# How do I integrate $\sqrt{x^2+81}$ using trig substitution?

How do I integrate $\sqrt{x^2+81}$ using trig substitution? Please be as specific as possible, thank you!

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–  Amzoti Sep 28 '12 at 2:09
Hyperbolic trig functions will work better in this case, probably. –  Javier Sep 28 '12 at 2:11
$x = 9 \sinh t$ –  Will Jagy Sep 28 '12 at 2:19
Alpha will show a step-by-step solution –  Ross Millikan Sep 28 '12 at 2:32
You can get by an easy substitution to the integral $\int_{0}^{1} \sqrt{1+x^2} \, dx$, which was solved (by several methods) in answers to this question: Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ dx}$. –  Martin Sleziak Sep 28 '12 at 5:41

The idea here is that you want to simplify the radical using an identity like $1-\cos^2t=\sin^2t$ or $1+\tan^2t=\sec^2t$. In your particular case, if you let $$x=9\tan t,$$ then your integrand is (assuming we restrict to a region where secant is positive) $$\sqrt{x^2+81}\,dx=9\sec t\,dx.$$ You have to also compute $dx=9\sec^2 t\,dt$ from the change of variable. Then $$\int dx\,\sqrt{x^2+81}=81\int dt\,\sec^3 t=\frac{81}{2}\left(\sec t\tan t+\ln\left|\sec t+\tan t\right|\right)+{\rm const}.$$ To put that back in terms of $x$, refer again to the change of variable. $\tan t=x/9$ and $\sec t=\sqrt{1+\tan^2 t}=\sqrt{1+x^2/81}$.

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Jonathan's answer is good, but he left a lot of the work to you when evaluating

$$81 \int \sec^3 t \ dt$$

So, I'll expand on that since it is frequently proved in elementary calculus courses and is a key part of this problem.

Preword: when you see integrals of the form $$\sqrt{a^2+x^2}$$

the substitution $x = a \tan \theta$ usually cleans up the integral quite nicely and makes it easier to work with. Note that the new integral may require another common integration method (such as integration by parts in this example.)

Similarly, if you see integrals of the form $$\sqrt{a^2-x^2}$$

try the sub. $x = a \sin \theta$. For integrals of the form $$\sqrt{x^2-a^2}$$ try the sub. $x = a \sec \theta$.

Our original integral is $$\sqrt{x^2+81}$$

Since it fits our $\sqrt{x^2+a^2}$ format above, I'll sub. $x = 9 \tan \theta$. Then, $dx = 9 \sec^2 \theta$. We need to find what $\sqrt{x^2+81}$ is now. $$x^2+81 = 81 \tan^2 \theta + 81 = 81(\tan^2 \theta +1) = 81\sec^2 \theta \implies \sqrt{x^2+81} = 9 \sec \theta$$

Now we can rewrite our integral as $$\int 9\sec \theta \cdot 9\sec^2 \theta \ d\theta$$ which is the same as $$81\int\sec^3 \theta \ d\theta$$

For ease later on, I will now define our integral

$$I := \int\sec^3 \theta \ d\theta \tag{1}$$

Note that I DID drop the $81$ here on purpose so I won't have to carry it down for further calculations. I can simply multiply through by $81$ at the end.

Here comes the fun integral! I'll use integration by parts for this guy and it will turn out to not be so bad after all. Recall the integration by parts formula:

$$\int u \ dv = uv - \int v \ du \tag{2}$$

We want to pick a $dv$ so that, upon integrating, we get a simple $v$. Thus, it's wise to choose a $dv$ that is an exact differential (i.e. $dv = \sec^2 \theta \implies v = \tan \theta$ is a good choice).

I will let $u = \sec \theta, dv = \sec^2 \theta \implies du = \sec\theta\tan\theta \ d\theta, v = \tan \theta$. Plugging this into the integration by parts formula, we have

$$I = \sec\theta\tan\theta - \int \tan\theta \cdot \sec\theta\tan\theta \ d\theta \tag{3}$$ $$I = \sec\theta\tan\theta - \int \sec\theta \ \tan^2\theta \ d\theta \tag{4}$$

Since $1 + \tan^2 x = \sec^2 x, \tan^2 x = \sec^2 x - 1$ which allows us to rewrite (4):

$$I = \sec\theta\tan\theta - \int \sec\theta \cdot (\sec^2 \theta - 1) \ d\theta \tag{5}$$

Expand $\sec\theta \cdot (\sec^2 \theta - 1)$ to get $\sec^3 \theta - \sec \theta$.

$$I = \sec\theta\tan\theta - \left( \int \sec^3 \theta \ d\theta - \int \sec \theta \ d\theta \right) \tag{6}$$

Distribute the negative sign through the integrals to get:

$$I = \sec\theta\tan\theta - \int \sec^3 \theta \ d\theta + \int \sec \theta \ d\theta \tag{7}$$

From here, recall that $I = \int \sec^3 \theta \ d\theta$. So, we have:

$$I = \sec\theta\tan\theta - I + \int \sec \theta \ d\theta \tag{8}$$

$$2I = \sec\theta\tan\theta + \int \sec \theta \ d\theta \tag{9}$$

Now, I will go ahead and evaluate $\int \sec \theta \ d\theta$ since in the end, we would like an expression for $I$ that does not involve any other integrals.

$$\int \sec \theta \ d\theta = \ln |\sec\theta + \tan \theta| \tag{10}$$

$$2I = \sec\theta\tan\theta + \ln |\sec\theta + \tan \theta| \tag{11}$$

$$\implies I = \frac{1}{2} \sec\theta\tan\theta + \frac{1}{2} \ln |\sec\theta + \tan \theta| \tag{12}$$

Since I defined $I$ as the $\int \sec^3 \theta \ d \theta$, and we really wanted $81\int\sec^3 \theta \ d\theta$, we need to multiply line (12) through by $81$.

$$81\int\sec^3 \theta \ d\theta = \frac{81}{2} \sec\theta\tan\theta + \frac{81}{2} \ln |\sec\theta + \tan \theta| \tag{13}$$

Are we done yet? Almost! We need to get $\theta$ in terms of $x$ now. Since we let $x = 9 \tan \theta$, then $\frac{x}{9} = \tan \theta$. Draw a right triangle with the sides opposite and adjacent to the angle $\theta$ being $x$ and $9$ respectively. This means that the hypotenuse is equal to $\sqrt{x^2+81}$ by the Pythagorean theorem. Since we want $\sec \theta$, we need to figure out $\cos \theta$ because $\sec \theta = \frac{1}{\cos \theta}$.

$$\cos \theta = \frac{9}{\sqrt{x^2+81}} \implies \sec \theta = \frac{\sqrt{x^2+81}}{9} \tag{14}$$

Putting this all together with line (13), we have that

$$81\int\sec^3 \theta \ d\theta = \frac{81}{2} \frac{\sqrt{x^2+81}}{9} \cdot \frac{x}{9} + \frac{81}{2} \ln |\frac{\sqrt{x^2+81}}{9} + \frac{x}{9}| \tag{15}$$

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