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# Week 3 Monday 7/3 brief notes.
## Announcement.
Exam 1 will be this **Thursday 7/6** in class from 6:30 PM to 9:15PM. It should not take whole time but you may use the whole time if you wish. You may use one page both sides (regular printer paper size) of cheat sheet. You may not use any calculator nor any technology for the exam. The exam will cover everything we discussed up to and including today's lecture. In the text this corresponds to sections up to 7.5. Focus on the material we discussed in class, our class notes, and your homework. Use the text to help supplement your understanding and practice.
You can do it!
There is no class 7/4 (Tuesday), and on Wednesday we will do a brief review.
Last time I forgot to mention another trigonometric integration situation, let me mention it now:
## Integrals of the form $\int\sin(nx)\cos(mx)dx$ and friends (7.2 continued).
If instead of powers of $\sin(x)$ and $\cos(x)$ with the same arguments, we have one of $$
\begin{align*}
\int \sin(nx)\cos(mx)dx \\
\int \sin(nx)\sin(mx)dx \\
\int \cos(nx)\cos(mx)dx
\end{align*}
$$Then we can use the trigonometric **product to sum** identities: $$
\begin{align*}
\sin(A)\cos(B)=\frac{1}{2}[\sin(A-B)+\sin(A+B)] \\
\sin(A)\sin(B)=\frac{1}{2}[\cos(A-B)-\cos(A+B)] \\
\cos(A)\cos(B)=\frac{1}{2}[\cos(A-B)+\cos(A+B)]
\end{align*}
$$
**Example.** Find $$
\int\sin(4x)\cos(17x)dx
$$
$\blacktriangleright$ Use product to sum formula, $$
\begin{align*}
\int\sin(4x)\cos(17x)dx & =\int \frac{1}{2}[\sin(-13x)+\sin(21x)]dx \\
& =\int \frac{1}{2}[-\sin(13x)+\sin(21x)]dx\\
& = \frac{1}{2}\left[ \frac{1}{13}\cos(13x)-\frac{1}{21}\cos(21x) \right]+C.\quad\blacklozenge
\end{align*}
$$
## Trigonometric substitution (7.3).
Quite often we encounter the expressions $\sqrt{x^{2}+a^{2}}$, $\sqrt{x^{2}-a^{2}}$, or $\sqrt{a^{2}-x^{2}}$ for some constant $a$ in our integrand. For instance our favorite integral $\int \frac{1}{1+x^{2}} dx = \arctan(x)+C$. When dealing with these, one can consider a **trigonometric substitution** or **hyperbolic substitution**, as these functions have similar identities, namely: $$
\begin{align*}
\cos^{2}(t)+\sin^{2}(t)=1 \\
1+\tan^{2}(t)=\sec^{2}(t) \\
\cosh^{2}(t)-\sinh^{2}(t)=1
\end{align*}
$$
Let us see how. Here's a summary first.
$$
\begin{array}{}
\text{form} & \text{substitution} & \text{identity} & dx \\
\sqrt{a^{2}+x^{2}} & x=a \tan(t) & a^{2}+x^{2}=a^{2}\sec^{2}(t) & dx=a\sec^{2}(t)dt \\
\sqrt{a^{2}-x^{2}} & x=a \sin(t) & a^{2}-x^{2}=a^{2}\cos^{2}(t) & dx=a\cos(t)dt \\
\sqrt{x^{2}-a^{2}} & x=a \sec(t) & x^{2}-a^{2}=a^{2}\tan^{2}(t) & dx=a\sec(t)\tan(t)dt \\
\sqrt{x^{2}-a^{2}} & x=a\cosh(t) & x^{2}-a^{2}=a\sinh^{2}(t) & dx=a\sinh(t)dt
\end{array}
$$
You will also need basic trigonometric identities and their derivatives, as well as drawing a right triangle to help you figure out the corresponding quantities.
## The case of $\sqrt{x^{2}+a^{2}}$.
When encountered $x^{2}+a^{2}$ in your integrand, notice how we have two identities that are similar, where we have some function squared plus a constant (squared): $$
\begin{align*}
1+\tan^{2}(t) = \sec^{2}(t) \\
\end{align*}
$$
Then this indicates they possibly suitable substitutions, by following your nose. Let's see our first example.
