Derivative of $\sec^{-1}x$ and integral of $\frac{1}{x\sqrt{x^2-1}}$

Question

My attempt is as follows:-

1. Derivative of $\sec^{-1}x$

   Let $\theta=\sec^{-1}x,$ where $\theta\in [0,\pi]-{\dfrac{\pi}{2}}.$

   $$\sec\theta=x.$$

   Differentiating both sides with respect to $x:$

   $$\sec\theta\cdot\tan\theta\cdot\dfrac{\mathrm d\theta}{\mathrm dx}=1\\ \dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{1}{x\sqrt{x^2-1}}.$$

   As $\sec^{-1}x$ is a strictly increasing function, its derivative should be positive, hence we write $x$ as $|x|$ to ensure that $\dfrac{\mathrm d\theta}{\mathrm dx}$ will not be negative if $x$ is negative. But I wonder why I didn't get $\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{1}{|x|\sqrt{x^2-1}}$ in the above calculation?

2. Integral of $\dfrac{1}{x\sqrt{x^2-1}}$

   Case $1:x>0$

   Then the integral is definitely $\sec^{-1}x.$

   Case $2: x<0$

   Then the integral is $-\sec^{-1}x.$

   But many textbooks write that $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C.$

   Shouldn't $\displaystyle\int\frac{1}{|x|\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C\:?$ What am I missing here?

Answer 1

The reason you didn't get the absolute value when you differentiated is that $$\newcommand{\sgn}{\text{sgn}} \tan(\theta)=\tan(\sec^{-1}x)=\sgn(x)\sqrt{x^2-1} $$ so the derivative is $$\frac{1}{\sgn(x)x\sqrt{x^2-1}}=\frac{1}{|x|\sqrt{x^2-1}} $$

Sometimes the principal range of $\sec^{-1}x$ is assumed to be $[0,\frac \pi2)\cup [\pi, \frac{3\pi}{2})$. This convention is popular when doing integration with $\sec^{-1}x$ substitution and avoids the issue with the absolute value. Under that convention, $$\int \frac{1}{x\sqrt{x^2-1}}=\sec^{-1}(x)+C $$

If you don't like redefining the range of $\sec^{-1}(x)$, then $$\int \frac{1}{x\sqrt{x^2-1}}=\sec^{-1}(|x|)+C $$ as @YvesDaoust wrote.

Answer 2

But many textbooks write that $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C.$

This is indeed wrong, since differentiating the equation at $x=-5$ gives $\displaystyle\frac{\sqrt6}{60}=-\frac{\sqrt6}{60}.$

Case $1:x>0$

the integral is definitely $\sec^{-1}x.$

Case $2: x<0$

the integral is $-\sec^{-1}x.$

Indeed, $\displaystyle\frac1{x\sqrt{x^2-1}}$ always has an antiderivative $$\frac x{|x|}\sec^{-1}x.\tag1$$ bjorn93 has also pointed out that it always has an antiderivative $$\sec^{-1}|x|.\tag{2}$$

Using the substitution $\displaystyle u=\frac1x$ and noting that $\sec^{-1}x+\operatorname{cosec}^{-1}x\equiv\frac{\pi}2,$ we can also obtain \begin{align}\int\frac{\mathrm dx}{x\sqrt{x^2-1}} &= \int\frac{-|u|}{u\sqrt{1-u^2}}\,\mathrm du\\ &= \begin{cases} \sin^{-1} u+C_1, &-1<u<0;\\ -\sin^{-1} u+C_2, &0<u<1\end{cases}\\ &= \begin{cases} \operatorname{cosec}^{-1}x+C_1, &x<-1;\\ -\operatorname{cosec}^{-1}x+C_2, &x>1\end{cases}\\ &= \begin{cases} \operatorname{cosec}^{-1}x+C_1, &x<-1;\\ \operatorname{sec}^{-1}x-\frac\pi2+C_2, &x>1\end{cases}\\ &= \begin{cases} \operatorname{cosec}^{-1}x+C, &x<-1;\\ \operatorname{sec}^{-1}x+D, &x>1.\tag3\end{cases}\end{align}

$(1),(2),(3)$ are consistent with one another because for $x<-1$ the expressions differ from one another by $\frac{\pi}2$ while for $x>1$ the expressions are identically equal. (Remember, each integration domain overlaps with just one of these intervals. Also, it is worth noting that when $C_1\ne C_2,$ differentiating the above result still returns the given integrand.)

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P.S. It can be similarly derived that on $(-1,0)\cup(0,1)$ $$-\operatorname{sech}^{-1}|x| \text{ is an antiderivative of } \frac1{x\sqrt{1-x^2}},$$ and on $\mathbb R{\setminus}\{0\}$ $$-\operatorname{cosech}^{-1}|x| \text{ is an antiderivative of } \frac1{x\sqrt{1+x^2}}.$$

Answer 3

To avoid $\pm$ signs, proceed as follows:

1. Derivative of $\sec^{-1}x$. Let $\theta=\sec^{-1}x,$ with $\theta\in [0,\pi]-{\dfrac{\pi}{2}}$. Then, $\sin \theta >0$

$$\cos\theta=\frac1{\sec\theta}=\frac1x\implies \sin\theta\cdot\dfrac{\mathrm d\theta}{\mathrm dx}=\frac1{x^2}\\ \dfrac{\mathrm d\theta}{\mathrm dx}=\frac{1}{x^2\sin\theta} =\frac{1}{x^2\sqrt{1-\frac1{x^2}}}= \frac{1}{\sqrt{x^2(x^2-1)}} $$

2. Integral of $\dfrac{1}{x\sqrt{x^2-1}}$. For integration valid for the whole domain $|x|>1$

$$\int \frac{1}{x\sqrt{x^2-1}}dx = \int \frac{d(\sqrt{x^2-1})}{(x^2-1)+1}=\tan^{-1}\sqrt{x^2-1}+C $$

Answer 4

$$\left(\sec^{-1}(x)\right)'=\left(\cos^{-1}\left(\frac1x\right)\right)'=\frac1{x^2\sqrt{1-\dfrac1{x^2}}}.$$

This is obviously a positive function, and can indeed not be expressed as

$$\frac1{x\sqrt{x^2-1}}.$$

So the antiderivative of the latter is indeed $\text{sgn}(x)\sec^{-1}(x)$, or $\sec^{-1}(|x|)$. Note anyway that the integration interval may not overlap $(-1,1)$.