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# Week 2 Monday 6/26 brief notes.
## A few words on what is to come this week.
Last week we started with a discussion of the natural logarithm function and the natural exponential function. We saw how we can build new functions out of ones we know, in particular we defined natural logarithm as the **integral of some other function** using FTC, namely $\ln(x)=\int_1^t \frac{1}{t}dt$; and we defined the natural exponentiation function $\exp(x)=e^x$ as the **inverse** of natural logarithm. We also explored ideas on inverses, L'Hospital rule, hyperbolic functions.
What we will continue to do this week is to build a bit more inverse functions out of old ones (6.6, 6.7), such as the **inverse trigonometric functions** and **inverse hyperbolic functions** ones. An immediate issue that arises is that some of these functions are not one-to-one on their usual domain, so we will have to be careful what we mean by inverse (by restricting the domain). We will use use these inverses to help us find various antiderivatives.
Then we enter the first major topic of calculus 2: **Finding antiderivatives**. That is, given some function $f(x)$, can we find some $F(x)$ such that $F'(x)=f(x)$? Before we begin, you might say, hold on a minute, by FTC Existence form, one can easily just define $F(x)=\int_a^x f(t)dt$, and, provided that $f$ is continuous on the desired domain, we have an antiderivative $F(x)$ such that $F'(x)=f(x)$! This is indeed true. So the task now becomes, ok, but can this $F(x)$ be expressed without the integral symbol, and perhaps expressed only in terms of **elementary functions** (composition, sum, and product of polynomials, power functions, rational functions, logarithm function, exponential function, trigonometric functions, inverse trigonometric functions, hyperbolic functions, inverse hyperbolic functions)? As it turns out, as proven by Liouville in 1830s, functions like $e^{(x^2)}$ or $\frac{\sin(x)}{x}$ do not have antiderivatives expressible as elementary functions, despite having antiderivatives! In any case, we will work with cases that we can deal with.
## Inverse trigonometric functions (6.6)
The trigonometric functions $\sin(x),\cos(x)$ do not have inverses on their usual domain $\mathbb{R}$, since they are not one-to-one. But if we **restrict their domain** such that on the new domain they are one-to-one, then we can speak of an inverse on that domain. So inherently there is a **free choice** on how to do this!
By **convention**, however, we commonly choose to restrict $\sin(x)$ on $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, and write the inverse of $\sin(x)$ on this domain as $\sin^{-1}(x)$. I will also write $\arcsin(x)$ to denote this inverse function. **This will be our default choice (in this class and it is quite universally accepted).** This $\sin^{-1}(x)$ has domain $[-1,1]$ and range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, their graphs are as follows, observe $\arcsin(x)$ increases:
![[1 teaching/smc-summer-2023-math-8/notes/week-2/---files/week-2-monday-notes 2023-06-25 11.19.48.excalidraw.svg]]
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Similarly $\cos(x)$ also do not have inverse on their usual domain, so we would need to restrict it. The **conventional choice** is to restrict $\cos(x)$ on $[0,\pi]$, which we define its inverse to be $\cos^{-1}(x)$ with domain $[-1,1]$ and range $[0,\pi]$. I will also write $\arccos(x)$ for this inverse function, observe $\arccos(x)$ decreases:
![[1 teaching/smc-summer-2023-math-8/notes/week-2/---files/week-2-monday-notes 2023-06-25 11.37.39.excalidraw.svg]]
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For $\tan(x)$, again it is not invertible on its usual domain, so we adopt the conventional choice of restricting it on $(-\frac{\pi}{2}, \frac{\pi}{2})$, giving an inverse function $\tan^{-1}(x)$, or alternatively written as $\arctan(x)$, whose domain is $(-\infty,+\infty)$, and range $(-\frac{\pi}{2}, \frac{\pi}{2})$, observe $\arctan(x)$ increases:
![[1 teaching/smc-summer-2023-math-8/notes/week-2/---files/week-2-monday-notes 2023-06-25 17.41.19.excalidraw.svg]]
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Notice $\arctan(x)$ has two horizontal asymptotes: $\lim_{x\to+\infty}\arctan(x)= \frac{\pi}{2}$ and $\lim_{x\to-\infty} \arctan(x)=-\frac{\pi}{2}$.
