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

Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$ $$ \begin{align} & \ \ \ \int \frac{dx}{\sqrt{x^2-a^2}} \\ &=\int \frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}} \\ &=\int \frac{a\sec\theta\tan\theta d\theta}{a\sqrt{\tan^2\theta}}\\ &= \int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{\tan\theta}}\\ &=\int\sec\theta d\theta \\ &=\log \lvert \sec\theta+\tan\theta \rvert|+C_0 \\ & \mbox{where $C_0$ is an arbitrary constant of integration } \\ &=\log \left\lvert \frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1} \right\rvert + C_0 \\ &= \log \left\lvert \frac{x+\sqrt{x^2-a^2}}{a} \right\rvert +C_0 \\ &= \log \left\lvert x+\sqrt{x^2-a^2} \right\rvert -\log|a|+C_0 \\ &= \log|x+\sqrt{x^2-a^2}|+ C, & \mbox{ where $C = \log \lvert a \rvert + C_0$}. \end{align} $$ 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 ?

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  • $\begingroup$ @TrostAft how can i say $\tan\theta$ is $+$ve from $x^2-a^2>0$ ? $\endgroup$
    – Sooraj S
    Commented May 2, 2018 at 7:20
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    $\begingroup$ 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. $\endgroup$
    – jonsno
    Commented May 2, 2018 at 7:35
  • $\begingroup$ Check by differentiating your solution(s). $\endgroup$
    – user65203
    Commented May 2, 2018 at 7:35
  • $\begingroup$ Agree with @samjoe. My comment is untrue. $\endgroup$
    – TrostAft
    Commented May 2, 2018 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 ? $\endgroup$
    – Sooraj S
    Commented May 2, 2018 at 7:57

3 Answers 3

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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.

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  • $\begingroup$ I don't uunderstand how you conclude $\theta \in (\pi/2, \pi]$ because that is the key to the question. $\endgroup$
    – jonsno
    Commented May 2, 2018 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 $\endgroup$
    – jonsno
    Commented May 2, 2018 at 8:18
  • $\begingroup$ Yes thats what we have to show that tan cannot be positive. $\endgroup$
    – jonsno
    Commented May 2, 2018 at 8:20
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    $\begingroup$ 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
    Commented May 2, 2018 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
    Commented May 2, 2018 at 8:29
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This is an elementary integral, using an inverse hyperbolic function.

By the formula for inverse functions, $$\cosh'(x)=\sinh(x)=\sqrt{\cosh^2(x)-1}$$ lets you establish

$$\text{arcosh}'(x)=\frac1{\sqrt{x^2-1}}.$$

It is known that

$$\text{arcosh}(x)=\log(x+\sqrt{x^2-1})$$ though the form $\text{arcosh}$ is fine.

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Given the known pitfalls of trigonometric substitutions for this integral, substitute $y= \sqrt{x^2-a^2}$ instead. Then, $y\ dy=x\ dx$ \begin{align} \int \frac{dx}{\sqrt{x^2-a^2}} =& \int \frac{dx}y= \int \frac{(1+\frac xy)dx}{x+y} =\int \frac{d(x+y)}{x+y}= \ln|x+y|+C \end{align} valid all domain $x$.

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