**Example.** Our favorite $\int \frac{1}{1+x^{2}}dx$.
We already know the answer, but let's say we don't. The numerator is the square of $\sqrt{1+x^{2}}$. Notice if we let $x=\tan(t)$, then the numerator $1+x^{2}=1+\tan^{2}(t)=\sec^{2}(t)$, and that $dx = \sec^{2}(x)dx$. So we get $$
\begin{align*}
\int \frac{1}{1+x^{2}}dx & = \int \frac{1}{1+\tan^{2}(t)} \sec^{2}(t)dt\\
& = \int \frac{1}{\sec^{2}(t)}\sec^{2}(t)dt \\
& =\int 1 dt \\
& = t + C = \arctan(x)+C.
\end{align*}
$$
Done!
So the ideal is to match the expression $1+(\text{something)}^{2}$ to $1+\tan^{2}(t)$.
**Example.** Find $$
\int \frac{x}{\sqrt{9+4x^{2}}}dx
$$
$\blacktriangleright$ Ok, factor to make it look like $1+u^{2}$. So,
$$
\begin{align*}
\int \frac{x}{\sqrt{9+4x^{2}}}dx & =\frac{1}{3}\int \frac{x}{\sqrt{1+\frac{4}{9}x^{2}}}dx
\end{align*}
$$so we see we let $\frac{4}{9}x^{2}=\tan^{2}(t)$, whence $x = \frac{3}{2}\tan(t)$, and $dx = \frac{3}{2}\sec^{2}(t)dt$. Then the integral becomes $$
\begin{align*}
\frac{1}{3} \int \frac{\frac{3}{2}\tan(t) \frac{3}{2}\sec^{2}(t)dt}{\sqrt{1+\tan^{2}(t)}} & = \frac{1}{3} \frac{9}{4}\int \frac{\tan(t)\sec^{2}(t)}{\sec(t)}dt \\
& = \frac{3}{4}\int\tan(t)\sec(t)dt \\
& = \frac{3}{4} \sec(t) +C \\
& = \frac{3}{4} \sqrt{1+\tan^{2}(t)} +C \\
& = \frac{3}{4} \sqrt{1+ \frac{4}{9}x^{2}} +C
\end{align*}
$$
Note, we could have also done direct $u$-substitution, with $u=9+4x^{2}$ directly. Try it and verify you get the same result. So although possible, trigonometric substitution need not be the fastest way to work out an integral.
**Example.** Find $$
\int \frac{1}{x^{2}\sqrt{x^{2}+16}} dx
$$
$\blacktriangleright$ Let us factor to get $$
\frac{1}{4} \int \frac{1}{x^{2}\sqrt{\frac{x^{2}}{16}+1}}dx
$$
So we will use $\tan^{2}(t)= \frac{x^{2}}{16}$, namely $\frac{x}{4}=\tan(t)$. So $dx = 4\sec^{2}(t)dt$. So the integral becomes $$
\begin{align*}
\frac{1}{4} \int \frac{1}{16\tan^{2}(t)\sec(t)} 4\sec^{2}(t)dt = & \frac{1}{16} \int \frac{\sec(t)dt}{\tan^{2}(t)} \\
& = \frac{1}{16} \int \frac{\cos(t)dt}{\sin^{2}(t)} \quad \text{let }u=\sin(t),du=\cos(t)dt \\
& =\frac{1}{16} \int \frac{du}{u^{2}} \\
& = \frac{1}{16}\frac{-1}{u} +C \\
& = - \frac{1}{16}\frac{1}{\sin(t)} + C
\end{align*}
$$
But what is $\sin(t)=?$ We know $\tan(t) = \frac{x}{4}$, so drawing a right triangle ![[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 15.00.18.excalidraw.svg]]
%%[[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 15.00.18.excalidraw.md|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 15.00.18.excalidraw.dark.svg|dark exported image]]%%
shows $\sin(t)= \frac{x}{\sqrt{x^{2}+16}}$. So the final result is $$
-\frac{1}{16} \frac{\sqrt{x^{2}+16}}{x} + C \quad\blacklozenge
$$
The other cases employ a similar strategy.
## The case of $\sqrt{a^{2}-x^{2}}$.
Here our main identity is $1-\sin^{2}(t)=\cos^{2}(t)$.