One can similarly do the same for the other derived trigonometric functions: $\sec(x),\csc(x),\cot(x)$, and restrict their domains to give an inverse. However, there are no universally agreed upon definition for $\sec^{-1}(x)$ and $\csc^{-1}(x)$. I will cite your textbook, and give you an alternative. The slight difference will matter in the sense that their derivatives would be different.
Restrict $\csc(x)$ on $(0, \frac{\pi}{2}]\cup(\pi, \frac{3\pi}{2}]$, giving $\csc^{-1}(x)$ with domain $|x|\ge 1$.
- Note with this definition, $\csc^{-1}(x)$ increases then decreases.
Restrict $\sec(x)$ on $[0, \frac{\pi}{2})\cup[\pi, \frac{3\pi}{2})$, giving $\sec^{-1}(x)$ with domain $|x|\ge 1$.
- Note with this definition, $\sec^{-1}(x)$ decreases then increases.
Restrict $\cot(x)$ on $(0,\pi)$, giving $\cot^{-1}(x)$ with domain all reals.
- Note with this definition, $\cot^{-1}(x)$ decreases.
These choices are to make their derivatives to have "nice" expressions. If you are to encounter these inverses elsewhere, check how they are defined!. Here are their graphs (again, according to Stewart's textbook definition):
![[1 teaching/smc-summer-2023-math-8/notes/week-2/---files/week-2-monday-notes 2023-06-26 11.56.46.excalidraw.svg]]
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![[1 teaching/smc-summer-2023-math-8/notes/week-2/---files/week-2-monday-notes 2023-06-26 12.13.29.excalidraw.svg]]
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![[1 teaching/smc-summer-2023-math-8/notes/week-2/---files/week-2-monday-notes 2023-06-26 12.32.59.excalidraw.svg]]
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## The derivatives of these inverse trigonometric functions.
Let us now find their derivatives. Notice functionally, recall if we want the derivative of $f^{-1}$, one could compose it with the original function $f$, and apply chain rule, by noting $f(f^{-1})(x)=x$, so $$
f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}.
$$
For $\frac{d}{dx} \arcsin(x)$.
Put $\arcsin(x)$ into $\sin(x)$, we get $\sin(\arcsin(x))=x$. Differentiate both sides, we get $\cos(\arcsin(x))\cdot \frac{d}{dx}\arcsin(x)=1$. Now we need to figure out $\cos(\arcsin(x))$.
Algebraically: Using the identity $\cos^2(x)+\sin^2(x)=1$, so $\cos(x)=\pm\sqrt{1-\sin^2(x)}$, so $\cos(\arcsin(x))=\pm \sqrt{1-\sin^2(\arcsin(x))}=\pm \sqrt{1-x^2}$. Now if we note the range of $\arcsin(x)$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and we know $\cos(x)\ge 0 $on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we must take the positive sign.
Hence $\cos(\arcsin(x))=\sqrt{1-x^2}$, and hence $$
\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}.
$$
Another way to see this must be positive because $\arcsin(x)$ was chosen to be increasing.
For $\frac{d}{dx}\arccos(x)$:
Put $\arccos(x)$ into $\cos(x)$. Note we have $\cos(\arccos(x))=x$. Differentiate both sides and we get $$
-\sin(\arccos(x))\cdot \frac{d}{dx}\arccos(x)=1
$$
To figure out $\sin(\arccos(x))$, note algebraically $\cos^2(x)+\sin^2(x)=1$, so $\sin(x)=\pm\sqrt{1-\cos^2(x)}$, whence $\sin(\arccos(x))=\pm\sqrt{1-x^2}$. Whence $$
\frac{d}{dx}\arccos(x)=\frac{-1}{\sqrt{1-x^2}}
$$where the negative sign is chosen because $\arccos(x)$ is chosen to be decreasing.
For $\frac{d}{dx}\arctan(x)$:
Ok, let us try the same trick. Put $\arctan(x)$ into $\tan(x)$. Note $\tan(\arctan(x))=x$, and differentiating we get $\sec^2(\arctan(x)) \frac{d}{dx}\arctan(x)=1$. Now algebraically, $\sec^2(x)= \frac{1}{\cos^2(x)}= \frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=1+\tan^2(x)$. So $$
\sec^2(\arctan(x))=1+\tan^2(\arctan(x))=1+x^2.
$$So we have, very importantly, $$
\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}
$$
One can similarly do the same to find the rest, and in summary:: $$
\begin{align*}
\frac{d}{dx} \arcsin(x) &= \frac{1}{\sqrt{1-x^2}} & \text{for } x\in(-1,1)\\
\frac{d}{dx} \arccos(x) &= -\frac{1}{\sqrt{1-x^2}} & \text{for }x\in(-1,1)\\
\frac{d}{dx} \arctan(x) &= \frac{1}{1+x^2} & \text{for all }x \\
\frac{d}{dx}\csc^{-1}(x) & =\frac{-1}{x \sqrt{x^2-1}} & \text{for }|x| > 1\\
\frac{d}{dx}\sec^{-1}(x) & =\frac{1}{x \sqrt{x^2-1}} & \text{for }|x| > 1\\
\frac{d}{dx}\cot^{-1}(x) & =\frac{-1}{1+x^2} & \text{for all } x
\end{align*}
$$
**Integration involving these inverse trigonometric functions.**
A list of derivatives is really giving a list of antiderivatives (reading it the other way). For instance, $$
\int \frac{1}{1+x^2}dx = \arctan(x)+C.
$$
And with a combination of **recognition of form** and $u$-substitution, we can deal with slightly more exotic looking antiderivatives.
Example. Find $$
\int \frac{1}{7+2x^2}dx
$$
$\blacktriangleright$ Note this is almost like $\frac{1}{1+u^2}$ in the integrand, we just need to manipulate it a bit: $$
\begin{align*}
\int \frac{1}{7+2x^2}dx & = \frac{1}{7}\int \frac{1}{1+\frac{2}{7}x^2}dx, \text{\quad let } u = \frac{2}{7}x \\
& = \frac{1}{7} \int \frac{1}{1+u^2} \frac{7}{2}du \\
& = \frac{1}{2} \arctan(u)+C \\
& =\frac{1}{2} \arctan\left( \frac{2}{7}u \right) +C. \quad\blacklozenge
\end{align*}
$$
Example. Find $$
\int \frac{3}{\sqrt{6-5x^2}} dx
$$
$\blacktriangleright$ This looks like it could be related to $\arcsin(x)$, whose derivative is $\frac{1}{\sqrt{1-x^2}}$. We will manipulate it and do $u$-substitution to get this form: $$
\begin{align*}
\int \frac{3}{\sqrt{6-5x^2}}dx & = \frac{3}{\sqrt{6}} \int \frac{1}{\sqrt{1-\frac{5}{6}x^2}}dx , \text{\quad let } u = \sqrt{\frac{5}{6}}x \\
& =\frac{3}{\sqrt{6}} \sqrt{\frac{6}{5}} \int \frac{1}{\sqrt{1-u^2}}du \\
& =\frac{3}{\sqrt{5}} \arcsin(u) + C \\
& = \frac{3}{\sqrt{5}} \arcsin\left( \sqrt{\frac{5}{6}} x\right) + C. \quad\blacklozenge
\end{align*}
$$
Example. Find $$
\int \frac{5}{x\sqrt{7x^2-4}}dx
$$
$\blacktriangleright$ This looks like $\sec^{-1}(x)$ could help, as it has derivative $\frac{1}{x\sqrt{x^2-1}}$. So we do a similar factor/$u$-sub trick: $$
\begin{align*}
\int \frac{5}{x\sqrt{7x^2-4}}dx & = \frac{5}{2} \int \frac{1}{x\sqrt{\frac{7}{4} x^2-1}}dx, \text{\quad let }u=\frac{\sqrt{7}}{2}x \\
& = \frac{5}{2} \frac{\sqrt{7}}{2} \frac{2}{\sqrt{7}}\int \frac{1}{u\sqrt{u^2-1}} du \\
& =\frac{5}{2} \sec^{-1}(u) +C \\
& = \frac{5}{2} \sec^{-1}\left( \frac{\sqrt{7}}{2}x \right) + C. \quad\blacklozenge
\end{align*}
$$(let us also write $\operatorname{arcsec(u)}$ for $\sec^{-1}(u)$).
