Definite integral of trig function | AP Calculus AB | Khan Academy

So let's see if we can evaluate the definite integral from ( \frac{11\pi}{2} ) to ( 6\pi ) of ( 9 \sin(x) , dx ).

The first thing, let's see if we can take the anti-derivative of ( 9 \sin(x) ). We could use some of our integration properties to simplify this a little bit. So this is going to be equal to; this is the same thing as ( 9 ) times the integral from ( \frac{11\pi}{2} ) to ( 6\pi ) of ( \sin(x) , dx ).

And what's the anti-derivative of ( \sin(x) )? Well, we know from our derivatives that the derivative with respect to ( x ) of ( \cos(x) ) is equal to negative ( \sin(x) ). So can we construct this in some way?

This is a negative ( \sin(x) ). Well, what if I multiplied on the inside? What if I multiplied it by a negative ( 1 )? Well, I can't just multiply only one place by negative ( 1 ); I need to multiply by negative ( 1 ) twice so I'm not changing its value. So what if I said negative ( 9 ) times negative ( \sin(x) )? Well, this is still going to be ( 9 \sin(x) ). If you took negative ( 9 ) times negative ( \sin(x) ), it is ( 9 \sin(x) ).

And I did it this way because now negative ( \sin(x) ) matches the derivative of ( \cos(x) ). So we could say that this is all going to be equal to; it's all going to be equal to, you have your negative ( 9 ) out front, negative ( 9 ) times; and I'll put it in brackets, negative ( 9 ) times the anti-derivative of negative ( \sin(x) ). Well, that is just going to be ( \cos(x) ), and we're going to evaluate it at its bounds.

We're going to evaluate it at ( 6\pi ), and we do that in a color I haven't used yet. We're going to do that at ( 6\pi ), and we're also going to do that at ( \frac{11\pi}{2} ). So this is going to be equal to; this is equal to negative ( 9 ) times; I'm going to create some space here, so actually, that's probably more space than I need; it's going to be ( \cos(6\pi) ) minus ( \cos\left(\frac{11\pi}{2}\right) ).

Well, what is ( \cos(6\pi) ) going to be? Well, cosine of any multiple of ( 2\pi ) is going to be equal to ( 1 ). You could view ( 6\pi ) as going around the unit circle ( 3 ) times, so this is the same thing as ( \cos(2\pi) ) or the same thing as ( \cos(0) ), so that is going to be equal to ( 1 ).

If that seems unfamiliar to you, I encourage you to review the unit circle definition of cosine. And what is ( \cos\left(\frac{11\pi}{2}\right) )? Let's see, let's subtract some multiple of ( 2\pi ) here to put it in values that we can understand better.

So this is—let me write it here—( \cos\left(\frac{11\pi}{2}\right) ) that is the same thing as; let's see if we were to subtract this; this is the same thing as ( \frac{11\pi}{2} - 5\frac{\pi}{2} ) right? Yeah. So this is—so we could view this as we could subtract ( 4\pi ) which is going to be; we could write that as ( \frac{8\pi}{2} ). In fact, no, let's subtract ( 4\pi ), which is ( \frac{8\pi}{2} ).

So once again, I'm just subtracting a multiple of ( 2\pi ), which isn't going to change the value of cosine, and so this is going to be equal to ( \cos\left(\frac{3\pi}{2}\right) ).

And if we imagine the unit circle, let me draw the unit circle here; so it's my ( y )-axis, my ( x )-axis, and then I have the unit circle. So, whoops, all right, the unit circle just like that. So if we start at—this is ( 0 ), then you go to ( \frac{\pi}{2} ), then you go to ( \pi ), then you go to ( \frac{3\pi}{2} ). So that's this point on the unit circle, so the cosine is the ( x )-coordinate. So this is going to be zero. This is zero.

So we get ( 1 - 0 ), so everything in the brackets evaluates out to ( 1 ). And so we are left with, so let me do that; so all of this is equal to ( 1 ). And so you have negative ( 9 ) times ( 1 ), which of course is just negative ( 9 ), is what this definite integral evaluates to.