Problem using trigonometric substitution $x = a \sec θ$ and domain of $\theta$

Question

I'm studying calculus from Rogawski's Calculus. In trigonometric substitution $x=a\sec \theta$, he made a note:

In the substitution $x = a \sec θ$ , we choose $0\le θ \le \frac π 2$ if $x \ge a$ and $π \le θ < \frac{3π}2$ if $x \le −a$. With these choices, $a \tan \theta$ is the positive square root $\sqrt{x^2 − a^2}$.

When I work on the integral:

$$\int \frac {\mathrm{d}x}{x\sqrt{x^2-9}}$$ Using the substitution of $x=3\sec\theta$ with the domain of $\theta$ shown above, the integration will be : $$\int \frac {dx}{x\sqrt{x^2-9}}= \int \frac {3\sec\theta\tan\theta d\theta}{(3\sec\theta)\sqrt{9\sec^2-9}}=\int \frac {\tan\theta d\theta}{3\sqrt{\tan^2 \theta}}\\= \int \frac{d\theta}{3}= \frac\theta 3+ \mathrm{C}= \frac 13 \sec^{-1}\left(\frac x3\right)+ \mathrm{C}$$

which is very wrong in the negative part of the domain of $x$, as shown in the graph below. The slope of the blue function in the negative domain should be positive not negative!.

There are 2 thoughts with this substitution:

$1)$ The problem with the domain of $\theta \in (\pi,\frac{3\pi}2) $ is that inverse-substitution cannot be done, because $\theta =\sec^{-1}x\notin (\pi,\frac{3\pi}2) $.

$2)$ Given the problem in (1), we should choose $\theta \in (0,\pi)-\{\frac\pi 2\}$ which makes $\sqrt{\tan^2 \theta}=|\tan \theta|$. The integral is then re-written as a piecewise function:

$$\int \frac {dx}{x\sqrt{x^2-9}}= \int \frac {3\sec\theta\tan\theta d\theta}{(3\sec\theta)\sqrt{9\sec^2-9}}$$

$$\int \frac {\tan\theta d\theta}{3\sqrt{\tan^2 \theta}} = \begin{cases} = \int \frac {d\theta}{3} = \frac 13 \sec^{-1}(\frac x3)+ C & \text{if $\theta \in (0,\frac \pi 2)$} \equiv x>3 \\= \int \frac {-d\theta}{3} = \frac {-1}3 \sec^{-1}(\frac x3)+ C & \text{if $\theta \in (\frac \pi 2,\pi)$}\equiv x<-3 \end{cases}$$

My questions:

Is above thinking right? Is there a mistake in choosing the domain of $\theta$, as mentioned in the boom?

Mathematica gives the answer of $-\dfrac {1}{3} \tan^{-1} \frac{3}{\sqrt{x^2-9}}+\mathrm{C}$, which is right when graphed. But, how do I derive this result?

Thanks for help.

Answer 1

In the substitution $x = a \sec\theta$ , we choose $0 \leq \theta < \frac\pi2$ if $x \geq a$ and $\pi \leq \theta < \frac{3\pi}2$ if $x \leq −a$. With these choices, $a \tan\theta$ is the positive square root $\sqrt{x^2 − a^2}$.

I think what Rogawski may have in mind here is that $x = a \sec\theta$ is not quite the same thing as $\theta = \sec^{-1}\left( \frac xa \right)$. Since $\sec (-\theta) = \sec\theta$, the possible solutions for $x = a \sec\theta$ are:

$$\theta = \sec^{-1}\left( \frac xa \right) + 2n\pi \quad \text{or} \quad \theta = -\sec^{-1}\left( \frac xa \right) + 2n\pi, \quad \text{where $n \in \mathbb Z$.}$$

If $x \geq a > 0$, so that Rogawski's advice is to let $0 \leq \theta < \frac\pi2$, then $\frac xa \geq 1$ and the "positive" solution with $n = 0$ produces a value of $\theta$ in the correct interval:

$$0 \leq \theta = \sec^{-1}\left( \frac xa \right) < \frac\pi2.$$

But keep in mind that the range of $\sec^{-1}$ is $\left[0, \frac\pi2\right) \cup \left(\frac\pi2, \pi\right]$. If $x \leq -a < 0$, so that Rogawski's advice gives $\pi \leq \theta < \frac{3\pi}2$, then $\frac xa \leq -1$ and the "positive" solution with $n=0$ can give only values of $\theta$ in the interval $\left(\frac\pi2, \pi\right]$, not $\left[\pi, \frac{3\pi}2\right)$. And any other value of $n$ produces a value of theta even farther from the desired interval. The only way to produce a value in the desired interval, $\left[\pi, \frac{3\pi}2\right)$, is to take the "negative" solution with $n = 1$:

$$\pi \leq \theta = -\sec^{-1}\left( \frac xa \right) + 2\pi < \frac{3\pi}2,$$

which you can verify after observing that in this case (with $\frac xa < -1$), $-\pi \leq -\sec^{-1}\left( \frac xa \right) < -\frac\pi2$.

So the correct subsitution (and reverse substitution) is

$$\int \frac {dx}{x\sqrt{x^2-9}} = \int \frac {d\theta}{3} = \frac \theta 3 + C = \begin{cases} \dfrac13 \sec^{-1}\left( \dfrac x3 \right) + C & \text{if }\ x \geq 3, \\ -\dfrac13 \sec^{-1}\left( \dfrac x3 \right) + C & \text{if }\ x < -3. \end{cases}$$ (Note that the constant $2\pi$ in the "negative" case is incorporated in the constant of integration, $C$.)

