How do you integrate an equation in the form of

$\displaystyle \int$$\displaystyle \sqrt{x^2+a^2}$$\displaystyle dx$

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- Jun 8th 2011, 01:06 PMiPodIntegration Question
How do you integrate an equation in the form of

$\displaystyle \int$$\displaystyle \sqrt{x^2+a^2}$$\displaystyle dx$ - Jun 8th 2011, 01:24 PMTKHunny
Various ways to go about it. A nice tangent substitution appears to be called for at first glance. What have you tried?

- Jun 8th 2011, 01:31 PMiPod
I was actually using cos and sin, but i didnt get anywhere. I dont see how I can use tan

- Jun 8th 2011, 01:35 PMTKHunny
$\displaystyle x = a\cdot\tan(\theta)$?

Of course, this might lead to [sec(x)]^3 and give you other problems, but it could result in something useful. - Jun 8th 2011, 01:37 PMCroat
There are three ways, known to me, anyway.

First you can substitute $\displaystyle x=a \tan t$.

The second way is $\displaystyle t=a \sinh t$.

And the third would be to rewrite this and use a special rule for integrals of this type:

$\displaystyle \int \frac{P_n(X)}{\sqrt{ax^2+bx+c}}\,dx=Q_{n-1}(x)\sqrt{ax^2+bx+c}+\lambda \int \frac{\,dx}{\sqrt{ax^2+bx+c}}$

Note that Q is a polynomial with unknown coefficients in X and is for one degree smaller than the degree of P. Lambda is also unknown so you actually have the same amount of unknowns as the degree of the polynomial. Although the last way might seem complicated I think it's quite easy so I'll demonstrate it and recommend to you that you try any other way (maybe the 2.?).

Anyway, rewrite:

$\displaystyle \int \sqrt{x^2+a^2}\,dx=\int \frac{x^2+a^2}{\sqrt{x^2+a^2}}\,dx$

Now you actually have an integral of the type I mentioned. Set it equal to the right side:

$\displaystyle \int \frac{x^2+a^2}{\sqrt{x^2+a^2}}\,dx=(Ax+B)\sqrt{x^2 +a^2}+\lambda \int \frac{\,dx}{\sqrt{x^2+a^2}}$

Differentiate both sides, and solve for A, B, lambda:

$\displaystyle \frac{x^2+a^2}{\sqrt{x^2+a^2}}=A\sqrt{x^2+a^2}+(Ax +B)\frac{x}{x^2+a^2}+\frac{\lambda }{x^2+a^2}$

$\displaystyle x^2+a^2=Ax^2+Aa^2+Ax^2+Bx+\lambda $

So now we have:

$\displaystyle A=\frac{1}{2}$

$\displaystyle B=0$

$\displaystyle \lambda =\frac{a^2}{2}$

So our integral is actually equal to:

$\displaystyle \int \sqrt{a^2+x^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac {a^2}{2}\int\frac{\,dx}{\sqrt{x^2+a^2}}$

$\displaystyle \int \sqrt{a^2+x^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac {a^2}{2} \ln(x+\sqrt{x^2+a^2})+C$

Well, that's my preferred way...

Recommended substitution is the hyperbolic sine. Good luck.

EDIT: Also noted you can do this with partial integration... I really love how many ways there are to integrals :D. - Jun 8th 2011, 02:14 PMTheCoffeeMachine
$\displaystyle \begin{aligned}\displaystyle I & = \int(x)'\sqrt{a^2+x^2}\;{dx} \\ & = x\sqrt{a^2+x^2}-\int{x}\left(\sqrt{a^2+x^2}\right)'\;{dx}\\ & = x\sqrt{a^2+x^2}-\int \frac{x^2}{\sqrt{a^2+x^2}}\;{dx}\\ & = x\sqrt{a^2+x^2}-\int \frac{(a^2+x^2)-a^2}{\sqrt{a^2+x^2}}\;{dx}\\ & = x\sqrt{a^2+x^2}-\int \frac{a^2+x^2}{\sqrt{a^2+x^2}}\;{dx}+\int \frac{a^2}{\sqrt{a^2+x^2}}\;{dx}\\ & = x\sqrt{a^2+x^2}-\int {\sqrt{a^2+x^2}\;{dx}+\int \frac{a^2}{\sqrt{a^2+x^2}}\;{dx}\\ & = x\sqrt{a^2+x^2}-I+a^2\sinh^{-1}\left(\frac{x}{a}\right) \\ & = \frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\sinh^{-1}\left(\frac{x}{a}\right)+k. \end{aligned}$

- Jun 9th 2011, 04:59 AMtom@ballooncalculus
Just in case a picture helps...

First, noticing that

$\displaystyle \sqrt{a^2 + x^2}$

looks vaguely Pythagorean, we factor out the a^2...

http://www.ballooncalculus.org/draw/internal/nine.png

... and, yes, we can see that substituting tan or else sinh for x/a would allow some kind of Pythag. Hang on, though - the reason the sub is useful is the chain rule...

http://www.ballooncalculus.org/draw/internal/nineb.png

... where (key in spoiler) ...

