## 17Calculus - Calculus 2 - Practice Exam 1 (Semester A)

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This is the first exam for second semester single variable calculus. The exam covers advanced integration and applications including work, volume of revolution and centroids. Also parametrics are part of this exam.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

1.5 hours

Questions

10

Total Points

100

Tools

Calculator

see instructions

Formula Sheet(s)

none

Other Tools

none

Instructions:
- This exam is in two main parts, labeled parts A and B, with different instructions for each part.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

Part A - Questions 1-4

Instructions for Part A - - You have 30 minutes to complete this part of the exam. No calculators are allowed. Show all your work, justify your conclusions and give exact, completely factored answers.

Section 1

Evaluate the following integrals. Each question in this section is worth 9 points.

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } }$$

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } = \frac{e^4-13}{32e^4} }$$

Problem Statement

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } }$$

Solution

Use integration by parts: $$\int{u~dv} = uv - \int{ v ~du }$$

 $$u=x^2 \to du = 2x ~dx$$ $$dv=e^{-4x}dx \to \displaystyle{v=\frac{e^{-4x}}{-4}}$$

$$\displaystyle{\left. \frac{-x^2}{4}e^{-4x} \right|_0^1 - \int_{0}^{1}{\left(\frac{-1}{4} e^{-4x} \right) 2x~dx} =}$$ $$\displaystyle{\frac{-1}{4}e^{-4} + 0 + \frac{1}{2} \int_{0}^{1}{xe^{-4x}dx}}$$
Use integration by parts again.

 $$\begin{array}{lcl} u=x & \to & du=dx \\ dv = e^{-4x}dx & \to & \displaystyle{ v=\frac{e^{-4x}}{-4} } \end{array}$$

$$\displaystyle{\frac{-1}{4e^4} + \frac{1}{2}\left[ \left. \frac{-x}{4}e^{-4x} \right|_0^1 - \int_{0}^{1}{\frac{-1}{4}e^{-4x}dx} \right]}$$
$$\displaystyle{\frac{-1}{4e^4} + \frac{1}{2}\left[ \frac{-1}{4}e^{-4} +0 \right] + \frac{1}{8} \int_{0}^{1}{e^{-4x}dx}}$$
$$\displaystyle{\frac{-3}{8e^4} + \left. \frac{1}{8}\frac{e^{-4x}}{4} \right|_0^1}$$
$$\displaystyle{\frac{-3}{8e^4} - \frac{1}{32}\left[ e^{-4} - 1 \right] = \frac{e^4-13}{32e^4}}$$

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } = \frac{e^4-13}{32e^4} }$$

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$$\int{ \sin^3(\pi t) \cos^3(\pi t) ~dt }$$

Problem Statement

$$\int{ \sin^3(\pi t) \cos^3(\pi t) ~dt }$$

$$\displaystyle{ \int{ \sin^3(\pi t)\cos^3(\pi t) ~dt } = }$$ $$\displaystyle{ \frac{\cos^4(\pi t )}{12}(2\cos^2(\pi t)-3) + C }$$

Problem Statement

$$\int{ \sin^3(\pi t) \cos^3(\pi t) ~dt }$$

Solution

$$\int{\sin(\pi t)\sin^2(\pi t)\cos^3(\pi t)~dt}$$
$$\int{\sin(\pi t)(1-\cos^2(\pi t))\cos^3(\pi t)~dt}$$

 $$u=\cos(\pi t)$$ $$du=-\sin(\pi t)dt$$ $$-\int{(1-u^2)u^3du}$$ $$\int{u^5-u^3 du}$$ $$\displaystyle{\frac{u^6}{6}-\frac{u^4}{4}+C}$$ $$\displaystyle{\frac{u^4}{12}(2u^2-3)+C}$$ $$\displaystyle{\frac{\cos^4(\pi t)}{12}(2\cos^2(\pi t)-3)+C}$$

$$\displaystyle{ \int{ \sin^3(\pi t)\cos^3(\pi t) ~dt } = }$$ $$\displaystyle{ \frac{\cos^4(\pi t )}{12}(2\cos^2(\pi t)-3) + C }$$

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$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } }$$

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } = \frac{3\sqrt{2}}{8} }$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } }$$

