Question Video: Using De Moivre’s Theorem to Integrate a Trigonometric Functions

Use de moivre’s theorem to find the exact value of ∫_(0) ^(πœ‹/2) cosβ΅πœƒ dπœƒ.

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Video Transcript

Use De Moivre’s theorem to find the exact value of the definite integral between zero and πœ‹ by two of cos πœƒ to the fifth power dπœƒ.

De Moivre’s theorem tells us that 𝑒 to the power of π‘–πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ. Now, there is an extra result. Suppose we let 𝑧 be equal to cos πœƒ plus 𝑖 sin πœƒ, then 𝑧 plus one over 𝑧 is equal to two cos πœƒ and 𝑧 minus one over 𝑧 is two 𝑖 sin πœƒ. And we can generalize further and say 𝑧 to the 𝑛th power plus one over 𝑧 to the 𝑛th power is two cos π‘›πœƒ, whereas 𝑧 to the 𝑛th power minus one over 𝑧 to the 𝑛th power is two 𝑖 sin π‘›πœƒ. In fact, in this video, we’re just going to use the identity for two cos π‘›πœƒ.

Now, let’s go back to our question. We’re looking to evaluate a definite integral of cos πœƒ to the fifth power. We know that two cos πœƒ is 𝑧 plus one over 𝑧. And we can raise both sides of this equation to the fifth power. By distributing the five over our parentheses, we could see that we can write this as 32 cos πœƒ the fifth power equals 𝑧 plus one over 𝑧 to the fifth power. We are going to distribute the parentheses on the right-hand side of this equation. And to do so, we recall the binomial theorem.

This says that π‘Ž plus 𝑏 to the 𝑛th power is equal to the sum from π‘˜ equals zero to 𝑛 of 𝑛 choose π‘˜ times π‘Ž to the power of 𝑛 minus π‘˜ times 𝑏 to the π‘˜th power. When 𝑛 is equal to five, we get π‘Ž to the fifth power plus five choose one π‘Ž to the fourth power 𝑏 plus five choose two π‘Ž cubed 𝑏 squared and so on. But of course, five choose one is equal to five. Five choose two is equal to 10. Five choose three is also equal to 10. And five choose four is equal to five.

This means that when we do distribute this parenthesis, the first term is 𝑧 to the fifth power. The second is five times 𝑧 to the fourth power times one over 𝑧, which simplifies to five 𝑧 cubed. The third is 10𝑧 cubed times one over 𝑧 squared. Well, that becomes 10𝑧 cubed times one over 𝑧 squared. So that’s 10𝑧. We then have 10𝑧 squared times one over 𝑧 cubed. And this simplifies to 10 times one over 𝑧. Our second to last term is five times 𝑧 times one over 𝑧 to the fourth power, which simplifies to five times one over 𝑧 cubed. Our very final term is one over 𝑧 to the fifth power, which is, of course, simply one over 𝑧 to the fifth power.

Now, we’re going to gather terms which have similar powers for 𝑧. We have 𝑧 to the fifth power plus one over 𝑧 to the fifth power. Then five 𝑧 cubed plus five times one over 𝑧 cubed, which is five times 𝑧 cubed plus one over 𝑧 cubed. And finally, 10𝑧 plus 10 times one over 𝑧. And factoring by this 10, we get 10 times 𝑧 plus one over 𝑧. And this step was really important because we can now go back to one of our earlier identities. That was 𝑧 to the 𝑛th power plus one over 𝑧 to the 𝑛th power equals two cos π‘›πœƒ.

In our first expression, 𝑛 is equal to five. And so, we can write this as two cos five πœƒ. In this second expression, 𝑛 is equal to three. So we have five times two cos three πœƒ. Now, whilst it’s not completely obvious, in this third expression 𝑛 is equal to one. So we get 10 times two cos πœƒ. We now almost have an expression for cos πœƒ to the fifth power purely in terms of cos of multiples of πœƒ. Our last step is to divide everything through by 32. When we do, we notice that each factor of two cancels. And so, we find that cos πœƒ to the fifth power is equal to a sixteenth times cos of five πœƒ plus five cos three πœƒ plus 10 cos πœƒ.

And we’re now ready to find the exact value of the definite integral between zero and πœ‹ by two of cos πœƒ to the fifth power dπœƒ. Let’s replace cos πœƒ to the fifth power with this new expression. We now see that we’re looking to evaluate the definite integral between zero and πœ‹ by two of a sixteenth times cos five πœƒ plus five cos three πœƒ plus 10 cos πœƒ with respect to πœƒ. Before we do, let’s take out the common factor of one sixteenth. And then, we can see we can simply focus on integrating term by term. We know that the indefinite integral of cos π‘›πœƒ, where 𝑛 is a real constant with respect to πœƒ, is one over 𝑛 times sin π‘›πœƒ plus some constant of integration 𝑐.

Now, of course, we’re working with a definite integral. So we first integrate cos five πœƒ and we get a fifth sin five πœƒ. But we no longer need the constant of integration. Then, when we integrate five cos three πœƒ, we get five times a third or five-thirds of sin three πœƒ. And finally, we integrate 10 cos πœƒ and we get 10 sin πœƒ. So we have a sixteenth times one-fifth sin five πœƒ plus five-thirds sin three πœƒ plus 10 sin πœƒ. We need to evaluate this between zero and πœ‹ by two. Now, in fact, sin of zero is zero. So we have a sixteenth times one-fifth sin of five πœ‹ by two plus five-thirds sin of three πœ‹ by two plus 10 sin of πœ‹ by two take away zero.

Now, sin of five πœ‹ by two is one, sin of three πœ‹ by two is negative one, and sin of πœ‹ by two is one. So we have one sixteenth times a fifth minus five-thirds plus 10. We’re going to add these fractions by creating a common denominator. And in this case, that needs to be 15. So we multiply the numerator and denominator of one-fifth by three. We multiply the numerator and denominator of five-thirds by five. And then, we write 10 as 10 over one. And we multiply both the numerator and denominator by 15. So this becomes three fifteenths minus 25 fifteenths plus 150 fifteenths. So we get a sixteenth times 128 over 15. But 128 can be divided by 16. And when we do, we get eight. This, therefore, simplifies quite nicely to eight fifteenths.

And so, we’re done. We’ve used De Moivre’s to find the exact value of the definite integral between zero and πœ‹ by two of cos πœƒ to the fifth power with respect to πœƒ. It’s eight fifteenths.

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