### Video Transcript

Evaluate the definite integral
between zero and π by four of sec of π‘ tan of π‘ π’ plus π‘ times cos of two π‘ π£
plus sin squared of two π‘ times cos of two π‘ π€ with respect to π‘.

Remember, to integrate a
vector-valued function, we simply integrate each component in the usual way. Since this is a definite integral,
we can use the fundamental theorem of calculus, which says that the definite
integral between π and π of π« of π‘ with respect to π‘ is equal to the
antiderivative capital π evaluated at π minus capital π evaluated at π. So essentially, all weβre going to
do is integrate each of our components with respect to π‘ and evaluate them
individually between the limits of zero and π by four.

Letβs complete this process for the
component for π’. Itβs the definite integral between
zero and π by four of sec π‘ tan π‘. We might recall the derivative of
sec of π‘ is equal to sec π‘ tan π‘. So this means the antiderivative of
sec π‘ tan π‘ must be sec π‘. So we use the fundamental theorem
of calculus. And we see that this is equal to
sec of π by four minus sec of zero. But of course, sec of π‘ is equal
to one over cos of π‘. So we need to work out one over cos
of π by four minus one over cos of zero. Thatβs root two minus one.

Weβve cleared some space so we can
evaluate the definite integral between zero and π by four of π‘ times cos of two
π‘. This time, we have the product of
two functions. So weβre going to use integration
by parts to evaluate. This says that the integral of π’
times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the integral of π£
times dπ’ by dπ₯ with respect to π₯. Weβll let π’ be equal to π‘. And thatβs because we know that dπ’
by dπ‘ is simply one. This makes the second integral much
simpler. This means dπ£ by dπ‘ must be cos
of two π‘. Then, the antiderivative of cos two
π‘ gives us π£. Itβs a half sin of two π‘.

Now, of course, weβre working with
π‘ as our parameter instead of π₯. But otherwise, the substitutions
are the same. We have π’ times π£. Thatβs π‘ times a half sin two π‘,
which we will shortly evaluate between zero and π by four minus the definite
integral between those limits of a half sin two π‘ times dπ’ by dπ‘, which is
one. We simplify this a little. And we see that weβre going to need
to work out the definite integral between zero and π by four of a half of sin two
π‘ with respect to π‘. Well, the antiderivative of a half
sin two π‘ is negative a quarter cos two π‘. And if we so wish, we can combine
these terms and then evaluate them between the limits of zero and π by four. When we substitute π by four, we
get π over eight plus zero. And when we substitute zero, we get
zero plus a quarter. So that gives us π by eight minus
a quarter.

Letβs clear some more space and
work out that final definite integral. This time, weβre going to use
integration by substitution. Weβre going to let π’ be equal to
sin two π‘. We know that dπ’ by dπ‘, the
derivative of sin two π‘ with respect to π‘, is two cos two π‘. And whilst dπ’ by dπ‘ isnβt a
fraction, we can equivalently rewrite this as a half dπ’ equals cos of two π‘
dπ‘. We then need to change our
limit. So we use our substitution to do
so. Our lower limit is when π‘ is equal
to zero. And when π‘ is equal to zero, π’ is
equal to sin of two times zero, which is zero. Then, when π‘ is equal to π by
four, π’ is equal to sin of two times π by four, which is one.

We then replace sin two π‘ with
π’. We can replace cos of two π‘ with a
half dπ’. And of course, we must make sure we
replace our limit. We might choose to take the
constant factor of a half outside our integral and focus on integrating π’ squared
itself. That gives us π’ cubed over
three. By substituting our limits, we get
a half times one cubed over three minus zero cubed over three, which is simply equal
to one-sixth. Now that weβve evaluated the
integrals of each of our component functions, weβre ready to return this into vector
form. Itβs root two minus one π’ plus π
over eight minus a quarter π£ plus one-sixth π€.