Computing $\int \sqrt{1+4x^2} \, dx$

Question

It was said that we did not learn how to take this integral in class yet, and that we should just use a graphing utility to find the integral, so being me i took that as a challenge and tried to take it using methods i knew.

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

$$ \sqrt{1+4x^2} = \sqrt{(2x+1)(2x-1)+2} $$

$u = 2x-1, du = \frac{1}{2} dx$

$$ \int \sqrt{u(u+2) + 2} du $$

$$ \int \sqrt{(u+1)^2 + 1} du $$ Nothing there

$$ \int\sqrt{1+4x^2}dx = x\sqrt{1+4x^2} - \int \frac{4x^2}{\sqrt{1+4x^2}} dx $$

Tried for a while to put the other integral into a workable form and got nowhere.

So what method is used to take this integral?

And once you get bored of trig substitutions, try Euler substitutions! — David H, Dec 07 '14 at 18:29
@SimonS That's the substitution I'd use too, but it does lead to the need to integrate $\int |\sec \theta| ,d\theta$, which can be done, but only with more work. — amWhy, Dec 07 '14 at 18:29
@DavidH That is actually one of the most beautiful substitutions i have ever seen — Eric L, Dec 07 '14 at 18:33
@SimonS Im guessing this would also have me change the differential? — Eric L, Dec 07 '14 at 18:35
@EricLawson Ain't it though? :) It's a crime that it's hardly ever taught in basic courses. — David H, Dec 07 '14 at 18:40
@SimonS do you end up with
$$ \frac{1}{2}\int \sec^3(\Theta) d\Theta $$ — Eric L, Dec 07 '14 at 18:49
@DavidH: Never heard of Euler substitution, they're really interesting! However they're a bit complicated, do you remember them without having to look them up? — rubik, Dec 07 '14 at 19:18
@rubik No, I almost always have to go back to the Wikipedia page and double check the formulas before I use them. — David H, Dec 13 '14 at 09:32

rubik · Accepted Answer · 2014-12-13T09:33:40.360

4

Substitute $x = \frac 12 \sinh \phi$ to get $$\int \sqrt{1+4x^2}\mathrm dx = \int \cosh^2 \phi\ \mathrm d\phi = \int \left(\frac 12 \cosh 2\phi + \frac 12\right)\mathrm d\phi = \frac 18\sinh 2\phi + \frac 12\phi + C = \boxed{\displaystyle\frac x2\sqrt{1+4x^2} + \operatorname{arcsinh}2x + C}$$

Note the similarity between this integral and $$\int \sqrt{1-4x^2} = \frac x2\sqrt{1-4x^2} + \arcsin 2x + C$$

If you didn't remember the identity $\cosh^2 x = \frac 12\cosh 2x + \frac 12$ you could have used its definition: $$\cosh^2 x = \frac {\left(e^x + e^{-x}\right)^2} 4$$

It's an easy integral then, but the final substitution is indeed ugly and does not work out as nicely as above.

edited Dec 13 '14 at 09:33

answered Dec 07 '14 at 19:13

rubik

9,344

I should probably study hyperbolic functions more in-depth, was never really taught them before. – Eric L Dec 07 '14 at 19:45
@EricLawson: that was the case for me as well. First thing to do is to understand their definition (how they're related to the hyperbola). Then the identities are very similar to the trigonometric ones (often some minuses are replaced with plusses). – rubik Dec 07 '14 at 19:47
Yea, it is interesting though,
$$ \sinh(x) = i^3 sin(ix) $$

$$ \cosh(x) = cos(ix) $$

Kinda a weird mix
– Eric L Dec 07 '14 at 20:11

Computing $\int \sqrt{1+4x^2} \, dx$

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