## 17Calculus - Calculus 1 - Practice Exam 3 (Semester B)

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This is the final exam for first semester single variable calculus. The topics covered are integration by substitution, both indefinite and definite, solution of separable differential equations, optimization, area between curves, exponential growth/decay and volumes of revolution using both the disc/washer method and the shell/cylinder method.

### 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

2.5 hours

Questions

10

Total Points

135

Tools

Calculator

not allowed

Formula Sheet(s)

one page

Other Tools

ruler

Instructions:
- 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).

[10 points] Evaluate $$\displaystyle{ \int{ \frac{e^{1/x}}{x^2} ~dx } }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int{ \frac{e^{1/x}}{x^2} ~dx } }$$

$$\displaystyle{ -e^{1/x} + C }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int{ \frac{e^{1/x}}{x^2} ~dx } }$$

Solution

Use integration by substitution.
Let $$u=1/x = x^{-1} ~~ \to ~~ du=-x^{-2}dx$$
Move the negative sign to the left in the last equation to get $$-du=x^{-2}dx$$.
Now the integral is
$$\displaystyle{-\int{e^u~du}=}$$ $$\displaystyle{-e^u+C =}$$ $$\displaystyle{-e^{1/x}+C}$$

$$\displaystyle{ -e^{1/x} + C }$$

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[10 points] Evaluate $$\displaystyle{ \int_{1}^{e^3}{ \frac{\ln\sqrt{x}}{x} ~dx } }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int_{1}^{e^3}{ \frac{\ln\sqrt{x}}{x} ~dx } }$$

$$\displaystyle{ 9/4 }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int_{1}^{e^3}{ \frac{\ln\sqrt{x}}{x} ~dx } }$$

Solution

 Before integrating, we simplify the integrand a bit, i.e. $$\ln\sqrt{x} = \ln(x^{1/2}) = (1/2)\ln x$$. $$\displaystyle{\int_{1}^{e^3}{ \frac{\ln\sqrt{x}}{x}~dx } =}$$ $$\displaystyle{\frac{1}{2}\int_{1}^{e^3}{ \frac{\ln x}{x}~dx}}$$ Use integration by substitution. Let $$u=\ln x \to$$ $$du=dx/x$$. We need to change the limits of integration. $$x=e^3 \to u=3$$ and $$x=1 \to u=0$$ $$\displaystyle{\frac{1}{2}\int_{1}^{e^3}{\frac{\ln x}{x}~dx}=}$$ $$\displaystyle{\frac{1}{2}\int_{0}^{3}{u~du}=}$$ $$\displaystyle{\left. \frac{1}{2}\frac{u^2}{2}\right|_0^3 =}$$ $$\displaystyle{\frac{1}{4}[3^2 - 0^2]=}$$ $$\displaystyle{\frac{9}{4}}$$

Another way
If you don't notice the initial simplification, this integral can still be solved with a different (more difficult) substitution, i.e. $$u=\ln x^{1/2}$$ as follows.
Let $$u=\ln x^{1/2} \to$$ $$du=(1/x^{1/2})(1/2)x^{-1/2}dx =1/(2x)dx \to$$ $$2du=dx/x$$
Change the limits of integration.
$$x=e^3 \to u=3/2$$ and $$x=1 \to u=0$$
So the integral in terms of $$u$$ is $$\displaystyle{\int_0^{3/2}{2u~du}=}$$ $$\displaystyle{\left. \frac{2u^2}{2}\right|_0^{3/2}=}$$ $$\displaystyle{\left(\frac{3}{2}\right)^2 - 0^2 = \frac{9}{4}}$$

$$\displaystyle{ 9/4 }$$

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[15 points] Evaluate $$\displaystyle{ \int{ x\sqrt{2x+3} ~dx } }$$

