I'm stuck integrating $\int \sqrt{x^2-a^2} dx$ using trigonometric substitution

Question

When I'm trying to integrate $\int \sqrt{x^2-a^2} dx$ using trigonometric substitution, I get stuck. Here's the complete solution so far:

$$ x(\theta)=a\sec{\theta}\\ x'(\theta)=a\tan{\theta}\sec{\theta}\\ \theta=\sec^{-1}\left(\frac{x}{a}\right)\implies \theta\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]\\ $$

$$ \begin{align} \int \sqrt{x^2-a^2} dx &=\int \sqrt{[x(\theta)]^2-a^2}x'(\theta)d\theta\\ &=\int \sqrt{a^2\sec^2{\theta}-a^2}a\tan{\theta}\sec{\theta}d\theta\\ &=a^2\int \sqrt{\tan^2{\theta}}\tan{\theta}\sec{\theta}d\theta\\ &=a^2\int |\tan{\theta}|\tan{\theta}\sec{\theta}d\theta \end{align} $$

I take it that at this point I end up with two integrals one for when $\tan{\theta}>0$ (on $\left[0,\frac{\pi}{2}\right)$) and another one for when $\tan{\theta}<0$ (on $\left(\frac{\pi}{2},\pi\right]$):

$$ \theta\in\left[0,\frac{\pi}{2}\right): \int \sqrt{x^2-a^2} dx = a^2\int \tan^2{\theta}\sec{\theta}d\theta\\ \theta\in\left(\frac{\pi}{2},\pi\right]: \int \sqrt{x^2-a^2} dx = a^2\int (-\tan{\theta}\tan{\theta}\sec{\theta})d\theta = -a^2\int \tan^2{\theta}\sec{\theta}d\theta $$

But that doesn't seem to be right because the integral of a function has one unique answer, as far as I know. What am I doing wrong?

Answer 1

Your substitution needs to be defined as follows:

$$ x=g(\theta)=a\sec{\theta}, \ \ \ \ \theta\in\left[0,\frac{\pi}{2}\right)\cup\left[\pi,\frac{3\pi}{2}\right) \\ \implies \ g'(\theta)=a\tan{\theta}\sec{\theta}$$

Because:

$$ g'(\theta)=\left(a\sec{\theta}\right)'=\left(\frac{a}{\cos{\theta}}\right)'=a\frac{\sin{\theta}}{\cos^{2}{\theta}}=a\frac{\sin{\theta}}{\cos{\theta}}\frac{1}{\cos{\theta}}=a\tan{\theta}\sec{\theta} $$

Defines an inverse function (one to one) that returns positive tangents for the specified interval, since the sine and the cosine have the same sign in said intervals.

Answer 2

You can do it with hyperbolic trig functions, but if you want to use normal ones, it's a bit more fun. Use the substitution $x = a \sec \theta$.

$$ \text{Let } x = a \sec \theta \Longrightarrow \text{d}x = a \sec \theta \tan \theta \text{d} \theta $$ $$ \Longrightarrow I = \int \sqrt{a^2 \sec^2 \theta - a^2} a \sec \theta \tan \theta \text{d} \theta $$ $$ = a^2 \int \tan^2 \theta \sec \theta \text{d} \theta $$

Integration by parts with $u$ = $\tan \theta$ and $\text{d} v$ = $\sec \theta \tan \theta \text{d} \theta$ yields

$$ \frac{I}{a^2} = \sec \theta \tan \theta - \int \sec^3 \theta \text{d} \theta $$ $$ \Longrightarrow \frac{I}{a^2} = \sec \theta \tan \theta - \int (1 + \tan^2 \theta) \sec \theta \text{d} \theta = \sec \theta \tan \theta - \int \sec \theta \text{d} \theta - \int \tan^2 \theta \sec \theta \text{d} \theta $$

But we know $\int \tan^2 \theta \sec \theta \text{d} \theta$ from earlier, so $$ \frac{I}{a^2} = \sec \theta \tan \theta - \int \sec \theta \text{d} \theta - \frac{I}{a^2} $$ $$ \Longrightarrow \frac{2I}{a^2} = \sec \theta \tan \theta - \ln(\sec \theta + \tan \theta) + C_0, C_0 \in \mathbb{R} $$ $$ \Longrightarrow I = \frac{1}{2} a^2 \sec \theta \tan \theta - \frac{1}{2} a^2 \ln(\sec \theta + \tan \theta) + \frac{1}{2} a^2 C_0 $$ A little more work, noting that $x = a \sec \theta$ and $\sqrt{x^2-a^2} = a \tan \theta$, yields $$ I = \frac{1}{2} x \sqrt{x^2-a^2} - \frac{1}{2} a^2 \ln \left( x + \sqrt{x^2-a^2}\right) + C, C \in \mathbb{R}. $$

Answer 3

Just for the sign problem...

Thanks to Marcus, for $\theta\in [0,\frac\pi 2):$ $$ I = \frac{a^2}{2} \sec \theta \tan \theta - \frac{a^2}2 \ln|\sec \theta + \tan \theta| + C $$ On this interval, since $\sec \theta=\frac xa$ and $\tan\theta=\frac{\sqrt{x^2-a^2}}a$ the solution is $$ I = \frac{1}{2} x \sqrt{x^2-a^2} - \frac{a^2}{2} \ln |x + \sqrt{x^2-a^2}| + C.\tag1 $$ For $\theta\in (\frac\pi 2,\pi]$, OP says negative signs comes giving $$ I = -\frac{a^2}{2} \sec \theta \tan \theta + \frac{a^2}2 \ln|\sec \theta + \tan \theta| + C $$ On this interval, it is different. We have $\sec \theta=\frac xa$ but $\tan\theta=-\frac{\sqrt{x^2-a^2}}a$. Then, keeping the same $+C$, the solution is $$ I = -\frac{1}{2} x(-\sqrt{x^2-a^2}) + \frac{a^2}{2} \ln|x-\sqrt{x^2-a^2}| + C$$ $$I=\frac{1}{2} x \sqrt{x^2-a^2} + \frac{a^2}{2} \ln \left| \frac{x^2-(x^2-a^2)}{x+\sqrt{x^2-a^2}}\right| +C$$ $$I=\frac{1}{2} x \sqrt{x^2-a^2}-\frac{a^2}{2} \ln|x+\sqrt{x^2-a^2}|+C\tag2 $$ $(1)$ and $(2)$ are same expressions in $x$.

Answer 4

$$x=a \cosh{y}\Rightarrow dx=a \sinh{y}\\ \int\sqrt{x^{2}-a^{2}}dx=\int (a \sinh{y})(a \sinh{y})dy\\ =a^{2}\int\frac{\cosh{2y}-1}{2}=\frac{a^{2}}{2}(\frac{\sinh{2y}}{2}-y)+c\\ =\frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\sinh^{-1}{(\frac{x}{a})}+c$$