Video: Evaluating the Definite Integral of a Trigonometric Function

Evaluate ∫_(πœ‹/6) ^(πœ‹) βˆ’4 sin πœƒ dπœƒ.

03:06

Video Transcript

Evaluate the definite integral of negative four sin πœƒ between the limits πœ‹ and πœ‹ on six.

So the first thing we’re gonna do is we’re gonna take the negative four, so our constant term, outside of our integral. And this is because this won’t affect the integral of sin πœƒ. So now, what we have is negative four multiplied by the definite integral of sin πœƒ between the limits of πœ‹ and πœ‹ 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 𝑏 and π‘Ž, then this is equal to the integral of that function with 𝑏 substituted in instead of π‘₯ minus the integral of that function with π‘Ž substituted in instead of π‘₯. So therefore, we’re gonna get negative four multiplied by β€” and then we’ve got negative cos πœƒ. And then, we still got our same limits. And that’s because if you integrate sin πœƒ, it’s one of our known integrals, we get negative cos πœƒ.

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

And then, this is minus negative root three over two. And that’s because cos πœ‹ over six is equal to root three over two. And that’s one of the standard trig values we should know. But if we want to think about it in degrees, because if that’s easier for you to remember, then πœ‹ 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, let’s see if we can simplify this. Well, we’re gonna get negative four. And that’s because negative four multiplied by one is negative four and then minus β€” and we’ve got four root three over two. And we’ve got minus four root three over two because we had negative four multiplied by positive root three over two. And that’s 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œƒ between the limits of πœ‹ and πœ‹ over six is equal to negative four minus two root three.

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