# Methods of Integration

It has been my experience with this material that students continue to struggle with methods of integration despite our best intentions, instructions and coaching. Let us all agree that not everyone learns the same amount at the same rate. Let us also understand that there are some teaching styles that don’t transmit all the necessary concepts and/or details to all the various learning styles that exist. In my mind, it is necessary to make the instruction as clear and simple as possible without being condescending or over simplified; however, we do have to assume some basic knowledge. I have seen 2^{nd} and 3^{rd} year students studying differential equations who are frustrated because they couldn’t complete the resulting integral. These are students who should be able to calculate most integrals with little or no difficulty but who haven’t yet mastered the techniques and this shows up at the worst of times: final exams. It is quite possible that the subtleties of the particular techniques weren’t grasped properly or practiced enough. There are more reasons of course, but these are some of them. In this text, we will attempt to rectify this situation by plainly putting into words those tricks and techniques that may alleviate some of the difficulties still experienced. All four analytical methods have little quirks that need to be clarified so that mastery can begin early and permanently.

For example, how would you integrate versus ? And how would you integrate versus? The principle concern whenever integrating is to know the best approach. Using the most efficient method saves a lot of work and frustration. Being able to spot the subtleties of each approach permits us to make better decisions. Therefore, I would suggest that the most difficult part of integration is deciding on the approach. As there are so many types of integrals, it is difficult at first to discern the various types from one another. In terms of pedagogy, my thinking is that we classify integrals fundamentally, get experience with these, then branch out to the more intricate questions. Hence, a strategy for beginners is to learn and master Substitution, Parts, Trig Substitution and Partial Fractions respectively. We will look at each method individually noting its subtleties as examples are demonstrated.

## Integration by Substitution:

This method is best used when the derivative of part of the integrand already appears in the integrand. This does appear confusing when we first read it; let us clarify then with a few examples:

**1**.

At the onset, this looks frightening. The length of the integrand alone is enough to scare most! However, we do notice that the derivative of (which is ) already appears in the integrand and so. We now know two things: we can use substitution and we know what to let u equal.

Letting

Therefore,and As you see, we multiply by dx in order to see clearly what portion will be replaced by du.

It is very important to remember that all portions of the given integral are to be put in terms of u and du. Now let us substitute u and du into the original equation. We get:

Now this is an easier equation to integrate! Integrating it we get:

Now remember that the original is not in terms of u but in terms of x, so don’t forget to substitute back. Our final form is:

As substitution is the easiest and most efficient method, we don’t need to try another method if this one works. Let us examine this technique applied to other functions.

**2.**

The derivative of is so we let where therefore,

Transforming we have:

Which integrates to

Now, putting the equation back in terms of x we get:

or

Let’s try another:

**3.**

Let so

Don’t worry that 2x isn’t in the original equation, as long as the variable is there (in this case the x) we can manipulate the integer. Going back to the original equation, we see that we can multiply the equation by 2 as long as we also multiply the equation by .

Therefore, we are really only multiplying the equation by 1.

So now our du is present in the equation so we can substitute in u and du.

When we integrate this through we end up with

Now simply substitute back to make the equation in terms of x and we get

Here’s another one to try:

**4**.

Let so

Now substitute u and du into the original equation and we get:

Doesn’t this look much easier to integrate!? When we do we get:

Now substitute back into terms of x and we get:

or

**5.**

Now here’s a tricky one. We’ll let so

Now we still have an x left unaccounted for so we use the equation and solve for x. We get

Now lets substitute in u and du and the value we found for x so that the entire equation is in terms of u.

This still looks scary so divide out and you get:

We can simplify further and write it as:

Now we can integrate the equation. We get:

Substituting back in for x we get:

## Integration by Parts:

This method is best used when we cannot use the substitution method. We tend to look for an integral made up of functions that seem to be unrelated. If we do not see the derivative of part of the integrand already appearing in the integrand then we look for algebraic and/or transcendental and/or trig functions, or a combination there of.

Recall our definition for integration by parts:

Integration of parts breaks down into one of **two situations**:

**1. **A single portion

**2. **A double portion

Let us start with an example of **situation 1**:

**1.**

Here we have **situation 1** because we have a single function whose derivative does not appear anywhere else in the integral. In terms of method for a **situation 1**-type integral, always let u equal the given function and always let dv equal the dx.

Let u=lnx, so and

Let dv=dx, so v=x

Now using the definition substitute in our values and we get:

Here is another example of a **situation 1 **type integral:

**2.**

Once again, we have a single function whose derivative does not appear anywhere else in the integral.

Let u=arctanx, so

Let dv=dx, so v=x

This gives:

Now lets try an example of **situation 2**:

**3.**

This is a **situation 2** because x and aren’t related as functions per se, and since we have more than one given function, it becomes a matter of best choices for both u and dv.

Let u=x, so du=1dx

Let , so

Now, substituting back using the definition we get:

**Choices for u and dv:** We tend to let u equal any function that would eventually disappear under repeated differentiation. These tend to be positive powers of x. As for dv, we tend to let this equal any function whose derivatives go in cycles such as trig functions or .

Unfortunately, we have an exception to all this and that is any integrand containing lnx. Whenever this happens we immediately let u=lnx and dv be whatever is left. The reasoning for this is that the beginner doesn’t necessarily know the anti-derivative of lnx is xlnx-x. Even if the beginner did know, letting du=lnx makes the process of solving the integral much longer. Here is an example of this:

**4.**

We would not use substitution because the derivative of does not appear in the given integrand. Instead we try integration by Parts (**situation 2**).

