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# How to Calculate Integrals of Trigonometric Functions

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1. 0:10 Review of Calculating Integrals
2. 0:43 Integrals of Sine and Cosine
3. 3:32 First Example
4. 5:32 Second Example
5. 7:23 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.

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.

## Review of Calculating Integrals

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.

## Integrals of Sine and Cosine

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 integral of cos(x)dx = sin(x) + C.

## First Example

So let's do an example. Let's say we want to integrate the function f(x) = sin(x) between x=0 and x=2pi. 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 2pi of sin(x)dx. Using the fundamental theorem, I know that's equal to the anti-derivative of sin(x) evaluated from 0 to 2pi. 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 2pi. Remember, this is like saying I'm evaluating this at 2pi and subtracting from it my evaluation of this at 0. So I've got -cos(2pi) - (-cos(0)). That's like -1 - -1, which is -1 + 1, which is just 0. Indeed, the integral from 0 to 2pi of sin(x) is 0; there's an equal amount of area above and below the x-axis.

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