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.
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 pseries \( \sum{ 1/n } \).
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.
video by Dr Chris Tisdell 

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

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

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.
Final Answer 

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the pseries 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 pseries with \( p=2 > 1\), so the series converges by the pseries 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.
Final Answer 

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the pseries 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 

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 

video by The Organic Chemistry Tutor 

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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 

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.
Final Answer 

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 

video by Michel vanBiezen 

Final Answer 

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.
Final Answer 

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.
video by turksvids 

video by Dr Chris Tisdell 

Final Answer 

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 

video by The Calculus Professor 

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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.
Final Answer 

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.
video by PatrickJMT 

video by Trefor Bazett 

Final Answer 

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 

video by The Calculus Professor 

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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 

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.
Final Answer 

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 

video by PatrickJMT 

Final Answer 

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.
Final Answer 

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 

video by Krista King Math 

Final Answer 

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.
Final Answer 

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.
video by blackpenredpen 

video by Educator.com 

Final Answer 

The series diverges by the integral test. 
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\(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n2}} } }\)
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n2}} } }\) using the integral test, if possible.
Solution 

video by The Organic Chemistry Tutor 

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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.
Final Answer 

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.
video by The Organic Chemistry Tutor 

video by MIP4U 

Final Answer 

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.
Final Answer 

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 

video by MIP4U 

Final Answer 

The series converges by the integral test. 
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{5n2} } }\)
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{5n2} } }\) using the integral test, if possible.
Solution 

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 

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 

video by The Organic Chemistry Tutor 

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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 

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 

video by blackpenredpen 

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Intermediate 

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^24n+5}}}\)
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^24n+5}}}\) using the integral test, if possible.
Final Answer 

The series converges by the integral test.
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^24n+5}}}\) 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.
video by PatrickJMT 

Final Answer 

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 

video by The Organic Chemistry Tutor 

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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.
Final Answer 

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 

video by PatrickJMT 

Final Answer 

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.
Final Answer 

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 

video by PatrickJMT 

Final Answer 

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 

video by blackpenredpen 

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\)
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\) using the integral test, if possible.
Final Answer 

The series diverges by the integral test.
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\) using the integral test, if possible.
Solution 

video by PatrickJMT 

Final Answer 

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.
Final Answer 

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.
video by turksvids 

video by MIP4U 

Final Answer 

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.
Final Answer 

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 

video by Math Cabin 

Final Answer 

The series diverges by the Integral Test. 
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Advanced 

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 

video by blackpenredpen 

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You CAN Ace Calculus
external links you may find helpful 

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 
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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Practice Instructions
Unless otherwise instructed, determine the convergence or divergence of the following series using the Integral Test, if possible.