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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Paul Bautista*

Ever feel like you are going around in circles? Like, periodically you have your ups and downs? Well, sines and cosines go up and down regularly too. In this lesson, learn how to integrate these circular functions.

Remember that you can calculate the definite integral of *f(x)* from *a* to *b* as being the anti-derivative of *f(x)* evaluated at *b* minus the anti-derivative of *f(x)* evaluated at *a*, which we write as the anti-derivative of *f(x)* from *a* to *b*. If you have an indefinite integral that is without limits, your indefinite integral is equal to the anti-derivative of *f(x)* plus some constant of integration *C*.

Let's take a look at trig functions, like *f(x)* = sin(*x*). If you recall, the derivative of sin(*x*) = cos(*x*), because the derivative is the slope of the tangent of the function. So here's sin(*x*) at *x*=0. The tangent has a slope of 1, which is the value of cos(*x*) evaluated at *x*=1. At *x* = *pi*/2, the slope of sin(*x*) is equal to 0. The value of cos(*x*) is equal to 0, because the derivative of sin(*x*) is equal to cos(*x*).

On the other side, the derivative of cos(*x*) with respect to *x* is equal to -sin(*x*). I can use these derivatives to determine what the integral, say of sin(*x*), is. The integral of sin(*x*) *dx* is equal to -cos(*x*) + *C*. How do you see this? Well, if I take the derivative of -cos(*x*) + *C*, I get minus the derivative of cos(*x*) plus the derivative of *C*. The derivative of a constant is zero because the slope of a line that has a constant value is zero, and the derivative of cos(*x*) is -sin(*x*). So my term *d/dx*(cos(*x*)) becomes - -sin(*x*), or just sin(*x*). So -cos(*x*) + *C* is an anti-derivative of sin(*x*). That is, if I take the derivative of -cos(*x*) + *C*, I end up with sin(*x*). You can make a similar argument for the integral of cos(*x*). Here, the integral of cos(*x*) is equal to sin(*x*) + *C*. You can see this by taking the derivative of sin(*x*) + *C*. That's just equal to cos(*x*).

There are a lot of trig functions out there, but really there are only two that you need to know the integral of off the top of your head, and those are sin(*x*) and cos(*x*). All of these other guys you'll generally look up in a table or you can determine just by knowing sin(*x*) and cos(*x*). So remember that **the integral of sin( x)dx = -cos(x) + C**, and the

So let's do an example. Let's say we want to integrate the function *f(x)* = sin(*x*) between *x*=0 and *x*=2*pi*. Remember that the integral is equal to the area under the curve. If you have a curve above the *x*-axis, that area is positive. But if you have something below the *x*-axis, this is actually a negative integral. So what you're really doing is adding this positive area and subtracting this negative area to find the integral. Just using your intuition, you know that if you're trying to find the integral - that is, this positive area minus this negative area - it might be zero, but let's see if we can calculate that exactly.

So let's calculate the integral from 0 to 2*pi* of sin(*x*)*dx*. Using the fundamental theorem, I know that's equal to the anti-derivative of sin(*x*) evaluated from 0 to 2*pi*. I also know that an anti-derivative is -cos(*x*), because the integral of sin(*x*)*dx* equals -cos(*x*) plus a constant. So let's plug in my anti-derivative, which is -cos(*x*). I've got -cos(*x*) evaluated from 0 to 2*pi*. Remember, this is like saying I'm evaluating this at 2*pi* and subtracting from it my evaluation of this at 0. So I've got -cos(2*pi*) - (-cos(0)). That's like -1 - -1, which is -1 + 1, which is just 0. Indeed, the integral from 0 to 2*pi* of sin(*x*) is 0; there's an equal amount of area above and below the *x*-axis.

What about a function like cos(*x*) + 1 between *x*=0 and *pi*? Again, let's use the fundamental theorem, and let's say that the integral from 0 to *pi* of (cos(*x*) + 1)*dx* is equal to the anti-derivative of this function evaluated from 0 to *pi*. Now let's break this integral up into two separate integrals so it's equal to the integral from 0 to *pi* of cos(*x*) plus the integral from 0 to *pi* of 1*dx*. This first term, the integral from 0 to *pi* of (cos(*x*))*dx*, is equal to the anti-derivative from 0 to *pi*. That anti-derivative is sin(*x*), so we plug that in here and I get sin(*x*) evaluated from 0 to *pi*, so that's sin(*pi*)-sin(0). Well, that's just equal to 0.

Okay, what about the second term, 0 to *pi* *dx*? 0 to *pi* *dx* is equal to the anti-derivative evaluated from 0 to *pi*. The anti-derivative of 1 is *x* + *C*, and remember we're ignoring *C* here because we're looking at a definite integral. So I'm going to plug *x* in for my anti-derivative and evaluate from 0 to *pi*, and I get *pi* - 0, which is just *pi*. So my total integral is equal to 0 + *pi*, and that's just *pi*, so the area under the curve here is equal to *pi*. That is, the integral from 0 to *pi* of cos(*x*) + 1 is equal to *pi*.

Let's review. There are a lot of trig functions, but really you just need to memorize two anti-derivatives. That is, you need to know the integral of sin(*x*)*dx* is equal to -cos(*x*) + *C*. That's because if you take the derivative of -cos(*x*) + *C* you get back sin(*x*). The integral of cos(*x*)*dx* is equal to sin(*x*) + *C*. And again, that's because if you take the derivative of sin(*x*) + *C* you end up getting back cos(*x*).

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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets

- Go to Continuity

- Go to Limits

- Calculating Integrals of Simple Shapes 7:50
- Anti-Derivatives: Calculating Indefinite Integrals of Polynomials 11:55
- How to Calculate Integrals of Trigonometric Functions 8:04
- How to Solve Integrals Using Substitution 10:52
- Substitution Techniques for Difficult Integrals 10:59
- Using Integration By Parts 12:24
- Partial Fractions: How to Factorize Fractions with Quadratic Denominators 12:37
- How to Integrate Functions With Partial Fractions 9:11
- Understanding Trigonometric Substitution 10:29
- How to Use Trigonometric Substitution to Solve Integrals 13:28
- How to Solve Improper Integrals 11:01
- Go to Integration and Integration Techniques

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