^2(\theta)d\theta \\ = \frac{1}{2}\int_0^{\arctan(2)}\sec^3(\theta)d\theta \end{align}$$

At this point, you will need to tangle with $\sec^3(\theta)$. This can be found in an integral table and you'll have your answer.

If you are now curious about how to tackle this integral without the help of a table, read on! It is actually not too hard to solve with integration by parts if you already know the integral of $\sec(\theta)$. The way to evaluate $\sec(\theta)$ is anything but intuitive, but it can be solved by multiplying by a clever form of $1$. Namely, $1 = \frac{\sec(\theta)-\tan(\theta)}{\sec(\theta)-\tan(\theta)}$. So you get $$\begin{align}\int \sec(\theta)d\theta =\int \sec(\theta) \frac{\sec(\theta)-\tan(\theta)}{\sec(\theta)-\tan(\theta)} d\theta \\ = \int \frac{\sec^2(\theta)-\tan(\theta)\sec(\theta)}{\sec(\theta)-\tan(\theta)}d\theta \end{align}$$ Now make the substitution $u = \sec(\theta)-\tan(\theta)$ and see if you have any luck! After that, you can solve $\int \sec^3(\theta)d\theta$ with integration by parts by rewriting it as $\int \sec^2(\theta)\cdot \sec(\theta)d\theta$ and using the substitutions $u =\sec(\theta), \space dv = \sec^2(\theta)$. You will need the identity $\tan^2(\theta) = \sec^2(\theta)-1$, but if you pay close attention, you will see that you only need to integrate by parts once. This is all probably far outside the scope of your original question, but it is great practice nonetheless.


Use the substitution $2x = \sinh u$. This gives you $\frac{dx}{du} = \frac{1}{2}\cosh u$. So you get: $$ \int_0^1 (1+4x^2)^\frac{1}{2} dx= \int_0^{\sinh^{-1}2} (1+\sinh^{2} u)^\frac{1}{2} \frac{1}{2} \cosh udu = \frac{1}{2} \int_0^{\sinh^{-1}2}\cosh^{2} udu = \frac{1}{2} \int_0^{\sinh^{-1}2} \frac{1}{2} (\cosh2u +1)du = \frac{1}{4}[\frac{1}{2}\sinh 2u + u]_0^{\sinh^{-1}2}$$ This substitution is often very useful when you have an expression of the sort $(1+x^2)^{\frac{1}{2}}$ because of the hyperbolic identity $\cosh^{2} u = 1+\sinh^{2} u$.


If you are not familiar with the hyperbolic identities, you may use $2x=tant$ You integral then becomes of the form $sec^3t$ which is a standard integral or otherwise can be done through integration by parts. Since the limits are 0 and 1, you can tag them along as well.


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