### Video Transcript

The integral between zero and π
over two of cos π divided by the square root of sin π with respect to π is
convergent. What does it converge to?

Straight away, we can notice that
the question has told us our integral is convergent. This gives us a hint that weβre
looking at an improper integral. Now, since we donβt see any
infinities as our limits of integration, itβs also implied that our integrand has a
discontinuity on the closed interval between zero and π over two, which are the
limits of the integral. Letβs look for this
discontinuity. Since the integrand that weβve been
given is a quotient, we should try and find where its denominator is equal to
zero. The square root of sin π is zero,
where sin π itself is also equal to zero. This occurs when π is equal to
zero plus ππ, which is of course just ππ. And π here is an integer. So here we see that our denominator
is equal to zero when π is equal to any of the following. The only one of these values which
occurs over the interval of our integration is zero itself.

Okay, for due diligence, letβs now
check what happens to our numerator when π is equal to zero. Our numerator is cos π. So this becomes cos of zero, which
is equal to one. We do this to check that when π
equals zero, our function is not equal to the indeterminate form of zero over
zero. Since we would then be looking at a
removable discontinuity. In actual fact, when π is equal to
zero, our function is equal to one over zero, which is the case of an infinite
discontinuity. Okay. Weβve now confirmed that we have an
improper integral. And we have an infinite
discontinuity at the lower limit of integration. The definition of an improper
integral specifically for this case tells us that if π is a continuous function on
the interval which is open at π and closed at π and discontinuous at π. Then the integral between π and π
of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the
right of the integral between π‘ and π of π of π₯ with respect to π₯. If this limit exists and is
finite.

Letβs now apply this to our
question. The integral that weβve been given
is equal to the limit as π‘ approaches zero from the right of the same integral, but
now between the limits of π‘ and π over two. Now, at this point, we might notice
that our integral is not entirely trivial. And weβll need to put in some work
to evaluate it. To move forward, weβre gonna be
using a π’-substitution, specifically π’ is equal to sin π. From this, we differentiate to find
that dπ’ by dπ is equal to cos π. An equivalent statement to this is
that dπ’ is equal to cos π dπ. We now do the following with our
substitution. We replace sin π with π’. And we replace cos π dπ with
dπ’. Weβre then left with the integral
of one over the square root of π’ with respect to π’. Another way of expressing this
integrand is π’ to the power of negative a half. Evaluating this, we get two π’ to
the power of a half, which, is of course, two times the square root of π’. We can now go back to our original
substitution and replace our π’ with sin π. We therefore have that our integral
is equal to two times the square root of sin π, of course, plus the constant of
integration.

Now that weβve done the legwork on
our integral, letβs use this result to move forward with our question. We use the antiderivative that
weβve just found for our integral. And we get the following. We now input our limits of
integration π‘ and π over two. And weβre left with the following
limit. We can now take a direct
substitution approach to our limit. And we can evaluate since sin of π
over two is equal to one. And sin of zero is equal to
zero. Weβre then left with two times the
square of one minus two times the square root of zero. Of course, the square root of one
is just one. And this entire term is just
zero. The answer that weβre left with is
therefore two. Upon reaching this result, we have
answered our question. We used our definition to find that
the integral given in the question converges to a value of two.