**Example.** Find $$
\int \frac{\sqrt{4-x^{2}}}{x^{2}}dx
$$
$\blacktriangleright$ Ok I will factor out a 2 first, to get $$
\int \frac{2\sqrt{1-\frac{x^{2}}{4}}}{x^{2}}dx
$$This prompts me to set $\sin^{2}(t)=\frac{x^{2}}{4}$, or $\sin(t) = \frac{x}{2}$, with $2\cos(t)dt=dx$. So substituting everything in, we get $$
\int \frac{2 \cos(t) 2\cos(t)dt}{4\sin^{2}(t)} = \int \frac{\cos^{2}(t)}{\sin^{2}(t)}dt = \int\cot^{2}(t) dt
$$
Ok, now we need to find the antiderivative of cotangent. Now we do have a list of related trigonometric identities and their calculus:
$$
\begin{align*}
\frac{d}{dt}\cot(t)= - \csc^{2}(t) \\
\frac{d}{dt}\csc(t) = -\cot(x)\csc(t)
\end{align*}
$$
and that $1+\cot^{2}(t)=\csc^{2}(t)$. So we see that if we convert this $\cot^{2}(t)$ into $\csc^{2}(t)$, then we have a corresponding antiderivative. So $$
\begin{align*}
\int\cot^{2}(t) dt & = \int (\csc^{2}(t) -1)dt \\
& = -\cot(t)-t+C
\end{align*}
$$
Now, with $\sin(t) = x/2$ we have $t=\arcsin(\frac{x}{2})$, and using a right triangle ![[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 15.32.22.excalidraw.svg]]
%%[[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 15.32.22.excalidraw.md|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 15.32.22.excalidraw.dark.svg|dark exported image]]%%
we see that $\cot(t) = \frac{\sqrt{4-x^{2}}}{x}$. Hence the integral is $$
- \frac{\sqrt{4-x^{2}}}{x}-\arcsin\left( \frac{x}{2} \right)+C. \quad\blacklozenge
$$
## The case of $\sqrt{x^{2}-a^{2}}$.
In this situation, the corresponding identity is $\sec^{2}(t)-1=\tan^{2}(t)$. You can also use the hyperbolic identity $\cosh^{2}(t)-1=\sinh^{2}(t)$.
**Example.** Find $$
\int \frac{1}{x^{3}(x^{2}-4)^{3/2}}dx
$$
$\blacktriangleright$ Here we recognize the part of $x^{2}-4$, so we will use the identity $\sec^{2}(t)-1=\tan^{2}(t)$ somehow. We will follow our nose. First, factor
$$
\begin{align*}
\int \frac{1}{x^{3}(x^{2}-4)^{3/2}} dx & = \frac{1}{8} \int \frac{1}{x^{3}\left( \frac{x^{2}}{4}-1 \right)^{3/2}} dx
\end{align*}
$$which suggests we set $\sec^{2}(t) = \frac{x^{2}}{4}$, or $\sec(t)=\frac{x}{2}$, which also gives $dx=2\sec(t)\tan(t)dt$. Substituting everything in gives $$
\begin{align*}
\frac{1}{8}\int \frac{2\sec(t)\tan(t)dt}{8\sec^{3}(t) \tan^{3}(t)} & = \frac{1}{32}\int \frac{1}{\sec^{2}(t)\tan^{2}(t)}dt \\
& = \frac{1}{32} \int\cos^{2}(t)\cot^{2}(t)dt
\end{align*}
$$
Now this looks better but still scary. The idea is to use trigonometric identities to turn them into things we know how to integrate. For instance $$
\begin{align*}
\frac{1}{32} \int(1-\sin^{2}(t))\cot^{2}(t)dt & =\frac{1}{32} \int[\cot^{2}(t)-\sin^{2}(t)\cot^{2}(t)]dt\\
& = \frac{1}{32}\int [\cot^{2}(t) - \cos^{2}(t)]dt
\end{align*}
$$
Now, we don't know how to integrate $\cot^{2}(t)$ but we know how to integrate $\csc^{2}(t)$, namely $\int \csc^{2}(t)dt = -\cot(t)+C$. And from last week we know how to integrate $\cos^{2}(t)$ using double-angle formula. So, we get $$
\begin{align*}
\frac{1}{32}\int[\csc^{2}(t)-1-\cos^{2}(t)]dt & = \frac{1}{32} \left[ -\cot(t) - t - \int\cos^{2}(t)dt \right] \\
& = \frac{1}{32}\left[ -\cot(t)-t-\frac{1}{2}\int[1+\cos(2t)]dt \right] \\
& =\frac{1}{32}\left[ -\cot(t)-t-\frac{1}{2}t-\frac{1}{4}\sin(2t) \right]+C \\
& =\frac{1}{32}\left[ -\cot(t)- \frac{3}{2}t-\frac{1}{4}\sin(2t) \right]+C