Problem Statement

[15 points] Evaluate $$\displaystyle{ \int{ x\sqrt{2x+3} ~dx } }$$

$$\displaystyle{ \frac{1}{5}(2x+3)^{3/2}(x-1) + C }$$

Problem Statement

[15 points] Evaluate $$\displaystyle{ \int{ x\sqrt{2x+3} ~dx } }$$

Solution

 Use integration by substitution. So let $$u=2x+3\to$$ $$du=2dx$$ When we substitute, the integrand becomes $$x\sqrt{u}$$. We need another factor to get rid of the $$x$$. To do so, we solve our choice of $$u$$ for $$x$$ to get $$x=(u-3)/2$$. $$\displaystyle{\int{x\sqrt{2x+3}~dx}=}$$ $$\displaystyle{\int{\left(\frac{u-3}{2}\right)u^{1/2}\frac{du}{2}}}$$ Now we combine the $$1/2$$'s and pull the constant outside the integral. $$\displaystyle{\frac{1}{4}\int{(u-3)u^{1/2}~du}}$$ Since there is no product rule in integration, we need to multiply the $$u^{1/2}$$ term through the $$(u-3)$$ term. $$\displaystyle{\frac{1}{4}\int{u^{3/2}-3u^{1/2}~du}}$$ Now integrate. $$\displaystyle{\frac{1}{4}\left[\frac{u^{5/2}}{5/2}-\frac{3u^{3/2}}{3/2}\right]+C}$$ Finally, we replace $$u$$ to get an answer in terms of $$x$$. $$\displaystyle{\frac{1}{2}\left[\frac{1}{5}(2x+3)^{5/2}-(2x+3)^{3/2}\right]+C}$$ Depending on how much simplification your instructor requires, this can be reduced to $$\displaystyle{\frac{1}{5}(2x+3)^{3/2}(x-1)+C}$$

There is another substitution that works here.
If we let $$u=\sqrt{2x+3}$$ then $$\displaystyle{ du = \frac{1}{2}(2x+3)^{-1/2}(2) dx }$$. This gives us the integral in terms of $$u$$ as $$(1/2)\int{u^4-3u^2~du}$$ and after integrating we get the same answer.

$$\displaystyle{ \frac{1}{5}(2x+3)^{3/2}(x-1) + C }$$

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[10 points] Evaluate $$\displaystyle{ \int_2^5{ x^2 \sqrt{x^3-4} ~dx } }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int_2^5{ x^2 \sqrt{x^3-4} ~dx } }$$

$$294$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int_2^5{ x^2 \sqrt{x^3-4} ~dx } }$$

Solution

 Use integration by substitution. So, let $$u=x^3-4 \to du=3x^2dx$$ We also need to change our limits of integration. $$x=5 \to u=121$$ and $$x=2 \to u=4$$ $$\displaystyle{\int_2^5{ x^2\sqrt{x^3-4}~dx}=}$$ $$\displaystyle{\int_4^{121}{ u^{1/2}\frac{du}{3}}}$$ Now we integrate. $$\displaystyle{\left. \frac{1}{3}\frac{u^{3/2}}{3/2}\right|_4^{121}}$$ Now substitute the limits of integration. $$\displaystyle{\frac{2}{9}\left[(121)^{3/2}-4^{3/2}\right]}$$ To simplify, we recognize that $$(121)^{3/2}=121(121)^{1/2}$$ and $$121=11^2$$ to get $$(121)^{3/2}=11^3$$. Similarly, $$4^{3/2}=4(4)^{1/2}$$ which gives us $$4^{3/2}=2^3$$. $$\displaystyle{\frac{2}{9}[11^3-8]}$$ Without a calculator, this answer is probably sufficient. However, if your instructor requires a more simplified answer, we get $$\displaystyle{\frac{2}{9}(1331-8)=}$$ $$\displaystyle{\frac{2}{9}(1323)=}$$ $$2(147)=294$$

$$294$$

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[10 points] $$\int{ \text{_________} ~dx } = (ax^2-a^2)^4 + C$$
If a is a constant, fill in the blank.

