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# How to Use Trigonometric Substitution to Solve Integrals

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1. 0:05 Solving with Tables
2. 1:26 Solving by Substitution
3. 3:06 Solving by Parts
4. 4:30 Solving by Riemann Sums
5. 4:49 Example of Solving by Parts
6. 12:00 Lesson Summary
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### Erin Lennon

Erin has taught math and science from grade school up to the post-graduate level. She holds a Ph.D. in Chemical Engineering.

In this lesson, we use each of the common integration techniques to solve different integrals. It's not always obvious which technique will be the easiest, so being familiar with an arsenal of methods might save you a lot of work!

## Solving with Tables

Let's take a minute to review some techniques for integration. Here we're going to integrate the indefinite integral - that means it has no limits - of f(x)dx. We know that the integral of f(x)dx equals the anti-derivative as a function of x plus a constant of integration. If you take the derivative of the anti-derivative, you get back your original function. So what are some of the ways we know to find integrals?

First, we can use a table. This may be a table in a book, online or what you have memorized. For example, the integral of x^2dx = (1/3)x^3 + C. You know this because you know how to integrate polynomials. The integral of sin(x)dx = -cos(x) + C. The integral of e^(x)dx = e^(x) + C. For each of these cases, if you take the derivative of the right side, you end up with the integrand. This is true of all integrals; it's how you calculate an integral.

## Solving by Substitution

The second way we know to calculate integrals is by substitution. In this case, we're going to take an integral that depends on x, and we're going to make a substitution where u equals some new function of x. By plugging in u, we hope to end up with a simpler integral that we can integrate with respect to u. For example, we have the integral sin(2x)dx. I want to substitute u for 2x, so u=2x and du=2dx. So I can plug those in, both for 2x and for dx, and my integral becomes 1/2 sin(u)du. I can use a table to solve this, because the integral of sin(u) is -cos(u). I get -1/2 cos(u) + C. Now I want to plug in 2x where I have u - that's my original substitution - so I get x back in my final answer. I get -1/2 cos(2x) + C. If I take the derivative of this, I end up with sin(2x). That's solving by substitution, and that is by far what you are going to use the most when solving integrals by hand, but there are a couple of other methods that you should be aware of.

## Solving by Parts

One is integration by parts. Here you have the integral of udv = uv minus the integral of vdu. This is just a rearrangement of the product rule. An example would be the integral of xe^(x)dx. Here I'm going to set x equal to a new variable, u so that du=dx. I'm going to set e^(x)dx equal to dv, so v has to be equal to e^x because the derivative of e^x is e^(x)dx. If I plug u, v, du and dv into the right side of my equation for integration by parts, I end up with xe^x (that's uv) minus the integral of e^(x)dx (that's vdu). At this point, I can use a table in my head because I have memorized this integral. The integral of e^(x)dx is e^x, so if I plug that in, my integral becomes xe^x - e^x + C. If I take the derivative, I end up with xe^x.

## Solving by Riemann Sums

The last way you can solve an integral is by Riemann Sums. This is not an analytical way to solve it; that is, you aren't going to have numbers left. You're going to solve it numerically, on a computer or calculator - you're going to plug in actual numbers.

## Example of Solving by Parts

Let's do an example. Let's say you're given the integral of x^2(sin(x))dx. Your first step would be to see if you can remember this integral or look it up in a table. Most likely, any integral you come up with is either in a table or it isn't solvable. There are whole books on writing out integrals like this, all the possible integrals you can think of that actually have solutions. But let's say you don't have that book handy, so the first thing you try is substitution: u=sin(x) and du=cos(x)dx. This doesn't make much sense, because if you plug in sin(x) you get u but x^2 becomes arcsin^2(u). You just made life a little more complicated, so maybe that's not the best way to do it. Because substitution is still the first thing you want to go to, what about using u=x^2 and du=2xdx. That's good, but you still have sin(x), which would become sin(square root of u) and that again sounds really complicated. So maybe substitution isn't the method you want to look at.

Since you have one function multiplied by another function, maybe you could do integration by parts. In integration by parts, I'm going to set this first function, x^2, equal to u. That leaves sin(x)dx as dv, because remember, the integral of udv = uv minus the integral of vdu. So my integral has to be u times dv: if x^2=u, then sin(x)dx=dv. If u=x^2, then du=2xdx. If dv=sin(x)dx, then v= -cos(x), because if I take the derivative of -cos(x), I end up with sin(x)dx. So I have u, v, du and dv. If I plug all of these into my equation for integration by parts, I get -x^2(cos(x)) + the integral of cos(x)2xdx. (The '+' comes from the minus sign of v combined with the minus sign in our equation.) I can rewrite this as -x^2(cos(x)) + 2 times the integral of x(cos(x))dx. I don't actually know the integral of x(cos(x))dx off the top of my head. It looks a little simpler than x^2(sin(x)), but it's still not something I just know.

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