Let u=lnx, so

Let dv=xdx, so v=

Thus,

**The Vicious Cycle:** What do we do for a situation 2-type integral whereby neither of the given functions disappear under repeated differentiation, i.e. both functions have cyclical derivatives?

Here is an example of this type of integral:

**5.**

As we can see, the neither of the derivatives disappear so our choices for u and dv are arbitrary. Since we will have to do this process more than once for this integral, we should be consistent with our choices. That is, if we let the first time around, then we should let the second time around too.

Let , so

Let dv=cosxdx, so v=sinx

Using our definition, we get:

**Golden Rule of Parts**: If, after our initial choices of u and dv, the resulting integral is more difficult then the original, then STOP! Instead, make different choices for both u and dv, then begin again.

Here, however, is not more complicated but is as complex as the original, so begin again by solving .

Let , so

Let dv=sinxdx, so v=-cosx

By definition we have:

Now substituting this new equation into the original () we get:

In order to break the cycle, we simply bring over to the left-hand side, in other words, add it to the left-hand side.

Now we have:

+C

Which simplifies to:

## Integration by Trig Substitution:

This method is best used when we cannot use Substitution or Parts as seen previously. The key factors are the sum or difference of two perfect squares, usually under a square root sign.

Here is a table of the identities that you will encounter:

Here is an example that must be solved by using trigonometric substitution:

**1**. Looking at the table above, we see that we should use **identity (1)**.

We let so When we substitute in we get

Using identity (1) we get

(The definition of inverse functions)

Here is another example.

**2**. Looking at the table above, we see that we should use **identity (2)**.

We let so When we substitute in we get

By using identity (2) we get

The ‘s cancel and we are left with

(The definition of inverse functions)

Here is another example

**3**. Looking at the table above, we see that we should use **identity (3)**.

We let so When we substitute in we get

The ‘s cancel and we are left with

By using **identity (3)** we get

Which simplifies to

The ‘s cancel and we are left with

(The definition of inverse functions)

**4**.

Now here is a tricky one

First we want to change the 9 to a 1 so we take out a 9 from inside the brackets. We also take the 5 out for simplicity.

The 9 can now come out from under the square root sign so we get 3.

Now this looks more familiar! We recognize this from the table above as **identity (1).**

We let so and .

Now go ahead and solve the problem as we showed you in **example** **1.**

Here is another tricky one

**5**.

First we need to complete the square, so we split up 5 into a 4 and a 1.

Now this looks familiar! We recognize this from the table above as **identity (2).**

We let and substitute the into the equation. Now you can solve.

## Partial Fractions:

This method is best used when we cannot use Substitution, Parts, or Trig Substitution. In particular, we look for a rational expression whose denominator is factorable. For example, has a factorable denominator that becomes .

Note: If the denominator is not factorable, consider completing the square and using Trig Substitution.

### Three Golden Rules of Partial Fractions:

- The power of the variable in the numerator is to be 1 less than the highest power of the variable in the denominator.
- Always complete the polynomial to the constant term.
- Always build up to the power of the binomial term.

Let us go through an example.

- We factor out the denominator and we get

Where is really and 0 is 1 less than the power of x on the bottom ().

Now to find constants and , we multiply both sides of this identity by the left-hand, common denominator . The result is

Now we’ll group the x’s together and the numbers together and we get

Since we have no term, we write 1 as and then equate the components.

We now have and

Now we can put everything in terms of

Now substitute in for and the result is

And since , Now we substitute in the values for and in the original equation and we get

Therefore,

Here is another example

**2.**

Notice that the in Ax is to the power 1 which is 1 less than and we completed the polynomial to the constant .

To find the constants , and , we multiply both sides of this identity by the common denominator . The result is,

Since we have no or terms, we write 3 as . This implies

so so

so so

Putting everything in terms of we get so

Substituting in the values for , , and into the original equation we get

**3.**

Now multiply each side by the common denominator , the result is

Factoring out the numerator we get

therefore,

so and

so and and and

so and and

, and

Now substituting these values back into the original we get

Here is a tricky example

**4.**

Now multiply both sides by the common denominator

This is what the top looks like.

so and

so and

so and so therefore

so and and therefore

so and therefore

, , , , and

Now put these values back into the original,

We will solve each separately.

Let , so and .

Substituting in we get

Let so so and

Substituting in we get

The ‘s cancel out and we are left with

Let so and

Substituting in we get

Let so and

Substituting in we get

Since , and . So Thus,

And finally the last one,

So our final answer is

There is a new, faster way of solving the identities by Partial Fractions, here it is:

**5. **

First determine the values of x in each denominator which makes the equation undefined. Then ignoring that factor, plug in the number for x in the other factor and evaluate it. For example,

A, let x=-5 since this is the value for x that makes the equation undefined, and evaluate the original rational expression,

We ignore so we get so 1. This means that A=1

B, let x=-6 and evaluate the original rational expression,

We ignore (x+6) and we get so –1. This means that B=-1.

Now we can substitute in these values for A and B into the original equation and we get

## Summary:

The purpose of this booklet was to demonstrate these four methods of integration. We do recognize that there are methods that cannot be solved by using any one of the four methods mentioned. One could use numerical methods such as Riemann Sums, Trapizodal, and/or Simpsons. Another method of course involves the use of Series.