Problem Statement

[10 points] $$\int{ \text{_________} ~dx } = (ax^2-a^2)^4 + C$$
If a is a constant, fill in the blank.

$$8ax(ax^2-a^2)^3$$

Problem Statement

[10 points] $$\int{ \text{_________} ~dx } = (ax^2-a^2)^4 + C$$
If a is a constant, fill in the blank.

Solution

The key to this problem is to remember that integration is the opposite of differentiation. So all you need to do is to take the derivative of $$(ax^2-a^2)^4+C$$ to get the integrand of the integral. As stated in the problem, $$a$$ is just a constant.
$$\displaystyle{ \frac{d}{dx}[(ax^2-a^2)^4+C]= }$$ $$\displaystyle{ 4(ax^2-a^2)^3\frac{d}{dx}[ax^2-a^2]= }$$ $$4(ax^2-a^2)^3(2ax)=8ax(ax^2-a^2)^3$$

$$8ax(ax^2-a^2)^3$$

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[20 points] Find the particular solution of the differential equation $$f''(x) = e^{3x} + \cos(x)$$ that satisfies the initial conditions $$f(0)=10/9$$ and $$f'(0)=4/3$$.

Problem Statement

[20 points] Find the particular solution of the differential equation $$f''(x) = e^{3x} + \cos(x)$$ that satisfies the initial conditions $$f(0)=10/9$$ and $$f'(0)=4/3$$.

$$\displaystyle{ f(x) = \frac{e^{3x}}{9} - \cos x + x + 2 }$$

Problem Statement

[20 points] Find the particular solution of the differential equation $$f''(x) = e^{3x} + \cos(x)$$ that satisfies the initial conditions $$f(0)=10/9$$ and $$f'(0)=4/3$$.

Solution

Usually, with separable differential equations, we have to do some work to separate the variables. However, in this problem we already have all the x's on the right and the function on the left. Also, it helps to think back about position, velocity and acceleration to realize that you need to integrate twice and, for each integration, determine the values of the constants.

 Starting with $$f''(x)=e^{3x}+\cos(x)$$ we integrate to get $$f'(x)=\int{e^{3x}+\cos x~dx}=$$ $$\displaystyle{\frac{e^{3x}}{3}+\sin x+C}$$ Now it is time to use one of the initial conditions, $$f'(0)=4/3$$ to find $$C$$. When $$f'(0)=4/3$$, we know that $$x=0$$. $$\displaystyle{f'(0)=\frac{1}{3}+0+C=\frac{4}{3}\to C=1}$$ So, now we have $$\displaystyle{f'(x)=\frac{e^{3x}}{3}+\sin x+1}$$ To find the general solution $$f(x)$$, we integrate again. $$\displaystyle{f(x)=\int{\frac{e^{3x}}{3}+\sin x+1~dx}=}$$ $$\displaystyle{\frac{e^{3x}}{9}-\cos x+x+C}$$ Now we use the other initial condition to find $$C$$. $$\displaystyle{\frac{1}{9}-\cos 0+0+C=\frac{10}{9}\to C=2}$$

$$\displaystyle{ f(x) = \frac{e^{3x}}{9} - \cos x + x + 2 }$$

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[15 points] For the region bounded by the graphs $$y=9-x^2$$, $$x=2$$ and $$x=3$$ in the first quadrant, a. accurately sketch the region; b. using calculus, calculate the area of the region that you sketched in part a.

Problem Statement

[15 points] For the region bounded by the graphs $$y=9-x^2$$, $$x=2$$ and $$x=3$$ in the first quadrant, a. accurately sketch the region; b. using calculus, calculate the area of the region that you sketched in part a.

$$8/3$$

Problem Statement

[15 points] For the region bounded by the graphs $$y=9-x^2$$, $$x=2$$ and $$x=3$$ in the first quadrant, a. accurately sketch the region; b. using calculus, calculate the area of the region that you sketched in part a.

