# Substitution Techniques for Difficult Integrals

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• 0:06 Example of a Difficult…
• 2:36 Solving by Substitution
• 5:56 Checking Yourself
• 7:03 Tricky U Substitutions
• 9:33 Lesson Summary
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Lesson Transcript
Instructor: Kelly Sjol
Up, down. East, West. Opposites are everywhere. In this lesson, learn how you can think of substitution for integration as the opposite of the chain rule of differentiation.

## Example of a Difficult Integral

Every year it's the same thing. The second week of April I finally sit down to do my taxes. I take a look at my bank statement, I take a look at my taxes and I have a little bit of a 'panic' moment. What happens to all of my income?

Well, I like math, so I tried to graph my income and expenses over time. This is a rate; it is how much I make or spend at any given point in time. So when this graph is above 0, I'm making more money than I'm spending. When it's below 0, it's getting toward the end of the month. I'm not making any money right then, but I'm spending a lot of money. So when it's down here, I'm losing money, and when it's up here, I'm gaining money.

If I integrate my income and expenses (which I'll call f(t); f is a function of time, t) from some t=a to t=b, I can find out how much has been added or subtracted from the bank in that amount of time. For example, I know intuitively that if I integrate over this region here, the area is all above the curve, so it's all positive. And it's greater than 0 because there is some area here. The integral over this region will be positive, so I have net money coming into my bank account. Now this makes a lot of sense, because the rate is above 0 for this whole region. Over this whole time period, I'm still making more money than I'm spending. So obviously at the end of the time period, I'm going to have more money than what I started with.

What is the rate of my income over time? Let's say it's f(t)=cos(2 sin(t)) * cos(t). So this is a cyclical path that occurs every month. Just to be clear, let's write it like this: I've got the cosine of some inner function times the cosine of t. That inner function is 2 * sin(t). So because I've got these big parentheses, cos(2 sin(t)), and a function inside of another function, I'm already thinking this is a composite function: 2 sin(t) is input into the cosine. It's a composite of the inner function and the outer function.

## Solving by Substitution

Okay, so that's the background. Now let's try to find out how much I make between time a and time b. Am I net in or net out? Let's take the integral of cos(2 sin(t)) * cos(t) * dt. Because this is a composite function, I'm already thinking I'm going to need to use some kind of substitution.

So I'm going to find some function, u, that is my inner function. I'm going to find out what the derivative of u is, and then I'm going to substitute u and du into my integral. Hopefully this will make my integral easier to solve. I can find the anti-derivative of my u function, and then I can substitute back in and get rid of all the us and replace them with xs or, specifically, with g(x)s. Once I've done all that, I want to check my answer to make sure I didn't screw up somewhere.

Okay, so let's look at our big function. I know that it is a composite function, so let's call this inner function u. So u=2 sin(t). If I take the derivative of u, I get du=2 cos(t) * dt. This is like saying dt=du / (2 cos(t). So let's substitute dt into our original integral and let's substitute 2 sin(t) and replace that with u. I end up with the integral of cos(u) * cos(t) * du / (2 cos(t). The cos(t)s cancel out, and I end up with the integral of 1/2 cos(u) * du. Well, cos(u) * du is an easy integral. It's just sin(u) * du, because if I take the derivative of sin(u) I get cos(u). So I can solve this u-based integral, 1/2 cos(u) * du, and I get 1/2 sin(u) + a constant of integration, C.

My original equation had t in it, not u, so let's replace the u with 2 sin(t). After we've done this, we find that our integral equals 1/2 sin(2 sin(t)) + a constant of integration, C, because it is an indefinite integral; it has no limits.

So what does this mean for my income, my account balance, over time? If I just graph the anti-derivative and ignore the constant of integration, I know that my account balance is actually in the black. So I've got money up untilâ€¦right about here. At this point, let's just say I'm running a credit with the bank. Then I end up making enough money next month to get back into the black and repeat the whole process over again.

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