# Find the integral $\int\frac{dx}{\sqrt{x^2-a^2}}$

Evaluate the integral of $\frac{1}{\sqrt{x^2-a^2}}$

Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$ \begin{align} \int\frac{dx}{\sqrt{x^2-a^2}}&=\int\frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\sqrt{\tan^2\theta}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{\tan\theta}}\\&=\int\sec\theta d\theta=\log|\sec\theta+\tan\theta|+C=\log|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}|\\&=\log|\frac{x+\sqrt{x^2-a^2}}{a}|+C=\log|{x+\sqrt{x^2-a^2}}|-\log|a|+C\\&=\log|{x+\sqrt{x^2-a^2}}|+C \end{align} This is how it is solve in my reference. But, $\sqrt{\tan^2\theta}=|\tan\theta|$ right ? Then, does that imply $$\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{|\tan\theta|}}=\color{red}{\pm}\int\sec\theta d\theta$$ Why am I getting this confusion and is the first solution complete ?

• @TrostAft how can i say $\tan\theta$ is $+$ve from $x^2-a^2>0$ ? – ss1729 May 2 '18 at 7:20
• No first solution is not complete, they silently slipped in the || inside. You know, $\int \sec t = \ln(\sec t + \tan t)$ is valid for $\sec t > 0$ and for $\sec t \lt 0$ the integrand is $\ln(-\sec t - \tan t)$ so the solution $\ln |\sec t + \tan t|$ combines both. – jonsno May 2 '18 at 7:35
• Check by differentiating your solution(s). – Yves Daoust May 2 '18 at 7:35
• Agree with @samjoe. My comment is untrue. – TrostAft May 2 '18 at 7:35
• @samjoe i'm srry dont understand how ur point help me with the doubt in OP ?. Could u pls explain bit more ? – ss1729 May 2 '18 at 7:57

Suppose that $a>0$.

The work is just for the case when $x>a$. The case for $x<-a$ is different, but the finals result is the same.

Let $x=a\sec\theta$, where $\theta\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$. This is the domain of $\textrm{arcsec}$.

For $x< -a$, $\theta\in(\frac{\pi}{2},\pi]$ and so $\tan\theta\le0$.

$$\sqrt{x^2-a^2}=\sqrt{a^2\tan^2\theta}=-a\tan\theta$$

\begin{align*} \int\frac{dx}{\sqrt{x^2-a^2}}&=\int\frac{a\sec\theta\tan\theta}{-a\tan\theta}d \theta\\ &=-\int\sec\theta d\theta\\ &=-\ln|\sec\theta+\tan\theta|+C\\ &=\ln|\sec\theta-\tan\theta|+C\\ &=\ln\left|\frac{x}{a}-\frac{-\sqrt{x^2-a^2}}{a}\right|+C\\ &=\ln\left|x+\sqrt{x^2-a^2}\right|-\ln|a|+C \end{align*}

There are two minus signs and they cancel each other to reach the final result.

• I don't uunderstand how you conclude $\theta \in (\pi/2, \pi]$ because that is the key to the question. – jonsno May 2 '18 at 8:17
• I mean why it can't be $\theta \in (\pi, 3\pi/2)$ where sec is negative but tan is positive – jonsno May 2 '18 at 8:18
• Yes thats what we have to show that tan cannot be positive. – jonsno May 2 '18 at 8:20
• If we take $\theta\in(\pi,3\pi/2)$, then $\tan\theta>0$. The integral is $\int\sec\theta d\theta$. The final answer will be still $\ln|x+\sqrt{x^2-a^2}|-\ln|a|+C$. My point is that we can have $-\int\sec\theta d\theta$ in the work. But then we will have one more minus sign and obtain the same final answer. – CY Aries May 2 '18 at 8:26
• We either take a range so that the work holds for both $x>a$ and $x<-a$, or take a range so that in one case we will have two minus signs to cancel each other. – CY Aries May 2 '18 at 8:29