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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 6

Lesson 10: Finding antiderivatives and indefinite integrals: basic rules and notation: definite integrals- Definite integrals: reverse power rule
- Definite integrals: reverse power rule
- Definite integral of rational function
- Definite integral of radical function
- Definite integral of trig function
- Definite integral involving natural log
- Definite integrals: common functions
- Definite integral of piecewise function
- Definite integral of absolute value function
- Definite integrals of piecewise functions

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# Definite integral of trig function

Sal finds the definite integral of 9sin(x) between 11π/2 and 6π.

## Want to join the conversation?

- at4:02i don't understand why he can just subtract?(5 votes)
- That's a good question Vivian! See, he is not actually subtracting; he is putting it in terms that are easier to understand. Cosine is a periodic function with a period of 2π. This means that every 2π, the function repeats. This makes f(0) = f(2π) = f(4π) and etc. It is a ton easier to know the cosine of a more familiar value, namely those that are 0<x<2π. Sal simply rewrote 11π/2 as 3π/2 + 4π. Just like f(0) = f(2π), we can also say that f(3π/2) = f(7π/2) = f(11/2) and etc. I hope that helped!(20 votes)

- Is it acceptable to state that the antiderivative of sin(x) is -cos(x) instead of multiplying the sin(x) and 9 by -1? I believe it results in the same answer in this case.(7 votes)
- I can see no reason not to, as long as you keep careful track of the signs.

(I do it that way as well.)(7 votes)

- This feels wrong to me.

I understand that cos(6pi) = cos(0), and cos(11pi/2) == cos(3pi/2), but those are just the point values of the the functions. When we integrate, we're taking tiny slices and adding them together. Don't have to do this over the entire range 6pi to 11pi/2?

It seems like this shortcut is disregarding the fact that this is a sum.(3 votes)- Because cos(x) is a constant oscillating function, the areas will keep on being the same because the area under a curve is - and above is + so because in intervals of pi/2, it is equal to each other if the intervals stay the same even if the bounds change(7 votes)

- So you can integrate in radians too?(3 votes)
- Actually it is easier to differentiate and integrate using radians instead of degrees. The formulas for derivatives and integrals of trig functions would become more complicated if degrees instead of radians are used (example: the antiderivative of cos(x) is sin(x) + C if radians are used, but is (180/pi)sin(x) + C if degrees are used). This is one of the main reasons why radian measurement is taught in trigonometry.

The simplicity of using radians instead of degrees should come as no surprise. Whereas the choice of 360 degrees in a circle is arbitrary, the fact that there are 2*pi radians (radiuses) in a circle is a consequence of the fact that the circumference is 2*pi*r. So radians, unlike degrees, is a natural unit of angle measurement.

Have a blessed, wonderful day!(4 votes)

- When do we know to use either radian or degree mode on our calculator when calculating integrals?(1 vote)
- When calculating integrals, always use radians.(3 votes)

- In this example Sal evaluates from 11/2pi to 6pi. In earlier videos on integration with non trig functions we saw that evaluating an integral from, for example 10 to 2 (where 10 is at the buttom and 2 at the top) required us to swap the signs for the final result.

Does this not apply to integrals of periodic functions?(1 vote)- I think you only need to do that if you're using Riemann sums.(2 votes)

- Not sure why at4:30he takes the answer to cos(3pi/2) and plugs that in for cos(11pi/2). I don't see how that works considering that he's ignoring the 4pi he subtracted to put the cosine in terms of 2pi. Wouldn't it be easier just to use a calculator instead of converting?(1 vote)
- The 4pi can be ignored there, as the cosine of even multiples of pi is always 0. So, if you write 11pi/2 as (4pi + 3pi/2), you'll see that the 4pi throws you back on the positive x axis, and then the 3pi/2 moves you into the first quadrant, where cosine is positive. So, it boils down to just cos(3pi/2)

Yes, it would be easier to use a calculator. But, you won't have a calculator handy everytime. It's good to know how to break down big angles into smaller, calculable ones(2 votes)

- Ya feel me, right?(1 vote)
- at1:00why did he just multiply by -1 since I know that the antiderivative of sin x = - cos x? The answer is still the same(1 vote)
- Where in real life do we use these kinds of things?(1 vote)

## Video transcript

- [Voiceover] Let's see if we can evaluate the definite integral from 11 pi over two to six pi of nine sine of x dx. So the first thing,
let's see if we can take the antiderivative of nine sine of x, and 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
nine times the integral from 11 pi over two to
six pi of sine of x dx. And what's the
anitderivative of sine of x? Well we know, from our
derivatives, that the derivative with respect
to x of cosine of x is equal to negative sine of x, negative sine of x. So can we construct this in some way so this is a negative sine of x? What if I multiplied, on the inside, what if I multiplied it by negative one? Well I can't just
multiply in only one place by negative one, I need to multiply by negative one twice so I'm
not changing its value. So what if I said negative
nine times negative sine of x? Well this is still
gonna be nine sine of x. If you took negative nine
times negative sine of x, it is nine sine of x,
and I did it this way because now negative sine of x, it matches the derivative of cosine of 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 nine out front,
negative nine times, times, I'll put it in brackets, negative nine times, the antiderivative of negative sine of x, well that is just going to be cosine of x, cosine of x, and we're going to
evaluate it at its bounds. We're going to evaluate it at six pi, let me do that in a
color I haven't used yet, we're gonna do that at six pi, and we're also going to do that at 11 pi over two, 11 pi over two, so this
is going to be equal to, this is equal to negative nine times, I'm going to create some space here, so actually that's probably
more space than I need, it's going to be cosine of six pi, cosine of six pi, minus cosine of 11 pi over two, cosine of 11 pi over two. Well what is cosine of six pi going to be? Well, cosine of any multiple of two pi is going to be equal to one. You could view six pi as, we're going around the unit circle three times. So this is the same thing
as cosine of two pi, or the same thing as cosine of zero, so that is going to be equal to one. If that seems unfamiliar to you I encourage you to review the unit circle definition of cosine. And what is cosine of 11 pi over two? Let's see, let's subtract some, let's subtract some
multiple of two pi here to put it in values that
we can understand better. So this is, so let me write it here, cosine of 11 pi over 2,
that is the same thing as, let's see, if we were to subtract, this is the same thing as cosine of 11 pi over two minus,
this is the same thing as five and one half pi, right? Yeah, so this is, so
we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two. In fact, no, let's subtract five, no, let's subtract four pi,
which is eight pi over 2. So once again, I'm just subtracting a multiple of two pi,
which isn't gonna change the value of cosine, and
so this is going to be equal to cosine of 3 pi over 2. And if we imagine the unit circle, and 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. So the unit circle just like that. So if we start at, this is zero, then you go to pi over two, then you go to pi, then you go
to three pi over two, 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 this
is zero, and so we get one minus zero, so
everything in the brackets evaluates out to one,
and so we are left with, so let me do that, so all
of this is equal to one. And so you have negative nine times one, which of course is just negative nine, is what this definite
integral evaluates to.