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Video transcript

let's see if we can evaluate the definite integral from 11 PI over 2 to 6 PI of 9 sine of X DX so the first thing let's see if we can take the antiderivative of 9 sine of X so 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 11 PI over 2 to 6 pi of sine of X DX and what's the antiderivative sine of X well we know from our derivatives that the derivative with respect to X of cosine of X is equal to 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 it on the inside what if I multiplied it by a negative 1 well I can't just multiply it 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 sine of X well this is still going to be 9 sine of X if you took negative 9 times negative sine of X it is 9 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 9 out front negative 9 times x and I'll put it in put it in brackets negative 9 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 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 11 PI over 2 11 PI over 2 and 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 I probably more space than I need it's going to cosine of 6 pi cosine of 6 6 pi cosine of 6 PI - minus cosine of 11 PI over 2 cosine of 11 PI over 2 well what is cosine of 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 we're going around the unit circle 3 times so this is the same thing as cosine of 2 pi or the same thing as cosine of 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 cosine of 11 PI over 2 let's see let's subtract some let's subtract some multiple of 2 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 2 - let's see this is the same thing as 5 and 1/2 pi right yeah so this is so we could view this as we could subtract let's subtract 4 pi which is going to be we could write that as 8 PI over 2 in fact now let's subtract 5 let's subtract now let's subtract 4 pi which is 8 PI over 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 cosine of 3 PI over 2 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 two 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 9 times 1 which of course is just negative 9 is what this definite integral evaluates to
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