## 17Calculus Infinite Series - Integral Test

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The Integral Test is easy to use and is good to use when the ratio test and the comparison tests won't work and you are pretty sure that you can evaluate the integral. The idea of this test is to evaluate the improper integral $$\displaystyle{ \int_{k}^{\infty}{f(x)~dx} }$$.

If you want a complete lecture on this test, we recommend this video.

### Prof Leonard - Calculus 2 Lecture 9.3: Using the Integral Test for Convergence/Divergence of Series, P-Series

video by Prof Leonard

Integral Test

For a series $$\displaystyle{\sum_{n=1}^{\infty}{a_n}}$$ where we can find a positive, continuous and decreasing function $$f$$ for $$n > k$$ and $$a_n = f(n)$$, then we know that if $\int_{k}^{\infty}{f(x) ~ dx}$ converges, the series also converges. Similarly when the integral diverges, the series also diverges.

Two Things To Watch For

1. The value of k
First, you need to find a constant k such that the function satisfies all of these conditions for all $$n > k$$:

 continuous positive decreasing

One of the favorite tricks that teachers like to put on exams (which I fell for when I first took the class) is to tell you to use the Integral Test but then not give you k. Many books just show this integral with $$k=1$$, which is not always valid. So be careful.
How To Find k:
The best way is to calculate the critical values of the function and then check that the derivative is negative to the right of the largest critical value. Then, if you have access to a graphing calculator, do a quick plot to check your answer. If everything looks good, choose k to be greater than the largest critical value. Any value will do, so choose one that will be easy to use in the integration.
There is no one value that will always work. It depends on the function.

2. The final value of the integration
Secondly, if you get a finite value for the integral and determine that the series converges, the finite value you got from the integral is NOT what the series converges to. The number itself has no meaning in this context (ie. we don't use the value of the number to tell us anything about the series). The significance of it lies in whether it is finite or not. That's it. That's all the information you can get from that number. So do NOT assume that the series converges to that number.

Okay, let's watch some videos to see how this test works.

In this first video clip, he does a great job explaining the integral test. He uses the integral test to show the divergence of the p-series $$\sum{ 1/n }$$.

### Dr Chris Tisdell - Intro to series + the integral test [11min-23secs]

video by Dr Chris Tisdell

In this next video, the instructor explains the integral test in more detail by using it on the two series $$\sum{ 1/n }$$ and $$\sum{ 1/n^2 }$$ to show that one diverges and the other converges.

### Dr Chris Tisdell - Integral test for Series [13min-43secs]

video by Dr Chris Tisdell

Here is another good explanation of the integral test. He looks at the sum $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n^p}} }$$.

### PatrickJMT - Integral Test - Basic Idea [3min-26secs]

video by PatrickJMT

Here is a great video giving an intuitive understanding on why this works.

### PatrickJMT - Integral Test for Series: Why It Works [14min-46secs]

video by PatrickJMT

This last video discusses the remainder estimate for the integral test. Although not required to understand how to use the integral test, this video will help you understand more intuitively what is going on.

### PatrickJMT - Remainder Estimate for the Integral Test [7min-45secs]

video by PatrickJMT

Integral Test Proof

Here are 4 consecutive videos showing a proof of the Integral Test. You do not need to watch these in order to understand and use the Integral Test but we include them here for those of you who are interested.

video by Chau Tu

video by Chau Tu

video by Chau Tu

### Chau Tu - Integral Test Proof - Part 4 of 4

video by Chau Tu

Practice

Unless otherwise instructed, determine the convergence or divergence of the following series using the Integral Test, if possible.

Basic

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ using the integral test, if possible.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ converges by the p-series test or the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ using the integral test, if possible.

Solution

This is a p-series with $$p=2 > 1$$, so the series converges by the p-series test.
We could also have used the Integral Test, as follows.

 $$\displaystyle{ \int_{1}^{\infty}{\frac{1}{x^2}dx} }$$ $$\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{\frac{1}{x^2}dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{x^{-2}dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \left[ -x^{-1} \right]_{1}^{b}} }$$ $$\displaystyle{ \lim_{b \to \infty}{-b^{-1} + 1^{-1}} }$$ $$0 + 1 = 1$$

Since the improper integral is finite, the series converges by the Integral Test.
Note: The value $$1$$, from the integral is NOT necessarily what the series converges to. The significance of this number is only that it is finite.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ converges by the p-series test or the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^3} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^3} } }$$ using the integral test, if possible.

Solution

### 3232 video

video by blackpenredpen

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{(n+2)^2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{(n+2)^2} } }$$ using the integral test, if possible.

Solution

### 3226 video

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$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{n}{\ln(n)} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{n}{\ln(n)} } }$$ using the integral test, if possible.

Solution

### 3230 video

video by UConnQCenter

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$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }$$ using the integral test, if possible.

Solution

This problem is solved by two different instructors.

video by Chau Tu

### 240 video

video by Michel vanBiezen

The series converges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }$$ using the integral test, if possible.