This is the same result you got by setting $\theta \in \left[0, \frac\pi2\right) \cup \left(\frac\pi2, \pi\right]$ and using the fact that $\sqrt{\tan^2 \theta} = |\tan\theta\,|$. Personally, I see nothing wrong with your method; in fact it may be easier to keep track of the need to do a "sign change" for the integral when $x \leq -3$ when you have an explicit invocation of the absolute value function, $|\tan\theta\,|$, rather than rules that implicitly change the sign of $\sec^{-1}\theta$ (that is, the requirement that $\pi \leq \theta < \frac{3\pi}2$, whose application seems a bit obscure to me).

In short, both methods give the same result, which is the correct result, but only if applied exactly as required at every step. Your method seems less prone to error than the method Rogawski recommends, so everything considered, I think I prefer yours.

(Of course if you are currently taking a class in calculus and have to write solutions of integrals like this on homework or exams, it's a good policy to write your solution in a way that will be easy for the grader to grade correctly. So if you use a technique not shown in class or in the textbook, make sure you explain it clearly, like the way you explained that $\sqrt{\tan^2 \theta} = |\tan\theta\,|$. It just makes life easier for everyone in the end.)

Answer 2

To get to the answer Mathematica gives, you can let$$ u = \sqrt{x^2 -9}.$$ You should get $$ \frac{1}{3} \arctan \left( \frac{\sqrt{x^2-9}}{3} \right) + \text{constant} $$ and use the fact that $$\arctan(\frac{1}{x}) = \frac{\pi}{2} - \arctan(x)~~~,x\gt 0$$ and $$\arctan(\frac{1}{x}) = -\frac{\pi}{2} - \arctan(x)~~~,x\lt 0 $$

Answer 3

I am not going to answer your problem on the domain part. Instead, I am suggesting an alternate way to integrating the very last integral.

Edited

Claim-1) $P = \int { \frac {dx}{\sqrt {1 - x^2}}} = ... = \sin ^{-1} x$.

Let $x = \sin u$. Then, $dx = du$.

∴ $P = \int {\frac {(\cos u)du}{\sqrt {1 - \sin ^2 u}}} = u = \sin ^{-1} x$

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[The whole Claim-2 can be ignored. I just don't want to delete it.]

Claim-2) $Q = \int {\frac {dx}{\sqrt {a^2 – x^2}}} = … = \sin ^{-1} \frac {x}{a}$.

Let $x = av$. Then, $dx = (a)dv$.

∴ $Q = \int {\frac {(a)dv}{\sqrt {a^2 – (av)^2}}}$

$ = \int {\frac {dv}{\sqrt {1 - v^2}}}$, which is essentially $P$.

$ = \sin ^{-1} v$

$ = \sin ^{-1} (\frac {x}{a})$

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Claim-3) $R = \int {\frac {dx}{x \sqrt {x^2 – a^2}}} = … = \frac {-1}{a} \sin ^{-1} \frac {a}{x}$.

Further edited

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Let $w = \frac {a}{x}$. Then, $dx = \frac {(-a) dw}{w^2}$

∴ $R = \int {\frac {-a (dw)/(w^2)}{(a/w) \sqrt {(a/w)^2 - a^2}}}$

$= - \int {\frac {dw}{w \sqrt{(a/w)^2 -a^2}}}$

$= - \int {\frac {dw}{aw \sqrt {(1/w)^2 - 1^2}}}$

$= \frac {-1}{a} \int {\frac {dw}{\sqrt {1^2 - w^2}}}$

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$= \frac {-1}{a} P$

$= \frac {-1}{a} \sin ^{-1} w$

$= \frac {-1}{a} \sin ^{-1} \frac {a}{x}$

Answer 4

The domain choice $\theta \in (0,\frac\pi2)\cup(\pi,\frac{3\pi}2)$ is perfectly fine. The corresponding back-substitution in the last step of

$$I=\int \frac {dx}{x\sqrt{x^2-9}}=\cdots = \frac\theta 3= \frac 13 \sec^{-1}\frac x3$$

is only correct for $\theta \in(0,\frac\pi2)$, but wrong for $\theta\in(\pi,\frac{3\pi}2)$. Instead, it should be \begin{align} I_{\theta\in(\pi,\frac{3\pi}2)}=\frac13\theta &=\frac13[(\theta-\pi)+\pi] = \frac13[\sec^{-1}\sec(\theta-\pi)+\pi]\\ &= \frac13[\sec^{-1}(-\sec\theta)+\pi]= \frac13[\sec^{-1}(-\frac x3)+\pi]\\ &= \frac13\sec\frac{|x|}3+\frac\pi3 \end{align} Thus, the combined result for both $x>1$ and $x<-1$ is $$\int \frac {dx}{x\sqrt{x^2-9}} =\frac13\sec\frac{|x|}3+ C$$

Alternatively, the integral can be evaluated as follows $$\int \frac {dx}{x\sqrt{x^2-9}} = \int \frac {d(\sqrt{x^2-9})} {(x^2-9)+9} =-\frac13 \tan^{-1} \frac{3}{\sqrt{x^2-9}}+\mathrm{C} $$ which is given by WA.