## So what forms can we deal with here?
By looking at the derivatives of these inverse trigonometric functions, we can deal with the following kinds of antiderivatives $$
\begin{align*}
\int \frac{1}{A+Bx^2} dx &\quad...\arctan(u)?& \text{(if both $A,B > 0$, all $x$)}\\
\int \frac{1}{\sqrt{A-Bx^2}}dx &\quad ... \arcsin(u)? & \text{(if both $A,B > 0$, $|x|\le \sqrt{\frac{A}{B}}$)}\\
\int \frac{1}{x\sqrt{Ax^2-B}}dx &\quad ... \operatorname{arcsec}(u)? & \text{(if both $A,B > 0$, $|x|\ge \sqrt{\frac{B}{A}}$)}
\end{align*}
$$
Now looking at these forms, some "obvious ones" are missing: What if the integrand is $\frac{1}{A-Bx^2}$, $\frac{1}{\sqrt{A+Bx^2}}$, $\frac{1}{\sqrt{Ax^2-B}}$, $\frac{1}{x\sqrt{Ax^2+B}}$, or $\frac{1}{x\sqrt{A-Bx^2}}$?
Here, **inverse hyperbolic functions** can help.
## Inverse hyperbolic functions (6.7)
Like the inverse trigonometric functions, we need to restrict the domain of some of the hyperbolic functions to get an inverse. We summarize them below
$\sinh(x)$ domain on all of $\mathbb{R}$, with inverse $\operatorname{arcsinh(x)}$ with domain $\mathbb{R}$ and range $\mathbb{R}$.
- $\operatorname{arcsinh(x)}$ is increasing.
$\cosh(x)$ restrict domain on $[0,\infty)$, with inverse $\operatorname{arccosh(x)}$ with domain $[1,\infty)$, range $[0,\infty)$.
- $\operatorname{arccosh(x)}$ is increasing.
$\tanh(x)$ domain on $(-\infty,\infty)$, with inverse $\operatorname{arctanh(x)}$ with domain $(-1,1)$, range $(-\infty,\infty)$.
- $\operatorname{arctanh(x)}$ is increasing.
$\operatorname{csch(x)}=\frac{1}{\sinh(x)}$ domain on $x\neq 0$, with inverse $\operatorname{arccsch(x)}$ with domain $x\neq 0$, range $y\neq 0$.
- $\operatorname{arccsch(x)}$ is decreasing.
$\operatorname{sech}(x)=\frac{1}{\cosh(x)}$ domain on $\mathbb{R}$, with inverse $\operatorname{arcsech(x)}$ with domain $(0,1]$, range $[0,\infty)$.
- $\operatorname{arcsech(x)}$ is decreasing.
$\operatorname{coth}(x) = \frac{1}{\tanh(x)}$ domain on $x\neq 0$, with inverse $\operatorname{arccoth(x)}$ with domain $|x| > 1$, range $y\neq 0$.
- $\operatorname{arccoth(x)}$ is decreasing.
