When you initially learned integration, you were always given functions that were well-behaved in the sense that they were continuous over the interval you were integrating. Also, definite integrals were always between two finite values. Integration will work only under those conditions, i.e. you can't integrate over an interval that contains even one point where the function is discontinuous and, for definite integrals, the limits of integration must be finite. These are the requirements of the Fundamental Theorem of Calculus (FTC).
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However, under certain conditions we can use limits to be able to integrate over discontinuous functions or where the limits of integration are not finite. These types of integrals are called improper integrals. The way to handle these kinds of integrals is to convert the integral to a finite integral on a continuous finite interval, integrate and then take the limit of the result. We can handle only one type of discontinuity at a time, so we may end up with multiple integrals but, as long as each of them are continuous and finite, that shouldn't be a problem.
Before we get into the details, let's watch a quick overview video that will help get us started.
video by Krista King Math |
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Before we get into the details of the techniques, here is a list of keys that you need to keep in mind as you learn these techniques. These keys will not make much sense at first but as you learn how to evaluate improper integrals, these keys will help keep you on track.
Improper Integral Keys
There are several keys to handling improper integrals.
1. We can handle only one 'problem' per integral.
2. The 'problem' must occur at one of the limits of integration or we need to break the integral to get it at one of the limits of integration.
3. Use a limit to set up the integral so that it is finite and continuous and then deal with the discontinuity after the integral is evaluated.
In summary, in all three cases, we use limits to rewrite the integral. We can do this because of the Limit Key. We can also have a combination of cases, in which case we rewrite the definite integral by breaking it into pieces so that we have only one case to handle for each integral.
The first thing we need to do is to determine the domain of the integrand and compare the domain with the limits of integration. Any points included in the limits of integration that are NOT in the domain are considered 'problem points'. Also, if one or either of the endpoints of is/are \(\pm\)infinity, they are also considered problems.
Let's start with the endpoints. The simplest case is where we have a continuous interval and one of the endpoints is infinite. This is discussed on the next page.
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external links you may find helpful |
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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