Solution

This problem is solved by two different instructors.

### 241 video

video by turksvids

### 241 video

video by Dr Chris Tisdell

The series converges by the integral test.

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$$\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-k} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-k} } }$$ using the integral test, if possible.

Solution

### 3239 video

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$$\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }$$ using the integral test, if possible.

Solution

The first video solves the given problem. The second video solves a very similar problem, the same as given in the problem statement but multiplied by a constant. Since multiplication by a constant does not change the result, the solution to the second one is very similar.

### 242 video

video by PatrickJMT

### 242 video

video by Trefor Bazett

The series converges by the integral test.

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$$\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-2k^2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-2k^2} } }$$ using the integral test, if possible.

Solution

### 3238 video

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$$\displaystyle{ \sum_{n=1}^{\infty}{ n^2 e^{-n^3} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ n^2 e^{-n^3} } }$$ using the integral test, if possible.

Solution

### 3233 video

video by blackpenredpen

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }$$ using the integral test, if possible.

The series diverges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }$$ using the integral test, if possible.

Solution

### 243 video

video by PatrickJMT

The series diverges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }$$ using the integral test, if possible.

Solution

### 248 video

video by Krista King Math

The series converges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }$$ using the integral test, if possible.

The series diverges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }$$ using the integral test, if possible.

Solution

This problem is solved by two different instructors.

### 249 video

video by blackpenredpen

### 249 video

video by Educator.com

The series diverges by the integral test.

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$$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n-2}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n-2}} } }$$ using the integral test, if possible.

Solution

### 3229 video

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }$$ using the integral test, if possible.

The series diverges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }$$ using the integral test, if possible.

Solution

This problem is solved by two different instructors.

### 250 video

video by MIP4U

The series diverges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }$$ using the integral test, if possible.

Solution

### 251 video

video by MIP4U

The series converges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{5n-2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{5n-2} } }$$ using the integral test, if possible.

Solution

### 1213 video

video by MIP4U

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$$\displaystyle{ 1 + 1/3 + 1/5 + 1/7 + 1/9 + . . . }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ 1 + 1/3 + 1/5 + 1/7 + 1/9 + . . . }$$ using the integral test, if possible.

Solution

### 3236 video

video by blackpenredpen

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n}{3n^2+4} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n}{3n^2+4} } }$$ using the integral test, if possible.

Solution

### 3227 video

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(\sqrt{n})}{n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(\sqrt{n})}{n} } }$$ using the integral test, if possible.

Solution

### 3231 video

video by William Lindsey

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$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n\sqrt{\ln n}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n\sqrt{\ln n}} } }$$ using the integral test, if possible.

Solution

### 3235 video

video by blackpenredpen

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Intermediate

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}$$ using the integral test, if possible.

Solution

Although this problem would be more easily solved using one of the comparison tests, this is a great example using the integral test.

### 244 video

video by PatrickJMT

The series converges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2+6n+10} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2+6n+10} } }$$ using the integral test, if possible.

Solution

### 3228 video

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }$$ using the integral test, if possible.

Solution

### 245 video

video by PatrickJMT

The series converges by the integral test.

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$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }$$ using the integral test, if possible.

Solution

### 246 video

video by PatrickJMT

The series converges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^4+1} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^4+1} } }$$ using the integral test, if possible.

Solution

### 3234 video

video by blackpenredpen

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt{n}}+\frac{5}{n}\right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt{n}}+\frac{5}{n}\right] } }$$ using the integral test, if possible.

The series diverges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt{n}}+\frac{5}{n}\right] } }$$ using the integral test, if possible.

Solution

### 247 video

video by PatrickJMT

The series diverges by the integral test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }$$ using the integral test, if possible.

The series converges by the integral test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }$$ using the integral test, if possible.

Solution

This problem is solved by two different instructors.

### 252 video

video by turksvids

### 252 video

video by MIP4U

The series converges by the integral test.

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$$\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }$$ using the integral test, if possible.

The series diverges by the Integral Test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }$$ using the integral test, if possible.

Solution

### 2443 video

video by Math Cabin

The series diverges by the Integral Test.

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Find $$p$$ so that $$\displaystyle{ \sum_{n=1}^{\infty}{ n(1+n^2)^p } }$$ will converge.

Problem Statement

Find $$p$$ so that $$\displaystyle{ \sum_{n=1}^{\infty}{ n(1+n^2)^p } }$$ will converge.

Solution

### 3237 video

video by blackpenredpen

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### integral test 17calculus youtube playlist

You CAN Ace Calculus

 limits at infinity integration improper integrals infinite series basics

WikiBooks - Integral Test

### Integral Test Quick Notes

 used to prove convergence yes used to prove divergence yes can be inconclusive yes $$a_n$$ must be positive $$a_n$$ must be decreasing requires that the integrand must be integrable (not always possible) requires the evaluation of infinite limits (after integration) if the result of the limit (after integration) does not exist (different than diverges), this test is inconclusive

### 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

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

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