### Video Transcript

Evaluate the definite integral of negative four sin π between the limits π and π on six.

So the first thing weβre gonna do is weβre gonna take the negative four, so our constant term, outside of our integral. And this is because this wonβt affect the integral of sin π. So now, what we have is negative four multiplied by the definite integral of sin π between the limits of π and π over six.

Now just remind us what we do with definite integrals is if we want to find the definite integral of a function between π and π, then this is equal to the integral of that function with π substituted in instead of π₯ minus the integral of that function with π substituted in instead of π₯. So therefore, weβre gonna get negative four multiplied by β and then weβve got negative cos π. And then, we still got our same limits. And thatβs because if you integrate sin π, itβs one of our known integrals, we get negative cos π.

So now, what we do is we use the rule that we described earlier. And weβre gonna substitute in π and π over six for π. And when we do that, we get negative four multiplied by negative cos π minus negative cos π over six. Well, these are known values because, first of all, we know that cos π is gonna be equal to negative one. So then, weβre gonna get negative four multiplied by one. And we get one because we had negative cos π. And cos π was equal to negative one. And negative negative one is positive one.

And then, this is minus negative root three over two. And thatβs because cos π over six is equal to root three over two. And thatβs one of the standard trig values we should know. But if we want to think about it in degrees, because if thatβs easier for you to remember, then π over six radians is equal to 30 degrees. And we know that cos 30 is equal to root three over two. Okay great.

So now, letβs see if we can simplify this. Well, weβre gonna get negative four. And thatβs because negative four multiplied by one is negative four and then minus β and weβve got four root three over two. And weβve got minus four root three over two because we had negative four multiplied by positive root three over two. And thatβs because we had minus and negative. So that became positive. So then, we can divide the numerator and the denominator of the second term by two.

So therefore, we can say that the definite integral of negative four sin π between the limits of π and π over six is equal to negative four minus two root three.