### Video Transcript

Use De Moivreβs theorem to find the
exact value of the definite integral between zero and π by two of cos π to the
fifth power dπ.

De Moivreβs theorem tells us that
π to the power of ππ is equal to cos π plus π sin π. Now, there is an extra result. Suppose we let π§ be equal to cos
π plus π sin π, then π§ plus one over π§ is equal to two cos π and π§ minus one
over π§ is two π sin π. And we can generalize further and
say π§ to the πth power plus one over π§ to the πth power is two cos ππ, whereas
π§ to the πth power minus one over π§ to the πth power is two π sin ππ. In fact, in this video, weβre just
going to use the identity for two cos ππ.

Now, letβs go back to our
question. Weβre looking to evaluate a
definite integral of cos π to the fifth power. We know that two cos π is π§ plus
one over π§. 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 the fifth power equals
π§ plus one over π§ 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 plus π to the
πth power is equal to the sum from π equals zero to π of π choose π times π to
the power of π minus π times π to the πth power. When π is equal to five, we get π
to the fifth power plus five choose one π to the fourth power π plus five choose
two π cubed π 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 π§ to the fifth power. The second is five times π§ to the
fourth power times one over π§, which simplifies to five π§ cubed. The third is 10π§ cubed times one
over π§ squared. Well, that becomes 10π§ cubed times
one over π§ squared. So thatβs 10π§. We then have 10π§ squared times one
over π§ cubed. And this simplifies to 10 times one
over π§. Our second to last term is five
times π§ times one over π§ to the fourth power, which simplifies to five times one
over π§ cubed. Our very final term is one over π§
to the fifth power, which is, of course, simply one over π§ to the fifth power.

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

In our first expression, π is
equal to five. And so, we can write this as two
cos five π. In this second expression, π is
equal to three. So we have five times two cos three
π. Now, whilst itβs not completely
obvious, in this third expression π is equal to one. So we get 10 times two cos π. We now almost have an expression
for cos π to the fifth power purely in terms of cos of multiples of π. 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 to the
fifth power is equal to a sixteenth times cos of five π plus five cos three π plus
10 cos π.

And weβre now ready to find the
exact value of the definite integral between zero and π by two of cos π to the
fifth power dπ. Letβs replace cos π to the fifth
power with this new expression. We now see that weβre looking to
evaluate the definite integral between zero and π by two of a sixteenth times cos
five π plus five cos three π plus 10 cos π with respect to π. Before we do, letβs 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 ππ, where π is a real constant with respect to π, is one over π
times sin ππ plus some constant of integration π.

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

Now, sin of five π by two is one,
sin of three π by two is negative one, and sin of π by two is one. So we have one sixteenth times a
fifth minus five-thirds plus 10. Weβre 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, weβre done. Weβve used De Moivreβs to find the
exact value of the definite integral between zero and π by two of cos π to the
fifth power with respect to π. Itβs eight fifteenths.