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In problem 5(iv) of Chapter 19 of Spivak's Calculus, "Integration in Elementary Terms", we are asked to integrate the following integral $$\int\frac{1}{\sqrt{1+e^x}}dx\tag{1}$$

It seems I have made a mistake in my attempt at solving this integral. I can't figure out where it is. I know because it differs from the solution manual, and when I differentiate my result in Maple I don't obtain the expected result.

I use two consecutive substitutions (now, obviously, the solution manual did something simpler, but for didactical purposes I'd really like to figure out what step is wrong below).

First let's use the substitution $$u=\sqrt{1+e^x}$$

$$x=\log{(u^2-1)}$$

$$dx=\frac{2u}{u^2-1}du$$

Thus $(1)$ becomes

$$\int\frac{1}{u}\frac{2u}{u^2-1}du=2\int\frac{1}{u^2-1}du$$

At this point, the solution manual uses partial fractions decomposition. Instead, I use another substitition.

$$u=\sec{\theta}$$

$$du=\tan{\theta}\sec{\theta}d\theta$$

$$\theta=\text{arcsec}{(u)}$$

Subbing in

$$2\int\frac{1}{\tan^2{\theta}}\tan{\theta}\sec{\theta}d\theta$$

$$=2\int\frac{1}{\cos{\theta}}\cdot \frac{\cos{\theta}}{\sin{\theta}}d\theta$$

$$=2\int\csc{\theta}d\theta\tag{2}$$

Now,

$$\int\csc{x}dx=-\log{(\csc{x}+\cot{x})}$$

So, from $(2)$,

$$=-2\log{(\csc{\theta}+\cot{\theta})}\tag{3}$$

$$=-2\log{(\csc{(\text{arcsec}{(u)})}+\cot{(\text{arcsec}{(u)}}))}\tag{4}$$

$$=-2\log{(u+\frac{1}{\sqrt{u^2-1}})}\tag{5}$$

To get from $(4)$ to $(5)$, I used the following

$$\tan^2{(\text{arcsec}{(u)})}=\frac{\cos^2{(\text{arcsec}{(u)})}}{\sin^2{(\text{arcsec}{(u)})}}=\sec^2{(\text{arcsec}{(u)})}-1=u^2-1$$

But

$$\cos^2{(\text{arcsec}{(u)})}=1-\sin^2{(\text{arcsec}{(u)})}$$

so

$$\frac{\cos^2{(\text{arcsec}{(u)})}}{\sin^2{(\text{arcsec}{(u)})}}=\frac{1-\sin^2{(\text{arcsec}{(u)})}}{\sin^2{(\text{arcsec}{(u)})}}=\frac{1}{\sin^2{(\text{arcsec}{(u)})}}-1=u^2-1$$

$$\frac{1}{\sin^2{(\text{arcsec}{(u)})}}=u^2$$

$$\csc{(\text{arcsec}{(u)})}=\frac{1}{\sin{(\text{arcsec}{(u)})}}=u\tag{6}$$

Which is one of the terms we need in $(4)$. The other is $\cot{(\text{arcsec}{(u)}})$, which is

$$\cot{(\text{arcsec}{(u)}})=\frac{1}{\tan{(\text{arcsec}{(u)}})}=\frac{1}{\sqrt{u^2-1}}\tag{7}$$

Thus, from $(4)$, we have

$$\int\frac{1}{\sqrt{1+e^x}}dx\tag{1}$$

$$=-2\log{(\csc{(\text{arcsec}{(u)})}+\cot{(\text{arcsec}{(u)}}))}\tag{4}$$

$$=-2\log{\left (u+\frac{1}{\sqrt{u^2-1}}\right )}$$

$$=-2\log{\left (\sqrt{1+e^x}+\frac{1}{\sqrt{e^x}}\right )}$$

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  • $\begingroup$ Maybe not a particularly helpful comment for you now, but I would have used partial fraction decomposition at step (1) instead of substituting secant. $\endgroup$ Commented Aug 7, 2022 at 21:39
  • $\begingroup$ @blacknapkins7 You know what, I said Spivak did "algebraic manipulations", but what he actually used was partial fraction decomposition, I just did the calculations here. Still, yes, even though the secant substitution is way more difficult than necessary, I would like to test my understanding of all the steps by finding the incorrect one. $\endgroup$ Commented Aug 7, 2022 at 21:44
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    $\begingroup$ I generally find it’s a good idea to graph everything on Desmos. That way you can emphatically see exactly which step went wrong, and if you’re still stuck, at least the MSE post could be made a good deal shorter $\endgroup$ Commented Aug 7, 2022 at 21:50

2 Answers 2

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Here are the mistakes I found, and I'm going to disregard domain restrictions for now. First,

$$\tan^{2}\left(\operatorname{arcsec}\left(u\right)\right) \neq \frac{\cos^{2}\left(\operatorname{arcsec}\left(u\right)\right)}{\sin^{2}\left(\operatorname{arcsec}\left(u\right)\right)},$$

it is

$$ \eqalign{ \tan^{2}\left(\operatorname{arcsec}\left(u\right)\right) &= \frac{\sin^{2}\left(\operatorname{arcsec}\left(u\right)\right)}{\cos^{2}\left(\operatorname{arcsec}\left(u\right)\right)} \cr &= \frac{1-\cos^{2}\left(\operatorname{arcsec}\left(u\right)\right)}{\cos^{2}\left(\operatorname{arcsec}\left(u\right)\right)} \cr &= \frac{1}{\cos^{2}\left(\operatorname{arcsec}\left(u\right)\right)}-1 \cr &= u^2-1. } $$

Also, I'm not sure how you got $\csc\left(\operatorname{arcsec}\left(u\right)\right) = u$ since $\operatorname{csc}$ and $\operatorname{arcsec}$ aren't inverses of each other. In fact, I'm not even sure how these equations help you. I agree that $\cot\left(\operatorname{arcsec}\left(u\right)\right) = \frac{1}{\sqrt{u^{2}-1}}$, but that's solved by doing that right triangle method. And also by using the right triangle method, $\csc\left(\operatorname{arcsec}\left(u\right)\right) = \frac{u}{\sqrt{u^{2}-1}}$.

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The issue with your attempt is from step (4) to step (5).

$-2\log{(\csc{(\text{arcsec}{(u)})}+\cot{(\text{arcsec}{(u)}}))}\tag{4} $

$= -2 \log(\frac{1}{(u \sqrt{(1 - 1/u^2))}} + \frac{1}{\sqrt{(1 - 1/u^2))}}$

instead of $$-2\log{(u+\frac{1}{\sqrt{u^2-1}})}\tag{5}$$

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