Solution

a. Here is a plot. We need to find the area of the shaded region.

$$\begin{array}{rcl} A & = & \displaystyle{\int_2^3{9-x^2~dx}} \\ & = & \displaystyle{\left[9x-\frac{x^3}{3}\right]_2^3} \\ & = & \displaystyle{\left[27-\frac{27}{3}\right]-\left[18-\frac{8}{3}\right]} \\ & = & \displaystyle{18-18+\frac{8}{3}=\frac{8}{3}} \end{array}$$

$$8/3$$

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[10 points] The mass of radioactive material in a sample has decreased by 30% since the decay began. Assuming a half-life of 1500 years, how long ago did the decay begin?

Problem Statement

[10 points] The mass of radioactive material in a sample has decreased by 30% since the decay began. Assuming a half-life of 1500 years, how long ago did the decay begin?

$$\displaystyle{ t = \frac{-1500\ln(0.70)}{\ln 2} }$$

Problem Statement

[10 points] The mass of radioactive material in a sample has decreased by 30% since the decay began. Assuming a half-life of 1500 years, how long ago did the decay begin?

Solution

 This is an exponential decay problem. So our general equation is $$y(t)=y_0 e^{kt}$$ where $$k<0$$. We can also think about this as $$y(t)=y_0 e^{-kt}$$ where $$k>0$$. We will use the first equation. First, we are going to use the information about the half-life at $$t=1500$$ to get a value for $$k$$. If the initial amount is $$y_0$$, then half of the initial amount is $$y_0/2$$. $$\begin{array}{rcl} \displaystyle{\frac{y_0}{2}} & = & y_0 e^{k(1500)} \\ \displaystyle{\frac{1}{2}} & = & e^{k(1500)} \\ -\ln 2 & = & k(1500) \\ k & = & \displaystyle{\frac{-\ln 2}{1500}} \end{array}$$ Now, we know that after $$t$$ years, the initial amount has decreased 30%. This tells us that $$70%$$ remains. So know we need to find $$t$$ when we have 70% of the initial amount. $$\displaystyle{0.70y_0=y_0 exp\left[\frac{-t\ln 2}{1500}\right]}$$ In this last equation, we have used the alternate notation for the exponential function, i.e. $$e^t=exp[t]$$. This makes it easier to see the exponent of $$e$$. Next, we take the natural log of both sides and solve for $$t$$. $$\displaystyle{\ln(0.70)=\frac{-t\ln 2}{1500}\to}$$ $$\displaystyle{t=\frac{-1500\ln(0.70)}{\ln 2}}$$

$$\displaystyle{ t = \frac{-1500\ln(0.70)}{\ln 2} }$$

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[10 points]

A zookeeper needs to add a rectangular outdoor pen to an animal house with a corner notch, as shown in the figure. If 85m of new fence is available, what dimensions of the pen will maximize its area? No fence will be used along the walls of the animal house.

Problem Statement

[10 points]

A zookeeper needs to add a rectangular outdoor pen to an animal house with a corner notch, as shown in the figure. If 85m of new fence is available, what dimensions of the pen will maximize its area? No fence will be used along the walls of the animal house.

25m X 25m

Problem Statement

[10 points]

A zookeeper needs to add a rectangular outdoor pen to an animal house with a corner notch, as shown in the figure. If 85m of new fence is available, what dimensions of the pen will maximize its area? No fence will be used along the walls of the animal house.

Solution

For this problem, we need to maximize the area of a rectangle, so we need to assign some variables in order to come up with an area equation. We will call the bottom length $$x$$ and the right side length $$y$$. Therefore our area equation is $$A=xy$$.
Now we need to use the information that we have $$85$$m of fence to come up with an equation for the perimeter of the rectangle including only the lengths where we will be using fence. So, the right and the bottom side area pretty obvious, they are $$y$$ and $$x$$. However, the top side that requires fence is $$x-10$$ and the left side is $$y-5$$. So the length of fence we need is $$x+(x-10)+y+(y-5)$$ and this is equal to how much fence we have, $$85$$m. So we have $$x+(x-10)+y+(y-5)=85$$. After simplification we have $$x+y=50$$. We will solve this for $$y$$ to be substituted back into the area equation before taking the derivative.
$$y=50-x$$ and since $$A=xy$$, then $$A=x(50-x)$$.
Now we can take the derivative of the area equation with respect to $$x$$ since we have only one variable there now.
$$\displaystyle{\frac{dA}{dx}=50-2x}$$
Setting this last equation to zero and solving for $$x$$ gives us $$x=25$$.