__Spoiler__:

So we're going to integrate up the straight dashed line, i.e. with respect to the dashed balloon. But to make sure that doing that is the same as integrating the original with respect to x, we have to take account of the derivative of the inner function...

http://www.ballooncalculus.org/draw/internal/ninea.png

... just as the chain rule demands. Now, though, we can indeed remove the inner function (apply lubricant and turn carefully anti-clockwise, and the whole thing should come out)...

http://www.ballooncalculus.org/draw/internal/nineb.png

... and replace with sinh...

http://www.ballooncalculus.org/draw/internal/ninec.png

... which we had a hunch would be useful. And now we can see why - the hyperbolic counterpart of Pythag gives us cosh squared, which we can integrate with respect to theta.

http://www.ballooncalculus.org/draw/internal/nined.png

But then we know F (as soon as we've re-expressed the function of theta as a function of sinh x) and F is a function of the dashed balloon, and doesn't care what's inside. So we can remove the inner function (i.e. sinh) again, and replace the original (x/a).

__Spoiler__:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jun 9th 2011, 05:24 AMProve It
I would use a hyperbolic substitution, because $\displaystyle \displaystyle \cosh^2{t} - \sinh^2{t} = 1 \implies \sinh^2{t} + 1 = \cosh^2{t}$, and hyperbolic integrals are often easier than trigonometric...

Anyway, after making the substitution $\displaystyle \displaystyle x = a\sinh{t} \implies dx = a\cosh{t}\,dt$ the integral $\displaystyle \displaystyle \int{\sqrt{x^2 + a^2}\,dx}$ becomes

$\displaystyle \displaystyle \begin{align*} \int{\sqrt{(a\sinh{t})^2 + a^2}\,a\cosh{t}\,dt} &= a\int{\sqrt{a^2\sinh^2{t} + a^2}\cosh{t}\,dt} \\ &= a\int{\sqrt{a^2(\sinh^2{t} + 1)}\,\cosh{t}\,dt} \\ &= a\int{a\sqrt{\cosh^2{t}}\,\cosh{t}\,dt} \\ &= a^2\int{\cosh{t}\,\cosh{t}\,dt} \\ &= a^2\int{\cosh^2{t}\,dt} \\ &= a^2\int{\frac{1}{2} + \frac{1}{2}\cosh{2t}\,dt} \\ &= a^2\left(\frac{1}{2}t + \frac{1}{4}\sinh{2t}\right) + C \\ &= a^2\left(\frac{1}{2}t + \frac{1}{2}\sinh{t}\sqrt{1 + \sinh^2{t}}\right) +C \\ &= a^2\left[\frac{1}{2}\sinh^{-1}{\left(\frac{x}{a}\right)} + \frac{1}{2}\left(\frac{x}{a}\right)\sqrt{1 + \left(\frac{x}{a}\right)^2}\right]+ C \\ &= a^2\left[\frac{1}{2}\sinh^{-1}{\left(\frac{x}{a}\right)} + \frac{x}{2a}\sqrt{\frac{a^2 + x^2}{a^2}}\right] + C \\ &= a^2\left[\frac{1}{2}\sinh^{-1}{\left(\frac{x}{a}\right)} + \frac{x\sqrt{a^2 + x^2}}{2a^2} \right]+ C \\ &= \frac{a^2}{2}\sinh^{-1}{\left(\frac{x}{a}\right)} + \frac{x\sqrt{a^2+ x^2}}{2} + C\end{align*}$ - Jun 9th 2011, 07:36 AMTheCoffeeMachine
Just for reference then, let's do this one too:

$\displaystyle \begin{aligned}\displaystyle I & = \int(x)'\sqrt{a^2-x^2}\;{dx} \\ & = x\sqrt{a^2-x^2}-\int{x}\left(\sqrt{a^2-x^2}\right)'\;{dx}\\ & = x\sqrt{a^2-x^2}+\int \frac{x^2}{\sqrt{a^2-x^2}}\;{dx}\\ & = x\sqrt{a^2-x^2}+\int \frac{a^2-(a^2-x^2)}{\sqrt{a^2-x^2}}\;{dx}\\ & = x\sqrt{a^2-x^2}+\int \frac{a^2}{\sqrt{a^2-x^2}}\;{dx}-\int \frac{a^2-x^2}{\sqrt{a^2-x^2}}\;{dx}\\ & = x\sqrt{a^2-x^2}+a^2\int \frac{1}{\sqrt{a^2-x^2}}\;{dx}-\int \sqrt{a^2-x^2}\;{dx}\\ & = x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\;{dx}-I\\ & \end{aligned}$

$\displaystyle \begin{aligned}\therefore \space & 2I = x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\;{dx}+k \\& \therefore I = \frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\right)+k. \end{aligned} $

Isn't it wonderful that I can use the [tex] [*/*MATH] tags! Magic! - Jun 9th 2011, 01:32 PMiPod
Sorry, i had posted a reply earlier on but my internet had crashed. I had successfully managed to integrate it, using x=sinh(t)

so i presume that integrating sqrt(x^2-a^2) requires the use of x=cosh(t) ?

and what about sqrt(a^2-x^2)?? - Jun 9th 2011, 07:31 PMTKHunny
No. As I said in the first place, there are various ways to go about it.

Why ask when you can explore? What do you think? Is this one easier or harder? Is hyperbolic trig warranted or maybe just $\displaystyle x = a\cdot\sin(\theta)$? - Jun 9th 2011, 07:39 PMiPod
ah true say.

for $\displaystyle \sqrt{a^2-x^2}$

i presume x=asin(t)

would be a nice one since we get nice and easy trig stuff.. - Jun 9th 2011, 07:41 PMiPod
ended up integrating a^2 cos^2(x), which is a^2*(1-2sin(t/2)).. worked out perfectly

- Jun 10th 2011, 03:42 AMTKHunny
Very good. Way to be up to it!