Their derivatives are as follows $$
\begin{align*}
\frac{d}{dx}\operatorname{arcsinh(x)} &= \frac{1}{\sqrt{1+x^2}} & \text{for all } x \\
\frac{d}{dx}\operatorname{arccosh(x)} &= \frac{1}{\sqrt{x^2-1}} & \text{for }x > 1\\
\frac{d}{dx}(-\operatorname{arccosh(-x)} ) & =\frac{1}{\sqrt{x^2-1}} & \text{for } x <1 \\
\frac{d}{dx}\operatorname{arctanh(x)} &= \frac{1}{1-x^2} & \text{for } |x| < 1\\
\frac{d}{dx}\operatorname{arccsch(x)} &= -\frac{1}{|x|\sqrt{x^2 + 1}} & \text{for } x\neq 0\\
\frac{d}{dx}\operatorname{arcsech(x)} &= -\frac{1}{x\sqrt{1-x^2}} & \text{for }x\in(0,1)\\
\frac{d}{dx}\operatorname{arccoth(x)} &= \frac{1}{1-x^2} &\text{for }|x| > 1
\end{align*}
$$
It is a bit tricky to memorize all of them, but knowing their pictures and derivations help.
## Integrals with these inverse hyperbolic.
**Example**. Find $$
\int \frac{1}{\sqrt{x^2+4x-7}}dx
$$
$\blacktriangleright$ This is of the type with integrand $\frac{1}{\sqrt{\text{quadratic}}}$, so our choices are $\arcsin$, $\operatorname{arcsinh}$, or $\operatorname{arccosh}$. We won't know until we put it in the right form. Let us complete the square: $$
x^2 + 4x - 7 = x^2 + 4x+4-4-7=(x+2)^2-11
$$
So we have $$
\begin{align*}
\int \frac{1}{\sqrt{x^2+4x-7}}dx & = \int \frac{1}{\sqrt{(x+2)^2-11}}dx \\
& = \frac{1}{\sqrt{11}}\int \frac{1}{\sqrt{\frac{(x+2)^2}{11}-1}} dx\quad\text{let } u= \frac{x+2}{\sqrt{11}} \\
& =\frac{1}{\sqrt{11}} \int \frac{1}{\sqrt{u^2-1}} \sqrt{11}du \\
& =\begin{cases}
\operatorname{arccosh}(u)+C & \text{if }u > 1 \\
-\operatorname{arccosh}(-u)+C & \text{if } u < 1
\end{cases} \\
& = \begin{cases}
\operatorname{arccosh}\left( \frac{x+2}{\sqrt{11}} \right) + C & \text{if } \frac{(x+2)}{\sqrt{11}} > 1 \\
-\operatorname{arccosh}\left( -\frac{x+2}{\sqrt{11}} \right) + C & \text{if } \frac{(x+2)}{\sqrt{11}} < 1
\end{cases}. \quad\blacklozenge
\end{align*}
$$
Example. Find $$
\int \frac{1}{x^2-6x+1} dx
$$
$\blacktriangleright$ This is of the type with integrand $\frac{1}{\text{quadratic}}$, so it could be $\arctan$, $\operatorname{arctanh}$, or $\operatorname{arccoth}$. Let us complete the square: $$
x^2 - 6x + 1 = x^2 - 6x + 9 - 9+1=(x-3)^2
-8$$So we have $$
\begin{align*}
\int \frac{1}{x^2-6x+1}dx &= \int \frac{1}{(x-3)^2-8} dx \\
& = \frac{1}{8}\int \frac{1}{\frac{(x-3)^2}{8}-1}dx\quad\text{let }u=\frac{x-3}{\sqrt{8}}\\
& =\frac{1}{8} \int \frac{1}{u^2-1} \sqrt{8}du \\
& =\frac{-1}{\sqrt{8}}\int \frac{1}{1-u^2}du \\
&= \begin{cases}
\frac{-1}{\sqrt{8}} \operatorname{arctanh}(u)+C & \text{if }|u| < 1\\
\frac{-1}{\sqrt{8}} \operatorname{arccoth}(u)+C & \text{if }|u| > 1
\end{cases}\\
&= \begin{cases}
\frac{-1}{\sqrt{8}} \operatorname{arctanh}(\frac{x-3}{\sqrt{8}})+C & \text{if }|\frac{x-3}{\sqrt{8}}| < 1\\
\frac{-1}{\sqrt{8}} \operatorname{arccoth}(\frac{x-3}{\sqrt{8}})+C & \text{if }|\frac{x-3}{\sqrt{8}}| > 1
\end{cases}
\end{align*}
$$
This means if we seek the antiderivative on the interval $|x-3| < \sqrt{8}$, then we have $$\frac{-1}{\sqrt{8}}\operatorname{arctanh}(\frac{x-3}{\sqrt{8}})+C$$and on the interval $|x-3| > \sqrt{8}$ we have $$
\frac{-1}{\sqrt{8}}\operatorname{arccoth}(\frac{x-3}{\sqrt{8}})+C \quad\blacklozenge
$$
**Later, we will develop an alternate way of dealing with these.**
## Alternate forms of $\operatorname{arcsinh}(x)$ and $\operatorname{arccosh}(x)$.
Since $\cosh(x) = \frac{e^x + e^{-x}}{2}$, we can try to invert it directly. Switch $x$ with $y$, we solve for $y\ge 0$ (since the range of $\operatorname{arccosh(x)}$ is $y\ge0$), where $$
x= \frac{e^y + e^{-y}}{2}
$$In last week's homework, you may have seen this is secretly a quadratic, if we let $u=e^y$, then $$
\begin{align*} \\
&x = \frac{\left( u+\frac{1}{u} \right)}{2} \\
\implies & 2xu=u^2 + 1 \\
\implies & u^2-2xu+1 = 0 \\
\implies & u = \frac{2x\pm \sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}
\end{align*}
$$
Since $y\ge0$, we have $u=e^y\ge1$, which means we have to pick $+$ sign (when $x\ge1$). So $$
u = e^y = x + \sqrt{x^2 - 1}
$$Hence $y=\ln(x+\sqrt{x^2-1})$, namely $$
\operatorname{arccosh}(x) = \ln(x+\sqrt{x^2-1}) \quad\text{for }x\ge 1.
$$
Similarly, one can show that $$
\operatorname{arcsinh}(x)=\ln(x+\sqrt{x^2 + 1})\quad\text{for all }x
$$and $$
\operatorname{arctanh}(x)=\ln \sqrt{\frac{1+x}{1-x}}.
$$
## A brief summary of the integral types $\int \frac{1}{\text{quadratic}}$ and $\int \frac{1}{\sqrt{\text{quadratic}}}$.
If we have $\int \frac{1}{\text{quadratic}}$, then by completing the square and factoring, we will get the following cases:
$$
\begin{align*}
\int \frac{1}{1+u^2}du &= \arctan(u) + C \quad \text{for all }u \\
\int \frac{1}{1-u^2} du &=\begin{cases}
\operatorname{arctanh} (u)+C & \text{for }|u| < 1 \\
\operatorname{arccoth}(u) + C & \text{for }|u| > 1
\end{cases}
\end{align*}
$$
If we have $\int \frac{1}{\sqrt{\text{quadratic}}}$, then by completing the square and factoring, we will get the following cases:
$$
\begin{align*}
\int \frac{1}{\sqrt{1+u^2}} du &= \operatorname{arcsinh}+C \quad\text{for all } u \\
\int \frac{1}{\sqrt{1-u^2}} du &= \arcsin(u)+C \quad\text{for } |u| < 1 \\
\int \frac{1}{\sqrt{u^2-1}} du &=\begin{cases}
\operatorname{arccosh} (u)+C & \text{for } u > 1 \\
-\operatorname{arccosh} (-u)+C & \text{for } u < 1
\end{cases}
\end{align*}
$$
///