3
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ @TrostAft how can i say $\tan\theta$ is $+$ve from $x^2-a^2>0$ ? $\endgroup$ – ss1729 May 2 '18 at 7:20
  • 1
    $\begingroup$ 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. $\endgroup$ – samjoe May 2 '18 at 7:35
  • $\begingroup$ Check by differentiating your solution(s). $\endgroup$ – Yves Daoust May 2 '18 at 7:35
  • $\begingroup$ Agree with @samjoe. My comment is untrue. $\endgroup$ – TrostAft May 2 '18 at 7:35
  • $\begingroup$ @samjoe i'm srry dont understand how ur point help me with the doubt in OP ?. Could u pls explain bit more ? $\endgroup$ – ss1729 May 2 '18 at 7:57
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't uunderstand how you conclude $\theta \in (\pi/2, \pi]$ because that is the key to the question. $\endgroup$ – samjoe May 2 '18 at 8:17
  • $\begingroup$ I mean why it can't be $\theta \in (\pi, 3\pi/2)$ where sec is negative but tan is positive $\endgroup$ – samjoe May 2 '18 at 8:18
  • $\begingroup$ Yes thats what we have to show that tan cannot be positive. $\endgroup$ – samjoe May 2 '18 at 8:20
  • 1
    $\begingroup$ 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. $\endgroup$ – CY Aries May 2 '18 at 8:26
  • $\begingroup$ 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. $\endgroup$ – CY Aries May 2 '18 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.