25m X 25m

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[25 points] The region bounded by the graphs of $$y=2x$$, $$y=6-x$$ and $$y=0$$ is revolved about the line $$y=-2$$ to produce a volume. Calculate the volume that is produced. Set up the integrals for both the disc/washer and shell/cylinder methods and evaluate one of them.
Notes: Make sure you plot and shade the region and show the example rectangles, as well as the distances R, r, p and h. Each volume should have its own plot, i.e. you should have one plot for the disc/washer method and another separate plot for the shell/cylinder method.
[10 points extra credit] Evaluate the other integral.

Problem Statement

[25 points] The region bounded by the graphs of $$y=2x$$, $$y=6-x$$ and $$y=0$$ is revolved about the line $$y=-2$$ to produce a volume. Calculate the volume that is produced. Set up the integrals for both the disc/washer and shell/cylinder methods and evaluate one of them.
Notes: Make sure you plot and shade the region and show the example rectangles, as well as the distances R, r, p and h. Each volume should have its own plot, i.e. you should have one plot for the disc/washer method and another separate plot for the shell/cylinder method.
[10 points extra credit] Evaluate the other integral.

$$80 \pi$$

Problem Statement

[25 points] The region bounded by the graphs of $$y=2x$$, $$y=6-x$$ and $$y=0$$ is revolved about the line $$y=-2$$ to produce a volume. Calculate the volume that is produced. Set up the integrals for both the disc/washer and shell/cylinder methods and evaluate one of them.
Notes: Make sure you plot and shade the region and show the example rectangles, as well as the distances R, r, p and h. Each volume should have its own plot, i.e. you should have one plot for the disc/washer method and another separate plot for the shell/cylinder method.
[10 points extra credit] Evaluate the other integral.

Solution

disc/washer method

basic integral $$\displaystyle{ V=\pi\int_a^b{ R^2-r^2~dx} }$$

We need two integrals with a break at $$x=2$$.

1. $$[0,2]$$: $$R=2x+2, ~~ r=2$$

$$V_1=\displaystyle{\pi\int_0^2{(2x+2)^2-2^2~dx}}$$

$$V_1=\displaystyle{\pi\int_0^2{4x^2+8x~dx}}$$

$$V_1=\displaystyle{\pi\left[\frac{4x^3}{3}+\frac{8x^2}{2}\right]_0^2}$$

$$V_1=80\pi/3$$

2. $$[2,6]$$: $$R=6-x+2, ~~ r=2$$

$$V_2=\displaystyle{\pi\int_2^6{(8-x)^2-2^2~dx}}$$

$$V_2=\displaystyle{\pi\int_2^6{60-16x+x^2~dx}}$$

$$V_2=\displaystyle{\pi\left[60x-\frac{16x^2}{2}+ \frac{x^3}{3}\right]_2^6}$$

$$V_2=160\pi/3$$

$$V=V_1+V_2=80\pi/3+160\pi/3=80\pi$$

shell/cylinder method

basic integral $$\displaystyle{V=2\pi\int_c^d{ph~dy}}$$

$$p=y+2$$ and $$h=(6-y)-(y/2)$$

$$V=\displaystyle{2\pi\int_0^4{(y+2)(6-3y/2)~dy}}$$

$$V=\displaystyle{2\pi\int_0^4{3y+(3/2)y^2+12~dy}}$$

$$V=\displaystyle{2\pi\left[\frac{3y^2}{2}-\frac{3}{2}\frac{y^3}{3}+12y\right]_0^4}$$

$$V=80\pi$$

$$80 \pi$$

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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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