Video: Evaluating the Definite Integral of a Trigonometric Function

Determine ∫_(βˆ’7)^(βˆ’5) 6 sin ((πœ‹π‘§)/4) d𝑧.

04:40

Video Transcript

Find the definite integral of six sin πœ‹π‘§ over four between the limits negative seven and negative five.

So the first thing we do is we can take our six outside of our integration. And we do that because that’s a constant term. And it wouldn’t affect our integration. So now, what we have is six multiplied by the definite integral of sin πœ‹π‘§ over four between the limits of negative seven and negative five. So now, to remind us what we do when we’re looking to find a definite integral. So if we want to find a definite integral between the limits π‘Ž and 𝑏 of a function, then what we do is we integrate that function, substituting 𝑏 for π‘₯ and π‘Ž for π‘₯. And then, we subtract the value with π‘Ž substituted in from the value with 𝑏 substituted in.

So now, what we’re gonna need to do is integrate sin πœ‹π‘§ over four. And to complete this, what I’m gonna do is use substitution. So I’m gonna say that 𝑒 is equal to πœ‹π‘§ over four. So now, if we differentiate this with respect to 𝑧, what we get is d𝑒 d𝑧 is equal to πœ‹ over four. And we get that because if we’re differentiating πœ‹π‘§ over four, it’s the same as πœ‹ over four multiplied by the derivative of 𝑧, where the derivative of 𝑧 is just gonna be one. So therefore, we get our πœ‹ over four.

And in integration by parts, we think of d𝑒 and d𝑧 as differentials. So then, we can alternatively write this as d𝑒 is equal to πœ‹ over four d𝑧. And we can see that although d𝑒 over d𝑧 is definitely not a fraction, we do treat it a little bit like one in this process. So then, what we do is we divide through by πœ‹ over four, which is the same as multiplying by four and dividing by πœ‹. So we get four over πœ‹. d𝑒 is equal to d𝑧.

So great, we now have d𝑧 in terms of d𝑒. So we can substitute this back into our integration. And what we’re gonna have is six multiplied by the definite integral of four over πœ‹ sin 𝑒 d𝑒 between the limits negative seven and negative five. So once again, we can make things easier by taking out our coefficient. So we got a constant term. So what we have is 24 over πœ‹. And we get that because six multiplied by four over πœ‹ is 24 over πœ‹ multiplied by the definite integral of sin 𝑒 d𝑒 between the limits negative seven and negative five. So if we integrate sin 𝑒, we get negative cos 𝑒. This is one of our standard integrations because we know if we integrate sin π‘₯, we get negative cos π‘₯.

And at this stage, we could also use the definition of our substitution to change our limits. However, I’m going to replace 𝑒 with πœ‹π‘§ over four after integration, which has the same effect. So now, what we need to do is substitute back in our 𝑒. So we’ve got negative cos πœ‹π‘§ over four instead of negative cos 𝑒. So this is gonna give us 24 over πœ‹ multiplied by. Then, we’ve got negative cos of negative five πœ‹ over four minus negative cos of negative seven πœ‹ over four. And we’ve got that cause we’ve substituted in our negative seven and negative five for 𝑧.

So now, before we calculate this, we need to make sure that our calculator is in the correct setting. And we need to make sure it’s in radians. It should have a little r or a little rad in the corner of the display. And that’s because we’re dealing with the angles that are in terms of πœ‹. So we know that we’re dealing with radians. So it’s gonna give us 24 over πœ‹ multiplied by root two over two minus negative root two over two.

So this brings us to our final answer, which is 24 root two over πœ‹. And we got that because if we had root two over two minus negative root two over two. Well, if we subtract a negative, it’s the same as adding. So we get two root two over two. Well, if we have two root two over two, the twos will cancel. So we just get root two. So then, we get 24 over πœ‹ multiplied by root two, which gives us 24 root two over πœ‹.

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