\end{align*}
$$
Great! But what are any of these? We have $\sec(t)=\frac{x}{2}$, so drawing a right triangle ![[1 teaching/smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 16.34.04.excalidraw.svg]]
%%[[1 teaching/smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 16.34.04.excalidraw|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 16.34.04.excalidraw.dark.svg|dark exported image]]%%
we see that $\cot(t)=\frac{2}{\sqrt{x^{2}-4}}$, $t=\operatorname{arcsec}(\frac{x}{2})$. But what about $\sin(2t)$? Well by double angle formula, $\sin(2t)=2\sin(t)\cos(t)$, so using the right triangle again we get $\sin(2t)=2 \frac{\sqrt{x^{2}-4}}{x} \frac{2}{x}= \frac{4\sqrt{x^{2}-4}}{x^{2}}$.
Now substituting everything in, we finally get $$
\frac{1}{32}\left[ -\frac{2}{\sqrt{x^{2}-4}}- \frac{3}{2}\operatorname{arcsec}\left( \frac{x}{2} \right) -\frac{\sqrt{x^{2}-4}}{x^{2}}\right]+C. \quad\blacklozenge
$$
Don't forget you can complete the square as well.
**Example.** Find $$
\int (x^{2}+4x+3)^{1/2}dx
$$
$\blacktriangleright$ First complete the square, $$
\begin{align*}
\int (x^{2}+4x+3)^{1/2}dx & = \int((x^{2}+4x+4)-1)^{1/2}dx \\
& =\int((x+2)^{2}-1)^{1/2}dx
\end{align*}
$$
So first let us do a preliminary substitution, $u=x+2$, $du=dx$, so we get $$
\int(u^{2}-1)^{1/2} du
$$
Now our trigonometric identities remind us that $\sec^{2}(t)-1=\tan^{2}(t)$, so we set $u=\sec(t)$, with $du = \sec(t)\tan(t)dt$. Let us put everything in and get $$
\begin{align*}
\int(\sec^{2}(t)-1)^{1/2}\sec(t)\tan(t)dt & = \int\tan (t)\sec(t)\tan(t)dt \\
& = \int\tan^{2}(t)\sec(t)dt
\end{align*}
$$
Using the identity $1+\tan^{2}(t)=\sec^{2}(t)$, we have then $$
\int (\sec^{2}(t)-1)\sec(t)dt = \int[\sec^{3}(t)-\sec(t)]dt
$$
And last week we showed $\int \sec(t)dt=\ln|\sec(t)+\tan(t)|+C$ and $\int\sec^{3}(t)dt=\frac{1}{2}[\sec(t)\tan(t)+\ln|\sec(t)+\tan(t)|]+C$.
So then we have$$
\begin{align*}
& \frac{1}{2}[\sec(t)\tan(t)+\ln|\sec(t)+\tan(t)|] - \ln|\sec(t)+\tan(t)| + C \\
& =\frac{1}{2}\sec(t)\tan(t)-\frac{1}{2}\ln|\sec(t)+\tan(t)|+C
\end{align*}
$$
And finally, note that $u=\sec(t)$, so with a right triangle ![[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 17.56.22.excalidraw.svg]]
%%[[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 17.56.22.excalidraw.md|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-03 17.56.22.excalidraw.dark.svg|dark exported image]]%%
we have $\tan(t) = \sqrt{u^{2}-1}$. So we now have $$
\frac{1}{2} u \sqrt{u^{2}-1} -\frac{1}{2} \ln|u+\sqrt{u^{2}-1}|+C
$$
And finally, $u=x+2$, so we have $$
\frac{1}{2}(x+2)\sqrt{(x+2)^{2}-1}-\frac{1}{2}\ln|x+2+\sqrt{(x+2)^{2}-1}|+C. \quad\blacklozenge
$$
**Example.** Area of a lune.
Consider a large circle of radius $R$ and a small circle of radius $r < R$ arranged as below, where the centers of the circles are vertically aligned on an axis, and the points of the intersections of the circles form a line that is perpendicular to this axis that also goes through the center of the small circle:
![[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-05 09.40.52.excalidraw.svg]]
%%[[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-05 09.40.52.excalidraw.md|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-05 09.40.52.excalidraw.dark.svg|dark exported image]]%%
Find the area of the shaded area, a lune.
$\blacktriangleright$ One main application of integrals is to find areas. Notice the lune is bounded between two upper semi-circle arcs. Let us set the origin to be at the center of the large circle. Then the large semi-circle arc has equation $y=\sqrt{R^{2}-x^{2}}$. The small semi-circle arc has equation given by $y=\sqrt{r^{2}-x^{2}}+ h$, where it is shifted up by some amount $h$. Using Pythagorean theorem, one can see that $h=\sqrt{R^{2}-r^{2}}$. So the equation of the small semi-circle arc is $y=\sqrt{R^{2}-x^{2}}+\sqrt{R^{2}-r^{2}}$.
Now the given configuration shows the area is trapped between $x=-r$ and $x=r$. So the area is given by $\int_{a}^{b}(\text{top curve)}-(\text{bottom curve})dx$, which gives $$
\int_{-r}^{r} \underbrace{\sqrt{r^{2}-x^{2}}}_{(A)}+\underbrace{\sqrt{R^{2}-r^{2}}}_{(B)}-\underbrace{\sqrt{R^{2}-x^{2}}}_{(C)} \,dx
$$Now if we examine this integral carefully, the terms $(A),(C)$ are of the same type of trigonometric substitutions, while term $(B)$ is just integrating a constant number.
Let us find the general form of the antiderivative for $(A),(C)$ at once, namely $\int \sqrt{a^{2}-x^{2}}dx$ for some positive $a > 0$. For this we use the trigonometric identity $1-\sin^{2}(x)=\cos^{2}(x)$. So, $$
\begin{align*}
\int\sqrt{a^{2}-x^{2}}dx & = \int a \sqrt{1-\frac{x^{2}}{a^{2}}}dx \\
& \text{let } \sin(t)=\frac{x}{a}, dx=a\cos(t)dt\\
& = a \int \cos(t)a\cos(t)dt \\
& =a^{2} \int \cos^{2}(t) dt \\
& =a^{2} \int \frac{1}{2}(1+\cos(2t))dt \\
& = \frac{a^{2}}{2}\left[ t+\frac{\sin(2t)}{2} \right] + C \\
\end{align*}
$$
Now to put everything back to $x$, note that $t=\arcsin\left( \frac{x}{a} \right)$. But what about $\sin(2t)$? Again by double angle formula $\sin(2t)=2\sin(t)\cos(t)$, which we can then work out each piece using a right triangle with $\sin(t)=\frac{x}{a}$:
![[1 teaching/smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-05 10.10.42.excalidraw.svg]]
%%[[1 teaching/smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-05 10.10.42.excalidraw|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-3/---files/week-3-monday-notes 2023-07-05 10.10.42.excalidraw.dark.svg|dark exported image]]%%
Hence $\sin(2t)=2 \frac{x}{a} \frac{\sqrt{a^{2}-x^{2}}}{a}=\frac{2x\sqrt{a^{2}-x^{2}}}{a^{2}}$. So, $$
\begin{align*}
\int \sqrt{a^{2}-x^{2}}dx & = \frac{a^{2}}{2}\left[ \arcsin\left( \frac{x}{2} \right)+ \frac{x\sqrt{a^{2}-x^{2}}}{a^{2}} \right]+C \\
& = \frac{a^{2}}{2}\arcsin\left( \frac{x}{2} \right) + \frac{x\sqrt{a^{2}-x^{2}}}{2} +C
\end{align*}
$$
Finally, using this antiderivative we can find the definite integral that we set up for our area of the lune. I will leave it to you to do the substitutions. $\blacklozenge$
**Remark.** Quite often when finding some parts of the area of a circle, one would encounter these integrals involving trigonometric substitutions. One may need to pay closer attention to the bounds, however.