Solution

Use $$1+\tan^2(\theta)=\sec^2(\theta)$$ and let $$r=2\tan(\theta) ~~ \to ~~ dr=2\sec^2(\theta)~d\theta$$.
The triangle based on $$r=2\tan(\theta)$$ is shown to the right. Using the Pythagorean Theorem, the hypotenuse is $$\displaystyle{\sqrt{r^2+4}}$$.
First, we will evaluate the integral without the limits of integration (the indefinite integral).

 $$\displaystyle{\int{\frac{3}{(r^2+4)^{3/2}}dr}}$$ $$\displaystyle{\int{\frac{3}{(4\tan^2(\theta) + 4)^{3/2}}2\sec^2(\theta)d\theta}}$$ $$\displaystyle{\int{\frac{6\sec^2(\theta)}{4^{3/2}(1+\tan^2(\theta))^{3/2}}d\theta}}$$ $$\displaystyle{\frac{3}{4}\int{\frac{\sec^2(\theta)}{(\sec^2(\theta))^{3/2}}d\theta}}$$ $$\displaystyle{\frac{3}{4}\int{\frac{1}{\sec(\theta)}d\theta}}$$ $$\displaystyle{\frac{3}{4}\int{\cos(\theta)~d\theta}}$$ $$\displaystyle{\frac{3}{4}\sin(\theta)+C}$$ Use the triangle to convert from $$\theta$$ to $$r$$. $$\displaystyle{\frac{3}{4}\frac{r}{\sqrt{r^2+4}}+C}$$

Now we will evaluate the integral using the limits of integration.
$$\displaystyle{\left. \frac{3}{4} \frac{r}{\sqrt{r^2+4}} \right|_0^2 = \frac{3}{4} \frac{2}{\sqrt{8}} - 0 = \frac{3}{2} \frac{1}{2\sqrt{2}} = \frac{3\sqrt{2}}{8} }$$

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } = \frac{3\sqrt{2}}{8} }$$

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$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)} dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)} dx } }$$

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)}dx} = }$$ $$\displaystyle{ \ln\abs{(x+1)^3(x^2+4)^3} + k }$$

Problem Statement

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)} dx } }$$

Solution

Use partial fraction expansion.
$$\displaystyle{\frac{9x^2+6x+12}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4}}$$
$$9x^2+6x+12=A(x^2+4)+(Bx+C)(x+1)$$
$$x=-1$$: $$~~ 9-6+12 = A(1+4)$$ $$\to$$ $$15=A(5)$$ $$\to$$ $$A=3$$
$$x=0$$: $$~~ 12=A(4)+C(1)$$ $$\to$$ $$12=12+C$$ $$\to$$ $$C=0$$
$$x=1$$: $$~~ 9+6+12=A(5)+B(1+1)$$ $$\to$$ $$27=15+2B$$ $$\to$$ $$(27-15)/2=B$$ $$\to$$ $$B=6$$
The integral can now be written
$$\displaystyle{\int{\frac{3}{x+1}+\frac{6x}{x^2+4}dx}}$$
The first fraction is easily integrated.
$$\displaystyle{\int{\frac{3}{x+1}dx}=3\ln\abs{x+1}+c_1}$$
Use substitution for the second fraction.
$$u=x^2+4 ~~ \to ~~ du=2x~dx$$
$$\displaystyle{\int{\frac{6x}{x^2+4}dx}=}$$ $$\displaystyle{\int{\frac{3~du}{u}dx}=}$$ $$\displaystyle{3\ln\abs{u}+c_2=}$$ $$\displaystyle{3\ln\abs{x^2+4}+c_2}$$
Putting the results together, we get
$$3\ln\abs{x+1}+3\ln\abs{x^2+4}+k=$$ $$3\ln\abs{(x+1)(x^2+4)}+k=$$ $$\ln\abs{(x+1)^3(x^2+4)^3}+k$$
We combined the constants $$k=c_1+c_2$$.

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)}dx} = }$$ $$\displaystyle{ \ln\abs{(x+1)^3(x^2+4)^3} + k }$$

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Part B - Questions 5-10

Instructions for Part B - - You may use your calculator. You have one hour to complete this part of the exam. Show all your work. Correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions. Give exact, simplified answers.

Section 2 [16 points]

A region $$R$$ is bounded by the graph of $$y = e^x$$, the line $$y=1$$ and the line $$x=3$$.
Sketch the region $$R$$. [ click here to show/hide the plot ]

A solid is generated by revolving the region $$R$$ about the line $$x=-1$$. Set up, but do not evaluate, a definite integral for the volume of the resulting solid of revolution.

