Integration methods.htm

INTEGRATION METHODS

Introduction - Antiderivatives

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.

Trig Powers

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

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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 $\cos \theta \;$ is considered a variable. Thus the substitution:
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If is odd, a similar substitution of yields Now, using the substitution:
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$\quad$

If both and $n\$ are even, then by using and we can change into a function of $\ 2\theta$ .

$\$ Remember that it may be necessary to treat each term of

as a separate integral.

Now to look at it is first necessary to note that since we have:
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If is a positive even integer then the substitution can be used to change into a function of specifically:
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The substitution:
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can then be used to solve the integral.

If $m\$ is odd, try changing into
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and use the substitution
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If $m\$ is even and odd then you must use integration by parts.

Now, integrate

by parts by substituting:

$\$

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.

Trig Substitution

When integrating
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we may find it necessary to allow or $\sec \theta$ and use the Pythagorean identities to simplify the function combination $gh\$ into a function of trig powers as talked about in the previous section.

First look at $1-x^{2}$ :
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which, in some cases, may be easier to handle.

Likewise, since , when the substitution
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is applied to $1+x^{2\text{ }}$ we get

Using

$x=\tan \theta$ as described above we see:

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Also, since the substitution
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will give

Using

$x=\sec \theta$ as described above we see:
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We can generalize these in the following manner, for $a^{2}-b^{2}x^{2}$ , we can substitute to get . A similar substitution can be used in the other cases.

To find the antiderivative of

$\quad$

Partial Fractions

If $P\left( x\right)$ and $Q\left( x\right)$ 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 $P\left( x\right)$ is less than the degree of $Q\left( x\right) ,$ 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 $x\$ 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)
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for example. (These are the more common examples.)

To understand partial fractions one must remember certain basic algebra concepts. They are:

There exists $k_{1},k_{2},$ and $k_{3}$ such that
When is a polynomial in of degree than $b,\;c,$ and are all polynomials in of degrees less that
If $\frac{a}{bcd}$ is a PARE then, when $b,\;c,$ and are polynomials, $k_{1},k_{2}$ and $k_{3}$ will all be polynomials of degrees less than the degrees of $b,\;c$ and , respectively. Thus, if $\frac{a}{bcd}$ is a PARE then and $\frac{k_{3}}{d}$ are also PARE's. Hence if (linear) then $k_{1}=m,$ a constant and if $b=kx^{2}+hx+l$ (quadratic) then $k_{1}=mx+n,$ a linear binomial, etc..
If then for every

To find

$k_{1}$ and

$k_{2},$ I prefer to use the method of equating the coefficients (see # 4 above). In this example after multiplying by the common denominator we have
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which leads to the system
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Therefore

$k_{1}=-1$ and

$k_{2}=3,$ so the integral becomes:
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Equating coefficients as above we can get:
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(Note that in

I used the substitution

$u=x^{2}+1$ to solve.)

Now to finish the opening example (Again equating coefficients to complete the partial fractions and disregarding the constant of integration for the moment):
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In the case of repeated linear factor(s) of the denominator, the partial breakdown will look like this:

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Notice that the partial breakdown includes a fraction with a constant numerator ( $k_{1}\ldots k_{n}$ are constants) for each power of the repeated factor from 1 to n. This is possible since the common denominator is giving:

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With repeated quadratic factors we get fractions with linear numerators:

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multiplying by the common denominator we have:

$\ \$

which leads to the system:
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therefore

$a=-3\$ ,

$b=3\$ and

which gives:
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Other Related Topics

Completing the Square

To integrate functions involving $\sqrt{ax^{2}+bx}$ or , $\$ you may find it advantageous to complete the square. Remembering your algebra you will get
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which can be solved using trig substitution. Likewise the same approach leads to
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which can be solved by using partial fractions since .

Completing the square is also useful on forms such as and where $f\left( x\right)$ is a non-factorable quadratic.

Different Angles

In order to find the antiderivative of
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try combinations ( $\pm )$ of the following trig identities to get an integral of terms of sine or cosine only.
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Definite integrals of Even and Odd functions

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:
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Since

is even and

is odd,

is odd. Therefore:
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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.
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