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# Week 2 Monday Problems
On Monday's lecture, we deduced much of the following antiderivatives to deal with integrals of the 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:
$$
\begin{align*}
\int \frac{1}{\sqrt{1+u^2}} du &= \operatorname{arcsinh}+C \quad\text{for all } u \\
\int \frac{1}{\sqrt{1-u^2}} du &= \arcsin(u)+C \quad\text{for } |u| < 1 \\
\int \frac{1}{\sqrt{u^2-1}} du &=\begin{cases}
\operatorname{arccosh} (u)+C & \text{for } u > 1 \\
-\operatorname{arccosh} (-u)+C & \text{for } u < 1
\end{cases}
\end{align*}
$$
So for these cases, remember: (1) Complete the square, (2) factor so it looks like the desired form, (3) $u$-substitution. You can do $\int \frac{1}{\text{quadratic}}$ and $\int \frac{1}{\sqrt{\text{quadratic}}}$ for any quadratic !! You got this !!
## Reading.
Read Ch. 6.6 and 6.7.
## Problems.
Problem 5 is "optional", but you should still read it to know the final result. It's not too terrible and you can do it! At the minimum, make sure you can do computation problems such as 2, 3, 4, 6.
1. Sketch out the following. Use our class notes and text as reference, and note the domain and range:
1. $\arcsin(x)$
2. $\arccos(x)$
3. $\arctan(x)$
4. $\operatorname{arcsinh(x)}$
5. $\operatorname{arccosh}(x)$
6. $\operatorname{arctanh}(x)$
7. $\operatorname{arccoth}(x)$
2. Compute the following antiderivatives.
1. $\displaystyle\int \frac{1}{x^2+6x+17} dx$
2. $\displaystyle \int \frac{1}{x^2+6x+7}dx$
3. $\displaystyle\int \frac{1}{\sqrt{10+6x^2}} dx$
4. $\displaystyle \int \frac{1}{\sqrt{5+6x+2x^2}} dx$
5. $\displaystyle \int \frac{1}{\sqrt{5+6x - 2x^2}} dx$
6. $\displaystyle \int \frac{1}{\sqrt{x^2 - 8x + 10}} dx$
3. You can now actually secretly do $\int \frac{\text{linear}}{\text{quadratic}}$ and $\int \frac{\text{linear}}{\sqrt{\text{quadratic}}}$, but you have to be sneaky. Let's see if you can figure it out:
1. $\displaystyle\int \frac{x}{x^2+1}dx$ (you can directly do $u$-sub here. Do you see how?)
2. $\displaystyle\int \frac{x+5}{x^2+1}dx$ (split this into two integrals...)
3. $\displaystyle\int \frac{3x-7}{x^2 + 6x + 10} dx$ (slightly tricky, but complete the square first...you can do it!!)
4. Show $\operatorname{arcsinh}(x) = \ln(x+\sqrt{x^2 + 1})$ for all $x$.
- Hint. Start with $\displaystyle y= \sinh(x) = \frac{e^x-e^{-x}}{2}$. Switch $x\leftrightarrow y$ to find the inverse. There will be a secret quadratic somewhere.
- This is why in some integration table, you would see instead $$ \int \frac{1}{\sqrt{1+u^2}} du= \ln(x+\sqrt{x^2+1}) + C $$
5. If you look at an integration table, you might see $\displaystyle\int \frac{1}{\sqrt{x^2-1}}dx = \ln|x + \sqrt{x^2-1}| +C$. But the way we have it so far are in terms of $\operatorname{arccosh}(x)$ and $-\operatorname{arccosh}(-x)$. Let us show they are actually the same!! That is, $$
\ln|x+\sqrt{x^2-1}|=\begin{cases}\operatorname{arccosh}(x) & \text{if } x > 1 \\
-\operatorname{arccosh}(-x) & \text{if } x < -1
\end{cases}
$$
1. Show $\operatorname{arccosh}(x) = \ln(x+\sqrt{x^2-1})$ for $x \ge 1$. Note we obtain $\operatorname{arccosh}(x)$ by restricting $\cosh(x)$ to the domain $[1,\infty)$.
- Hint. To do this, recall $\displaystyle y = \cosh(x) = \frac{e^x + e^{-x}}{2}$. So swapping $x\leftrightarrow y$ , we have $\displaystyle x= \frac{e^y + e^{-y}}{2}$ and solve for $y$. There is a secret quadratic. This $y=\operatorname{arccosh}(x)$.
2. Using this result, we should get $-\operatorname{arccosh}(-x)=-\ln(-x+\sqrt{x^2-1})$ for $x \le 1$. Using properties of log and conjugates, show $-\operatorname{arccosh}(-x) = \ln(-x-\sqrt{x^2-1})$.
3. Note that $$\ln|x+\sqrt{x^2-1}|=\begin{cases}\ln(x+\sqrt{x^2-1}) & \text{if } x+\sqrt{x^2-1} > 0 \\\ln(-x-\sqrt{x^2-1}) & \text{if } x+\sqrt{x^2-1} < 0\end{cases}$$So to finally establish this equality, we just need to know the condition $x+\sqrt{x^2-1} > 0$ is the same as $x > 1$; and the condition $x+\sqrt{x^2-1} < 0$ is the same as $x < -1$. Show this. If this is too hard, just use a graphing calculator like Desmos to verify this.
6. Show $\displaystyle\operatorname{arctanh}(x)=\ln \sqrt{\frac{1+x}{1-x}}$ for $|x|<1$.
7. Show $\displaystyle\operatorname{arccoth}(x)= \ln \sqrt{\frac{x+1}{x-1}}$ for $|x| > 1$.
By the way, combining the last two together we get $$
\int \frac{1}{x^2-1} dx = \ln \sqrt{\left|\frac{1+x}{1-x}\right|}+C = \frac{1}{2} \ln\left| \frac{1+x}{1-x}\right| + C
$$ for all $x\neq \pm1$.
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