Problem Statement

A solid is generated by revolving the region $$R$$ about the line $$x=-1$$. Set up, but do not evaluate, a definite integral for the volume of the resulting solid of revolution.

Solution

Since the problem statement did not specify which method, we can use either the disc or shell method. Both solutions are shown here.

Disc Method

$$\displaystyle{V=\pi\int_c^d{R^2-r^2~dy}}$$
$$R=4 ~~~~ r=x+1$$
$$y=e^x ~~ \to ~~ \ln(y)=x ~~ \to ~~ r=\ln(y)+1$$
$$\displaystyle{V=\pi\int_1^{e^3}{16-[1+\ln(y)]^2~dy}}$$

Shell Method

$$\displaystyle{V=2\pi\int_a^b{rh~dx}}$$
$$r=1+x ~~~~ h=y-1=e^x-1$$
$$\displaystyle{V=2\pi\int_0^3{(1+x)(e^x-1)~dx}}$$

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A second solid has base $$R$$ and cross-sections perpendicular to the x-axis are disks with a diameter in $$R$$. Set up, but do not evaluate, a definite integral for the volume of this region.

Problem Statement

A second solid has base $$R$$ and cross-sections perpendicular to the x-axis are disks with a diameter in $$R$$. Set up, but do not evaluate, a definite integral for the volume of this region.

Solution

Disks mean that we have circles with diameter $$d$$, so the radius is $$r=d/2$$. The area of a disk is $$\pi r^2$$. The radius is determined by cross-sections perpendicular to the x-axis, which are dependent on $$x$$.
$$d=y-1=e^x-1 ~~ \to ~~ r=(e^x-1)/2$$
$$\displaystyle{V=\int_0^3{\pi \left( \frac{e^x-1}{2}\right)^2 dx}}$$

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Section 3

(16 points) A thin metal plate in the plane occupies the triangle with vertices $$(0,2)$$, $$(0,0)$$ and $$(2,0)$$. Find the centroid of the plate. You may use symmetry and known area or volume formulas from geometry to evaluate integrals.

Problem Statement

(16 points) A thin metal plate in the plane occupies the triangle with vertices $$(0,2)$$, $$(0,0)$$ and $$(2,0)$$. Find the centroid of the plate. You may use symmetry and known area or volume formulas from geometry to evaluate integrals.

$$\displaystyle{\left(\frac{2}{3},\frac{2}{3}\right)}$$

Problem Statement

(16 points) A thin metal plate in the plane occupies the triangle with vertices $$(0,2)$$, $$(0,0)$$ and $$(2,0)$$. Find the centroid of the plate. You may use symmetry and known area or volume formulas from geometry to evaluate integrals.

Solution

From symmetry we can see that $$\bar{x}=\bar{y}$$.
The equation for centroid is
$$\displaystyle{\bar{x}=\frac{\text{moment about x}}{\text{area}}=\frac{\int_a^b{x(y_2-y_1)~dx}}{\text{area}}}$$
where $$y_1=0$$ and $$y_2=2-x$$
The area of the triangle is $$A=bh/2=(2)(2)/2=2$$

 $$\displaystyle{\int_0^2{x(2-x-0)~dx}}$$ $$\displaystyle{\int_0^2{2x-x^2~dx}}$$ $$\displaystyle{\left[ \frac{2x^2}{2}-\frac{x^3}{3}\right]_0^2}$$ $$\displaystyle{4-\frac{8}{3}=\frac{4}{3}}$$

$$\displaystyle{\bar{x}=\frac{4/3}{2}=\frac{2}{3}}$$

$$\displaystyle{\left(\frac{2}{3},\frac{2}{3}\right)}$$

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(16 points) An upright water tank has a flat base in the plane and is filled with water. The tank has height $$7$$ feet, and, for $$0\le y\le 7$$, the cross-section parallel to the base at height $$y$$ has area $$y^2+2$$. How much work is done in pumping all the water over the top edge of the tank? [Use $$\delta~ lb/ft^3$$ for the weight-density of water.]

Problem Statement

(16 points) An upright water tank has a flat base in the plane and is filled with water. The tank has height $$7$$ feet, and, for $$0\le y\le 7$$, the cross-section parallel to the base at height $$y$$ has area $$y^2+2$$. How much work is done in pumping all the water over the top edge of the tank? [Use $$\delta~ lb/ft^3$$ for the weight-density of water.]

$$\displaystyle{ W = \frac{2989\delta}{12} }$$ ft-lbs

Problem Statement

(16 points) An upright water tank has a flat base in the plane and is filled with water. The tank has height $$7$$ feet, and, for $$0\le y\le 7$$, the cross-section parallel to the base at height $$y$$ has area $$y^2+2$$. How much work is done in pumping all the water over the top edge of the tank? [Use $$\delta~ lb/ft^3$$ for the weight-density of water.]

