1. Introduction - Antiderivatives
2. Trig Powers
3. Trig Substitution
4. Partial Fractions
5. Other Related Topics
When you wish to integrate functions that are not in the basic list and a simple substitution will not work there are some algebraic methods that might work. This is intended to get you started and to point you in the direction of a solution method. If one of these methods does not work in a specific case, they will hopefully help you find something that will. To paraphrase Dennis Zill: Try something, you could get lucky.
At times it becomes desirable to calculate the antiderivative of such functions as and This section looks at problems of this type, especially those where and are non-negative integers.
We will first look at: To integrate this type of function you will normally use the trig identities
If either or is an odd positive integer (say m), write noting that since is an odd integer is an even integer. Thus Now use identity (a)
to make the substitution This should lead to the following: where is a polynomial type function
if is considered a
variable. Thus the substitution:
If is odd, a similar substitution of
yields Now, using the substitution:
If both and are even, then by using and we can change into a function of .
Now to look at it is first necessary to note
that since we have:
If is a positive even integer then
the substitution can be used to change into a function of specifically:
The substitution:
can then be used to solve the integral.
If is odd, try changing into
and use the substitution
If is even and odd then you must use integration by parts.
Cotangent and cosecant are approached in a similar manor to tangent and secant. In some cases negative and fractional powers can be handled very much like the above and in others observing the above methods and a little imagination can lead to desired results. Trig identities allow for some very creative work just remember to not mix functions of different angles in the same term. If they occur in different combinations than covered here (sin mixed with tan, for example), it is often useful to convert everything into sin and cos.
When integrating
we may find it necessary to allow or and use the Pythagorean
identities to simplify the function combination into a function of trig powers
as talked about in the previous section.
First look at :
which, in some cases, may be easier to handle.
Likewise, since , when the substitution
is applied to we get
Also, since the substitution
will give
We can generalize these in the following manner, for , we can substitute to get . A similar substitution can be used in the other cases.
To find the antiderivative ofIf and are polynomials in then is called a rational algebraic expression (RAE), i.e. an algebraic fraction. To find first be sure that the degree of is less than the degree of this makes a ''proper'' rational algebraic expression (PRAE). If this condition does not exist then it becomes necessary to use division to achieve this condition.
To finish this problem one must integrate a PRAE.
Theoretically, at least, every polynomial in can be factored into a set of
linear and quadratic factors with real coefficients. Realistically, it
is not always reasonable to do this; however, when you can factor the
denominator into this type of set then the method of breaking the RAE
into its partial fractions gives an algebraic breakdown into simpler
fractions that may be integrated by other methods such as: (ignoring
the constants of integration)
for example. (These are the more common examples.)
To understand partial fractions one must remember certain basic algebra concepts. They are:
There exists and such that
When is a polynomial in of degree than and are all polynomials in of degrees less that
If is a PARE then, when and are polynomials, and will all be polynomials of degrees less than the degrees of and , respectively. Thus, if is a PARE then and are also PARE's. Hence if (linear) then a constant and if (quadratic) then a linear binomial, etc..
If then for every
Now to finish the opening example (Again equating coefficients to
complete the partial fractions and disregarding the constant of
integration for the moment):
In the case of repeated linear factor(s) of the denominator, the partial breakdown will look like this:
Notice that the partial breakdown includes a fraction with a constant numerator ( are constants) for each power of the repeated factor from 1 to n. This is possible since the common denominator is giving:
With repeated quadratic factors we get fractions with linear numerators:
To integrate functions involving or , you may find it advantageous to
complete the square. Remembering your algebra you will get
which can be solved using trig substitution. Likewise the same
approach leads to
which can be solved by using partial fractions since .
Completing the square is also useful on forms such as and where is a non-factorable quadratic.
In order to find the antiderivative of
try combinations ( of the following trig identities
to get an integral of terms of sine or cosine only.
Now a word about the definite integral. Recall that it is the algebraic sum of the areas between the curve, the dependant axis, and the end lines. Thus it can be found using known area formulas or estimated using the Riemann sum. We also know that if the function has a known antiderivative the exact answer can be found by using the Fundamental Theorem of Calculus. Sometimes, the calculations by these methods may be difficult and we latch onto any tool that might shorten and/or simplify the work. One of the nice shortcuts occurs when the function to be integrated is either even or odd.
Remember that a function is even when and is odd when . It should be noted that not all functions fall in one of these two categories. A look at the graph of these functions leads to a very easily verified observation:
If is even then and
If is odd then
Since we know that is an even function. Therefore: I will close this discussion of methods of integration with one
last example. This example requires a little imagination and creative
use of several methods. (i.e. it is fun!) Integrals similar to this
appear often in Physics.
Letting
and we get:
Equating coefficients (also known as using undetermined coefficients) we have:
Thus, after completing the square on the denominator of the second
fraction and adjusting the numerator we get: