# Integration Techniques

Since only textbooks group integrals according to the method necessary for solving them, it is essential that students learn to recognize the different types quickly and accurately.  While it is impossible to cover all possibilities, the intent here is to try to cover some of the more basic types of methods of integration and how to know when to use them.

## Standard Integral Forms or Properties of Integrals

One should always assume an integral is easy until good evidence suggests otherwise.  What I mean by that is that we should first look to see if the integral in question is one of our standard forms.  By that, I mean any of the following that come immediately from basic differentiation rules.

 Table 1 - Standard Integral Forms or Properties of Integrals 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. The properties and forms above are the basic ones that should simply be recognized.  There are some other fairly basic forms that, with a little algebraic manipulation or a trigonometric identity, become standard forms.

### Example 1

ò (x2 + 3x + 2) dx

In this example, we are not dealing with exactly a standard form but using properties 2 and 3 from Table 1 gives us a sum of three standard forms.

ò (x2 + 3x + 2) dx = ò x2 dx + 3 ò x dx + 2ò dx

=    x3/3 + 3x2/2 +2x + C

### Example 2

ò cos (2x) dx

This is not exactly a standard form since the angle in the trigonometric function is not exactly the same as the variable of integration.  But, letting u = 2x, so du = 2 dx and dx = du/2 gives the necessary standard form.

ò cos (2x) dx = (1/2) ò cos u du = (1/2) sin u + C = (1/2) sin (2x) + C

### This is not one of our standard forms because tan x is not the derivative of any of the six trigonometric functions.  However, with the use of a trigonometric identity and a u substitution it will become one of our standard forms.  ### Example 4 This is not a standard form since sec x is not the derivative of any of the six trigonometric functions.  In this case, the key step will be to multiply by 1 in a creative way. A quick note about this solution.  Do not feel bad if you do not think you would ever have thought to do it this way.  Very few people would ever think of this.  It is one of those cases where you will simply need to remember the solution of this problem, or its counterpart ò csc x dx.

In Examples 3 and 4, once we made the substitution after the identity we end up with a simple integral.  That is always our goal.  Given any integral we want to ask a couple of questions.

### Question 4.If not, then what???

The answer to Question 4 is the focus of the remainder of these notes.  We will look at a number of types of integrals and see how to approach them so that all of them eventually end up in our growing list of standard forms.

## Integration by Parts

We will first consider the method of integration by parts even though it will virtually never be the first choice for a problem.  Integration by parts uses the product rule for differentiation to give a way to rewrite an integral in a form that we can handle.  Consider first the derivation of the technique.

Suppose u and v are both functions of x.

d(u v) = u dv + v du

u dv = d(u v) - v du

ò u dv = ò d(u v) - ò v du    (Integrating both sides)

ò u dv = u v - ò v du

Now, given an integral, we need to break the integrand up into two pieces, u and dv.  There are a few things to consider when making the choices.

• u times dv must give the original integrand
• The dx must be part of the dv.
• For u, you want to choose something that can be differentiated without getting "too messy."
• For dv, you generally want to pick the most complicated part of the integrand that you can that will integrate to something that is not "too messy."

The "too messy" part is ambiguous by necessity.  There are no hard rules for how to make the choices for u and dv.

Let us look at some examples.

### Example 5

ò ln x dx

Consider first the four questions.  The answers to the first three are "No."  This is not a standard form.  Nor is it easily rewritten, either algebraically or using an identity, in order to be able to do it nicely.  No other "reasonable" approach works so we are left with integration by parts.  This is one case where, once you have decided to do integration by parts, the choices for u and dv are fairly easy.  The dv must include dx.  Then, ln x is the only other part to consider.  It will not be part of the dv for if it was, we would have to integrate ln x dx in order to set up the integration by parts.  If we could do that, we would not be looking at doing it by parts. Integration by parts is basically a way to rewrite an integral as a function and another, simpler integral.

### Example 6

ò x ex dx

Always remember to first consider the four questions.  The answers to the first three are "No."  This is not a standard form.  Nor is it easily rewritten, either algebraically or using an identity, in order to be able to do it nicely.  No other "reasonable" approach works so, again, we'll try integration by parts.  As always, the dx will be part of dv.  The ex can be integrated or differentiated with equal ease.  The x can also be either integrated or differentiated easily but if we make it part of dv and integrate it then we'll have a higher power of x in our final integral.  That would not make it easier.

ò x ex dx = x ex - ò ex dx                                   u = x                dv = ex dx

= x ex - ex + C                                  du = dx            v = ex

### Example 7

ò x2 ex dx

Again, consider the four questions.  The answers to the first three are "No."  This is not a standard form.  Nor is it easily rewritten, either algebraically or using an identity, in order to be able to do it nicely.  No other "reasonable" approach works so, again, we'll try integration by parts.  As always, the dx will be part of dv.  The ex can be integrated or differentiated with equal ease.  The x2 can also be either integrated or differentiated easily but if we make it part of dv and integrate it then we'll have a higher power of x in our final integral.  That would not make it easier.

ò x2 ex dx = x2 ex - ò 2x ex dx               u = x2               dv = ex dx

du = 2x dx       v = ex

We have a slight problem here.  We again have an integral that must be done by parts.  We'll need to make a second set of choices for u and dv.

ò x2 ex dx = x2 ex - ò 2x ex dx

= x2 ex - 2 ò x ex dx             u = x                dv = ex dx

= x2 ex - 2 (x ex - ò ex dx)   du = dx            v = ex

= x2 ex - 2x ex + 2 ò ex dx  du = dx            v = ex

= x 2 ex - 2x ex + 2 ex + C

The pattern, if x is raised to even higher powers, is clear.  Fortunately, for cases like this, where a choice for u exists that will eventually differentiate down to 0, there is a short hand way to handle the multiple integrations by parts.  It is called tabular integration.  It is not a "new method."  It is integration by parts, just in a shorthand form.

### Example 8

ò x5 e2x dx

We see that this fits our previous pattern well so we'll try integration by parts.  We know it will involve five integrations by parts and all five will act exactly the same.  We can therefore use the following approach.  Set up a table.  In the first column put x5.  Below it, differentiate it repeatedly until it becomes 0.  The first entry is the first "u."  The second is the first "du" and second "u."  In the last column put e2x.  Integrate it exactly as many times as the differentiation was done.  The first entry is the first "dv."  The second is the first "v" and second "dv."  Hopefully you noticed in the previous example that the signs on the integral alternated, positive and negative.  This pattern continues and is set up between the two columns by way of downward arrows. Now, simply take the products of the expressions on each end of each arrow and string them together using the signs on their connecting arrows.

ò x5 e2x dx = x5 e2x/2 - 5x4 e2x/4 + 20x3 e2x/8 - 60x2 e2x/16 + 120x e2x/32 + C

= x5 e2x/2 - 5x4 e2x/4 + 5x3 e2x/2 - 15x2 e2x/4 + 15x e2x/4 + C

Again, this is not a new method, just a shorthand way of doing integration by parts multiple times when the pattern is obvious and repetitive.

We will now look at another variation on integration by parts.

### Example 9

ò ex sin x dx

Looking at our four questions we see that this does not apparently fit any of the first three.  We will try integration by parts.  Notice that neither part, the ex nor the sin x, poses any real problems regardless of whether we integrate or differentiate them.  I will choose to differentiate sin x.

u = sin x          dv = ex dx

ò ex sin x dx = ex sin x - ò ex cos x dx             du = cos x       v = ex

The integral we have now is the same kind with which we started.  We will need to do parts again.  We will again differentiate the trigonometric function.

ò ex sin x dx = ex sin x - ò ex cos x dx                         u = cos x         dv = ex dx

= ex sin x - (ex cos x + ò ex sin x dx)                  du = - sin x      v = ex

The integral we have now is the exact same one with which we started.  This is beginning to look like an infinite loop.  However, a little algebra will take care of it if we add the integral to both sides of the equation.

ò ex sin x dx = ex sin x - (ex cos x + ò ex sin x dx)

ò ex sin x dx = ex sin x - ex cos x - ò ex sin x dx

2 ò ex sin x dx = ex sin x - ex cos x

ò ex sin x dx = (1/2)(ex sin x - ex cos x ) + C

This kind of approach comes up when the integrand is a product of two functions whose derivatives repeat or run through cycles, like ex (ex, ex, ex, . . . ), sin x (cos x, -sin x, -cos x, sin x, . . . ), etc.  It can also occur in other cases where an identity is used.

### Example 10

ò sec3 x dx

Looking at our four questions we see that this does not apparently fit any of the first three.  We will try integration by parts.  Recall that it is generally a good idea to choose the most complicated part possible for dv, as long as it returns something that is not too messy.

u = sec x         dv = sec2 x dx

ò sec3 x dx = sec x tan x - ò sec x tan2 x dx   du = sec x tan x          v = tan x

We still have a problem with the integral.  In this case, a trigonometric identity will help.

ò sec3 x dx = sec x tan x - ò sec x tan2 x dx

ò sec3 x dx = sec x tan x - ò sec x (sec2 x - 1) dx

ò sec3 x dx = sec x tan x - ò sec3 x dx +ò sec x dx

2 ò sec3 x dx = sec x tan x +ò sec x dx

2 ò sec3 x dx = sec x tan x + ln | sec x + tan x |

ò sec3 x dx = (1/2) (sec x tan x + ln | sec x + tan x |) + C

## Trigonometric Integrals

In this portion of these notes, we will look at integrals of the following forms.

 Table 3 - Trigonometric Integrals ò sinm x cosn x dx where m and n are integers ò secm x tann x dx

Note that if you have an integral of the form ò cscm x cotn x dx you should follow the same ideas as for ò secm x tann x dx.

Doing these integrals will require using the following identities.

 Table 4 - Trigonometric Identities These are variations on the half-angle identities. 1 + tan 2 x = sec 2 x tan 2 x = sec 2 x - 1 sin 2 x = 1 - cos 2 x These both come from sin 2 x + cos 2 x = 1 cos 2 x = 1 - sin 2 x sin 2x = 2 sin x cos x

There is no need to memorize the following techniques.  It is much easier to simply try the appropriate substitutions.  If they work, great, if they don't then each type of problem will have only one "Plan B."

### Example 11

ò sin 3 x cos 2 x dx

If the power on either function is odd, let u be the OTHER function.  Take one from the one with the odd power and put it with the dx.  Then substitute for the remaining powers of the one that had the odd power.

u = cos x,  du = -sin x dx

ò sin 3 x cos 2 x dx = ò sin 2 x cos 2 x sin x dx

= - ò sin 2 x u2 du

= - ò (1 - u2) u2 du

= - ò (u2 - u4) du

= - (u3/3 - u5/5) + C

= - (cos 3 x) /3 - (cos 5 x) /5 + C

### Example 12

ò sin 4 x cos 5 x dx

u = sin x,  du = cos x dx

ò sin 4 x cos 5 x dx = ò sin 4 x cos 4 x cos x dx

= ò u4  (cos 2 x) 2 du

= ò u4 (1 - u2)2 du

= ò u4  (1 - 2 u2 + u4) du

= ò (u4 - 2 u6 + u8) du

= (u5/5 - 2 u7/7 + u9/9) + C

= (cos 5 x) /5 - 2 (cos 7 x) /7 + (cos 9 x) /9 + C

If both sine and cosine have odd powers then either substitution would work.  However, it would be best to let u be the function with the higher power so that the rewriting in terms of u will be less painful algebraically.

### Example 13 ### Example 14

ò sec 4 x tan 2 x dx

If the power on secant is even, then let u = tan x.  Take two of the secants and put them with the dx.  Then substitute for the remaining powers of secant.

u = tan x,  du = sec 2 x dx

ò sec 4 x tan 2 x dx = ò sec 2 x tan 2 x sec 2 x dx

= ò (1 + u2) u2 du

= ò (u2 + u4) du

= (u3/3 + u5/5) + C

= (tan 3 x) /3 - (tan 5 x) /5 + C

### Example 15

ò sec 3 x tan 5 x dx

If the power on tangent is odd, then let u = sec x.  Take one of the tangents and one of the secants and put them with the dx.  Then substitute for the remaining powers of tan.

u = sec x,  du = sec x tan x dx

ò sec 3 x tan 5 x dx = ò sec 2 x tan 4 x sec x tan x dx

= ò u2  (tan 2 x) 2 du

= ò u2 (u2 - 1)2 du

= ò u2  (u4 - 2u2 + 1) du

= ò (u6 - 2 u4 + u2) du

= (u7/7 - 2 u5/5 + u3/3) + C

= (sec 7 x) /7 - 2 (sec 5 x) /5 + (sec 3 x) /3 + C

If the power on secant is even and the power on tangent is odd, then either substitution would work.  However, it would be best to let u be the function with the higher power so that the rewriting in terms of u will be less painful algebraically.

If the power on secant is odd and the power on tangent is even, then we have problems and will have go to "Plan B."   "Plan B" will end up using integration by parts like in Example 10 above.

## Trigonometric Substitutions

The next method we will consider is the method of trigonometric substitution.  This method will be of value when the first three of our important questions do not yield a solution and we have one of the following forms within the integral.

 a2 + u2 where a is a constant and u is a function a2 - u2 u2 - a2

In each case, we will use a right triangle, the Pythagorean Theorem and a little trigonometry to come up with a substitution that will change the form of the integral.

### The Key

The key will be how to set up the triangle.  This will be determined by two factors.

2.      which trigonometric function ends up "nice"

### The Sign

Suppose r and s are two sides of a right triangle.  What the third side will be depends on whether either of r or s is the hypotenuse.

1.      If not, then we have the following triangle. r  q

s

Note that we could have had r and s swapped.

2.  If so, assume it is r, then we have the following triangle.  r q

s

Note that we could have had s and .

### "Nice" trigonometric functions

What we will mean by "nice" trigonometric functions is twofold.

1.      It will not involve a negative sign in its derivative - Thus, we will mean either sine, tangent or secant.

2.      We will not want the square root from our triangle to be involved in either side necessary for the function.  In the first triangle that would mean we would want tangent since we can use only the sides opposite and adjacent to the angle.  In the second it will involve either sine or secant since we have the hypotenuse and either the side opposite or adjacent to the angle.

### Example 16 The minus sign indicates that the square root of the thing in front of the minus sign will be the hypotenuse.  We will choose the other side so that our trigonometric function will be 3x/2. 3x                    2 q sin q = Opposite/Hypotenuse = 3x/2.  This leads to our substitutions.  We want to find substitutions for ALL occurrences of the variable. Using these substitutions gives the following integral. It turns out that this was actually one of our standard forms.  However, this illustrates two things.  First, it shows how to use the method of trigonometric substitution.  Secondly, it shows that if we miss the fact that we have a standard form we can often get the answer anyway.

### Example 17 At first, this integral does not appear to have the right form.  However, it is a quadratic, so we can rewrite it using the method of completion of squares. It now fits the form for trigonometric substitution.  The plus sign indicates that the square root of the quadratic expression will be the hypotenuse.  We will choose the other side so that our trigonometric function will be u/3.  u q

3

tan q = Opposite/Adjacent = u/3.  This leads to our substitutions.  We want to find substitutions for ALL occurrences of the variable. Using these substitutions gives the following integral. ## Partial Fraction Decomposition

This next approach to integrals is, again, not really a new integration technique.  Rather, it is an algebraic rewriting that will transform something into a simpler, integratable, form.

Consider an integral in which the integrand is a rational function.  Recall, a rational function is a quotient in which both the numerator and denominator are polynomials.  If the denominator is a single term, than a simple rewriting of the fraction will put things in a form that we can integrate.  We will end up with power forms or log forms.

### Example 18

a.) b.) In both of these cases, the resulting integrals are all straightforward power forms or log forms.

If the denominator is not a single term the next step will depend on whether or not it is factorable.  If it is not factorable (by this we mean using only real numbers) then it is either first degree or even degree.  Problems with even degree greater than 2 will not fit any standard approach and will therefore not be dealt with here.

### Example 19 Now multiply out everything in the numerator and it will be like the problems in Example 18.  Something of this nature will always work if the denominator is first degree.

For cases of an unfactorable second degree denominator, we will first consider the case where the numerator has degree less than 2.

### Example 20 Now, the first is a log form and the second is an arctangent form.

### Example 21 What will need to be done in cases like this (and could have been done in Example 18) is to first do long division.  For cases where the denominator is degree 2 or more and the numerator is the same degree or higher we must first use long division to rewrite the problem.  Again, long division may be used when numerator and denominator are both degree 1. Thus, we can rewrite the integral as follows. The first term is a power form, the second a log form, and the third is an arctangent form.

### Example 22 For cases like this one, we will first complete the square in the denominator. Now, simply multiply the numerator out and proceed as in Example 21.

For cases where the denominator can be factored, we employ the method known as partial fraction decomposition to rewrite the integrand.  Partial fraction decomposition is essentially the reversal of the process of adding fractions.  In adding fractions with different denominators we get a common denominator and then add the terms.  In this process we begin with the common denominator and have to find how to decompose the sum into its original parts.  For the next few examples we will ignore integration and just look at the fractions apart from calculus.  As above, if the degree of the numerator is the same as or larger than the degree of the denominator then do long division.  The remainder portion will always have the numerator with smaller degree so we will look only at cases like that.

### Example 23 - Distinct Linear Factors The first step is to factor the denominator completely.  We then know that the fraction can be broken apart into separate fractions.  The number of fractions will be the same as the number of factors in the denominator.  The denominators will be the factors of the denominator.  The numerators will always be polynomials of one degree less than the degree of the denominator. Next we eliminate the fractions by multiplying both sides by the original denominator.

5x - 4 = A (x - 5) + B (x + 2)

Our task now is to find the values for A and B.  The thing to remember is that the same A and B will work for ALL values of x.  Often there will be "nice" values of x.  By "nice" we will mean values that will make some parts of the above expressions equal to zero.  We will then substitute those values in for x.

Let x = 5.                    5(5) - 4 = A (5 - 5) + B (5 + 2)

21 = 7B

3 = B

Let x = -2.                   5(-2) - 4 = A (-2 - 5) + B (-2 + 2)

-14 = -7 A

2 = A Now it would be time to integrate and both terms are log forms.

### Example 24 - Repeated Linear Factors The first step is to factor the denominator completely.  We then know that the fraction can be broken apart into separate fractions.  Again, the number of fractions will be the same as the number of factors in the denominator.  The denominators will be the factors of the denominator.  This time, though, we have a repeated factor.  In that case, the repeated factor will be the denominator of the same number of fractions as its multiplicity.  Each time it appears, it will be to one higher degree.  The numerators will always be polynomials of one degree less than the degree of the denominator.  The repeated factor, since it is linear, will still have just a constant as its numerator even though, if multiplied out, it would be second degree. Next we eliminate the fractions by multiplying both sides by the original denominator.

5x2 + 5x - 3 = A (x + 2) (x - 5) + B (x - 5) + C (x + 2)2

Our task now is to find the values for A, B and C.

Let x = 5.                    125 + 25 - 3 = 49C

147 = 49C

3 = C

Let x = -2.                   20 - 10 - 3 = -7 B

7 = -7 B

-1 = B

We have no other "nice" numbers.  We have two options.  One is to just pick another number.  For doing arithmetic 0 is usually "nice."

Let x = 0.                    -3 = -10 A - 5 B + 4 C

But we know B and C, so we have -3 = -10 A + 5 + 12 or -3 = 17 - 10 A.

Thus -20 = - 10 A or A = 2.

Our second option would be use the fact that two polynomials are equally if and only if all of their corresponding coefficients are equal.  That is, the constant terms must be the same, the first degree coefficients must be the same, etc.

We have 5x2 + 5x - 3 = A (x + 2) (x - 5) + B (x - 5) + C (x + 2)2

The x2 coefficient on the left is 5.  If the right side were multiplied out we would have A x2 and C x2.  Thus the x2 coefficient on the right is A + C.  We already have C = 3 so this tells us 5 = A + 3 so A = 2.

In either case, we have the following. Now it would be time to integrate.  Two terms are log forms.   The middle term is a power form if we let u = x + 2.

### Example 25 - Quadratic Factors with No Linear Term The process is essentially the same as in the previous examples.  The number of fractions will be the same as the number of factors in the denominator.  The denominators will be the factors of the denominator.  The numerators will always be polynomials of one degree less than the degree of the denominator. Next we eliminate the fractions by multiplying both sides by the original denominator.

3x2 + 2x - 4 = (Ax + B) (x - 5) + C (x2 + 2)

Our task now is to find the values for A, B and C.

Let x = 5.                    75 + 10 - 4 = 27C

81 = 27C

3 = C

There are no more "nice" numbers.  We still have two variables for which we need to solve.  We'll use both methods we used in the previous examples.

Let x = 0.                    -4 = -5 B + 2 C

-4 = -5 B + 6

-10 = -5 B

2 = B

x2 coefficients            3 = A + C

3 = A + 3

0 = A

Now we have the following. Again, the integration would be straightforward.

### Example 26 - Quadratic Factors with a Linear Term The problem can be done exactly as the previous one was.  However, there is a little change that can really help when the integration rolls around.  If we just do it as the previous one, where the fraction with denominator x2 + 2x + 2 would have numerator Ax + B, the integration would require some creativity because, when broken into two separate fractions,  Ax/(x2 + 2x + 2) is a slight problem.  We'd complete the squares and then end up with another E/(x2 + 2x + 2) term.  There is something we can do to take care of that.  Instead of Ax + B, we'll use

A(2x + 2) + B.  Notice that 2x + 2 is the derivative of x2 + 2x + 2. Next we eliminate the fractions by multiplying both sides by the original denominator.

x2 + 18x - 4 = (A(2x + 2) + B) (x - 5) + C (x2 + 2x + 2)

Our task now is to find the values for A, B and C.

Let x = 5.                    25 + 90 - 4 = 37C

111 = 37C

3 = C

There are no more "nice" numbers.  We still have two variables for which we need to solve.  We'll use both methods we used in the previous examples.

Let x = 0.                    -4 = -10 A - 5 B + 2 C

-4 = -10 A - 5 B + 6

-10 = -10 A - 5 B

2 = 2 A + B    *

x2 coefficients            1 = 2 A + C

1 = 2 A + 3

-2 = 2 A

-1 = A

Thus we have (from *) 2 = -2 + B or B = 4.

Now we have the following. For the integration, the first term is a log form, the second (after completing the squares) is an arctangent form, and the last is a log form.

### Example 27 - Everything Together Here we have repeated factors, quadratic factors and linear factors.  The set up would be as follows. Multiplying both sides by the denominator gives the following. Let x = 5.        -10952 = G(25 + 10 + 2)2(4)

-10952 = G 372(4)

-10952 = 5476 G

G = -2

Let x = 1.        4800 = H 25 (-64)

4800 = -1600 H

H = -3

There are no more nice numbers.  At this point we would compare coefficients and pick a few more numbers to find 6 equations in order to find the other 6 unknowns.  The tedium of the work is left to the interested reader.