Solution

The picture to the right is meant to represent an upright tank filled with water. The $$dy$$ in the picture is a thin differential slice of water with cross sectional area is given in the problem statement as $$A=y^2+2$$.
To calculate the work, we know that the work, $$W$$ is force times distance, $$W=F\cdot d$$. To get force, we multiply the volume of one small slice of water times the weight-density.
The volume of one slice of water is $$dV=(y^2+2)dy ~ ft^3$$.
The force of one slice of water is $$dF=\delta\cdot dV=\delta(y^2+2)dy~lbs$$
Now, the distance $$d$$ is the distance one slice of the water is moved up over the top of the tank. The very top slice of water is moved a distance $$d=0~ft$$ while the slice at the bottom of the tank is moved $$d=7~ft$$. So our expression for the distance of any given slice of water is dependent on $$y$$ and is $$d=7-y$$. Now we are ready to set up our integral.
$$\displaystyle{\int_0^7{\delta (y^2+2)(7-y)dy}}$$
We integrate from the bottom of the tank ($$y=0)$$ to the top of the tank $$(y=7)$$ since the tank is filled to the top with water.

 $$\displaystyle{\int_0^7{\delta(y^2+2)(7-y)dy}}$$ $$\displaystyle{\int_0^7{7y^2-y^3+14-2y~dy}}$$ $$\displaystyle{\delta \left[ \frac{7y^3}{3}-\frac{y^4}{4}+14y-\frac{2y^2}{2}\right]_0^7}$$ $$\displaystyle{\delta \left[ \frac{7}{3}(7^3)-\frac{7^4}{4}+14(7)-7^2\right]-0}$$ $$\displaystyle{\frac{49(61)\delta}{12}}$$

$$\displaystyle{ W = \frac{2989\delta}{12} }$$ ft-lbs

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Section 4

Let G be the portion of the graph with parametric equations
$$x= 1+\sin(t)$$     $$y = 3+2\cos(t)$$    for $$0 \le t \le 2\pi$$.
Plot G and answer the following questions based on the graph.

(8 points) Set up, but do not evaluate, the integral for the length of G.

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the length of G.

$$\displaystyle{ s = \int_0^{2\pi}{ \sqrt{\cos^2t+4\sin^2t} ~dt } }$$

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the length of G.

Solution

The general equation is $$\displaystyle{s=\int_a^b{\sqrt{[x'(t)]^2+[y(t)]^2}~dt}}$$
In this question
$$x(t)=1+\sin(t) ~~ \to ~~ x'(t)=\cos(t)$$
$$y(t)=3+2\cos(t) ~~ \to ~~ y'(t)=-2\sin(t)$$
So our integral is $$\displaystyle{s=\int_0^{2\pi}{\sqrt{\cos^2t+4\sin^2t}~dt}}$$

$$\displaystyle{ s = \int_0^{2\pi}{ \sqrt{\cos^2t+4\sin^2t} ~dt } }$$

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(8 points) Set up, but do not evaluate, the integral for the area of the surface generated by revolving G about the x-axis.

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the area of the surface generated by revolving G about the x-axis.

$$\displaystyle{ A_x = 2\pi \int_0^{2\pi}{ (3+2\cos(t))\sqrt{\cos^2t+4\sin^2t} ~dt } }$$

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the area of the surface generated by revolving G about the x-axis.

Solution

 The general equation is $$\displaystyle{ A_x = 2\pi \int_a^b{ y(t)\sqrt{(dx/dt)^2+(dy/dt)^2} ~dt } }$$

From the previous question, we know $$dx/dt$$ and $$dy/dt$$. So we just need to write down the integral.
$$\displaystyle{A_x=2\pi \int_0^{2\pi}{(3+2\cos(t))\sqrt{\cos^2t+4\sin^2t }~dt}}$$

$$\displaystyle{ A_x = 2\pi \int_0^{2\pi}{ (3+2\cos(t))\sqrt{\cos^2t+4\sin^2t} ~dt } }$$

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 integration by parts trig integration integration using trig substitution partial fractions volume of revolution centroids work parametrics

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### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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