3 Important Methods of Integration To Know
One of the most essential parts of Calculus, integration details the techniques of finding an antiderivative, F(x), of the function f(x). It has many applications both in mathematics and physics that involve volume, area, and mechanics. In this article, we’ll examine three of the key methods for integration to have in your arsenal.
Integration is essentially the opposite of differentiation. The variety of difficulty within the subject of integration is extremely extensive, with some problems taking 5 seconds to solve and some problems taking 5 minutes. For example, we know that the derivative of x³ is 3x². Therefore, when we integrate 3x², we get x³+C. It is always important to remember the +C, since the derivative of a constant is always zero, so any constant value can be tacked on to the end of a derivative expression and differentiating will yield the same result.
However, not all problems are as simple as finding the integral of 3x². What is the integral of sin(2x)? One may jump to write down -cos(2x) + C, but sadly, this would be incorrect. The reasoning behind why this solution is incorrect leads us into our first method of integration: U-Substitution.
As I was taught, U-Substitution is a way of dealing with the chain rule from differentiation: It reverses it! The chain rule deals with derivatives of composite functions. In examples like this, we say that the derivative of the function f(g(x)) is f’(g(x))*g’(x). This is why the derivative of -cos(2x) isn’t just sin(2x): we are missing an extra factor of 2 from the derivative of the inside function 2x. Now it should be apparent to us why integrating sin(2x) doesn’t simply yield -cos(2x). It is absolutely necessary to “account” for the chain rule in both differentiation and integration problems. Let’s look at an example problem together.
Our first step should be to locate any familiar functions and derivatives, going term by term down the integral. Because we are reversing the chain rule, we want to set our u = sin(x), leaving our du = cos(x)dx.
First we change our integral to fit into the u-world, after which we integrate in that same u-world. However, we want our final answer in terms of x, so it is necessary to substitute whatever we set u equal to back into the expression. Don’t forget the +C!
Before moving on to integration by parts, let’s examine U-Substitution in the context of definite integrals, integrals with bounds on them. For consistency, we will use the same problem from before.
The substitution remains the same: u = sin(x) and du = cos(x)dx. However, when we move into the u-world, the bounds must also be properly converted. Since the upper and lower bounds of the integral are pi/2 and 0, respectively, their corresponding upper and lower u-bounds are 1 and 0, respectively (found by plugging in the bounds into the sin(x) function).
From there, the steps are relatively simple to follow. After finding the antiderivative and evaluating the difference between the upper and lower bounds in the u-world, we arrive at our answer of 1/3.
Integration By Parts
The next technique of integration is integration by parts. The general formula is provided below.
The most effective use of integration by parts is to deal with the products of functions, especially when U-Substitution can not be performed.
Let’s dive into the integral shown above. Much like the example for U-Substitution, there is a product involved in the integrand. However, no function and its derivative can be simultaneously found. Therefore, the next method to consider should be integration by parts. Before jumping into this problem, let’s take a closer look as to why our little formula works.
The way of thinking about the formula is to first consider the product rule of functions u and v (these are really u(x) and v(x)). Next, integrate both sides of the equation. Finally, rearrange the integrals to obtain the initial form of the formula shown at the beginning of the section.
Now, let’s hop back into the example problem. For simplicity purposes, we will use Leibniz Notation for u, du, dv, and v. The key to solving our integral, and all other integration by parts problems, is to properly assign the functions to each “Leibniz Notation” part. In other words, what part of the integral should be u, du, dv, and v?
Although this seems rather tricky, there is a somewhat critical and effective method of selecting our functions properly: we must focus on what can be differentiated and broken down. In the case of xsin(x) dx, the proper way to assign our u’s and v’s is to first assign u = x and dv = sin(x) dx; the rest follows smoothly.
As we can see, by assigning them this way, our du simply becomes dx, effectively simplifying our second integral in the integration by parts formula. The next step and the final answer can be viewed below.
Say we had to complete the same integral xsin(x) dx, but this time there were bounds from 0 to pi. The key step to consider is to make sure that the bounds carry to both uv and the integral of vdu. For uv, evaluate the difference of the expression with inputs of the upper and lower bound, and for vdu, solve for the antiderivative and evaluate the difference of that antiderivative with inputs of the upper and lower bound as well. View the solution to our original problem below (look out for where the upper and lower bounds are visually).
By far the most difficult method of integration in this list, trigonometric substitutions are often slightly tedious, but I find them to be elegant and extremely satisfying to solve. The main situation in which this type of substitution arises is when there are expressions similar in form to those of the derivative of inverse trigonometric functions. I’ve listed three expressions below to look out for, as well as the appropriate subsitution.
We recognize that this is the form for sin. Follow through with the subsequent steps.
We equate our x based on the appropriate substitution, and then substitute in the thetas into our original integral. After simplification using identities and removing the square root sign (4 turns into two squared and one minus sin squared turns into cosine squared), we then integrate in the theta-world. HOWEVER, we need to have our answer in terms of x, so the job is not done yet.
Here is a problem that involves the second substitution with tangent.
You’ll notice that this is a slightly more difficult problem, as the theta we need to replace is within different functions. To deal with this problem, we must consider triangles.
First, we follow our normal procedure by rewriting theta in terms of x. However, when we plug x back in, we can see that we have inverse trigonometric functions within other functions. I’ve constructed a triangle to help visualize how the next step works. Tangent of inverse tangent of x/3 is simply x/3: no work needed there. However, having a secant as the outside function requires more thought.
What does inverse tangent actually mean? It represents the angle whose tangent gives us our “input ratio.” In our example, our input is x/3, so we can set up a triangle in which the tangent of some angle (a NEW theta), is x/3. From here, we can use the pythagorean theorem to calculate the hypotenuse, and now we find the secant of that same angle. Everything is now in terms of x, and we are happy! Note that since we have the constant ln3, we can just combine that with our +C as the new constant of integration.
Our last example problem involves substitution with secant.
As always, we need to rewrite theta in terms of x and substitute back into our original equation.
As we can see, we need to set up our little triangle again, much like the methodology in the previous example. We create an arbitrary angle such that its secant is x/5. Then, after solving for the missing side, we can substitute all expressions back in terms of x.
Like any other definite integral problem, we simply need to find the antiderivative and evaluate. Now we could change our bounds to fit the theta-world, but I recommend to just evaluate with the given bounds after finding the antiderivative in terms of x, at least for these types of problems. Let’s look at our problem from Example 1.
We have already solved for the given integral in terms of x using our sin substitution technique.
With this antiderivative, we can evaluate the difference between inputs at the upper and lower bound, arriving at our answer.
Perhaps my longest article yet, I’ve spend lots of time going through and examining certain problems that involved these integration techniques. I had learned U-Substitution and integration by parts last year, but I picked up trigonometric substitutions a little more recently. My favorite part was setting up those triangles, as I realized that drawing a simple figure could make problems a whole lot simpler. It really shows how clean math is.
I hope you have learned some new techniques that can be applied to different integration problems!