### Video Transcript

In this video, weβll learn how we
can use the antiderivative of inverse trigonometric functions to integrate more
complicated functions where itβs not immediately obvious how the substitution rule
and integration by parts can help. Itβs, therefore, important your
confident in differentiating inverse trigonometric functions of the form inverse
sine of π₯, inverse tan of π₯, and inverse secant π₯.

This entire video is made possible
thanks to the fundamental theorem of calculus. Remember, the first part of this
theorem says that if a function π is continuous, then the derivative of the
integral of a function π with respect to the variable π‘ over some interval π to
π₯ is equal to the function π with respect to π₯. Essentially, it describes the
derivative and integral as inverse processes. And luckily for us, this means that
if we can recognize the derivative of a function as our integrand, with a little
manipulation, we can quite easily integrate fairly nasty functions.

So at this stage, weβre going to
recall the general results for the derivatives of the key inverse trigonometric
functions that weβll be referring to throughout this video. The first weβre interested in is
the derivative of the function inverse sin of π₯ over π for real constants π. Itβs one over the square root of π
squared minus π₯ squared. And this is only valid for values
of inverse sin of π₯ over π greater than or equal to negative π by two and less
than π by two. With similar constraints on the
range of the function inverse tan of π₯ over π, we obtain its derivative to be π
over π squared plus π₯ squared.

And finally, we know that the
derivative of the inverse sec of π₯ over π is π over π₯ times the square root of
π₯ squared minus π squared. Now remember, when performing
calculus with trigonometric functions, we always work with radians. Weβre now going to have look at a
simple example of how these derivatives can help us to evaluate a function with a
simple inverse trigonometric function as its result.

Evaluate the definite integral
between the limits of one and root three of negative one over one plus π₯ squared
with respect to π₯.

Here, we have a definite integral
with limits of one and a square root of three. This means weβre going to need to
use the second part of the fundamental theorem of calculus to evaluate it. This tells us that if π is a real
value function on some closed interval π to π and capital πΉ is an antiderivative
of π in that closed interval such that capital πΉ prime of π₯ is equal to π of
π₯. Then if π is Riemann integrable on
the closed interval, then we can say that the definite integral between π and π of
π of π₯ is equal to capital πΉ of π minus capital πΉ of π. Essentially, we can evaluate the
integral here by finding the antiderivative of this function negative one over one
plus π₯ squared and evaluating it between root three and one.

Now, the integral of negative one
over one plus π₯ squared is not particularly nice. But actually we spot that we can
take out any constant factors and focus on the integral itself. So here, we take our constant
factor of negative one. And weβre now looking to evaluate
the negative integral of one over one plus π₯ squared between one and root
three. Then we spot that we have the
standard result for the derivative of the inverse tan of π₯ over π. Itβs π over π squared plus π₯
squared. And this, of course, means that the
antiderivative of π over π squared plus π₯ squared must be the inverse tan of π₯
over π.

Now, if we compare the function π
over π squared plus π₯ squared with our function one over one plus π₯ squared, we
can see that π is equal to one. So if we say that π of π₯ is equal
to one over one plus π₯ squared, then the antiderivative capital πΉ of π₯ must be
the inverse tan of π₯ over one, which is simply the inverse tan of π₯. By the second part of our theorem
then, we can say that the definite integral between one and root three of one over
one plus π₯ squared with respect to π₯ is equal to the inverse tan of root three
minus the inverse tan of one. And of course, we took out that
constant of negative one at the start.

We then know that the inverse tan
of root three is π by three, and the inverse tan of one is π by four. Weβre going to find the difference
between these two fractions by using a common denominator. We multiply the numerator and
denominator of our first fraction by four and the numerator and denominator of our
second fraction by three. And weβre looking to work out
negative four π over 12 minus three π over 12. Four π by 12 minus three π by 12
is π by 12. So our answer here is negative π
by 12. Now, we have just found the general
result for the indefinite integral of π over π squared plus π₯ squared with
respect π₯ for real constants π. Itβs the inverse tan of π₯ over π
plus some constant of integration π. And rather than jumping straight
into evaluating capital πΉ of π minus capital πΉ of π, we could, of course, have
included this extra step using those square brackets.

Find the most general
antiderivative capital πΊ of π£ of the function π of π£ equals four cos π£ plus
three over five root one minus π£ squared.

Remember, the antiderivative is
basically the opposite of the derivative. And another way of thinking about
it is to find the antiderivative capital πΊ of π£. We can find the indefinite integral
of this function. We, therefore, say that capital πΊ
of π£ is equal to the indefinite integral of lower case π of π£. Letβs replace π of π£ with the
function four cos π£ plus three over five times the square root of one minus π£
squared. We then recall a key property of
integrals; that is, the integral of the sum of two or more functions is equal to the
sum of the integral of each respective function. And we can, therefore, split our
integral up. And we see that capital πΊ of π£ is
equal to the integral of four cos π£ plus the integral of three over five times the
square root of one minus π£ squared.

Another key property we can apply
is that the integral of some constant times a function is equal to that constant
times the integral of the function. And so we can further rewrite this
as four times the integral of cos of π£ plus three-fifths times the integral of one
over the square of one minus π£ squared. Now, this is great because we can
use in general results for derivatives. Firstly, we know that the
derivative of sin π₯ is cos of π₯. So the antiderivative and,
therefore, the integral of cos of π£ is sin of π£. And of course, when dealing with
definite integrals, we add a constant of integration. Letβs call that π΄. So this first part becomes four
times sin of π£ plus π΄.

Next, we know that if the inverse
sin of π₯ over π is greater than or equal to negative π by two and less than or
equal to π by two, then its derivative is equal to one over the square root of π
squared minus π₯ squared. Now, in our example, π squared is
equal to one. So π must be equal to one
also. So the antiderivative of one over
the square root of one minus π£ squared and, therefore, the integral of this
function is the inverse sin of π₯ over one. And we add another constant of
integration π΅. Now, of course, the inverse sin of
π₯ over one can be written as the inverse sin of π₯.

Distributing our parentheses and
combining the constants into one new constant capital πΆ, we find that the general
antiderivative capital πΊ of π£ is equal to four sin of π£ plus three-fifths of the
inverse sin of π£ plus πΆ. And in this example, weβve seen
that the indefinite integral of one over the square root of π squared minus π₯
squared with respect to π₯ is the inverse sin of π₯ over π plus π.

Weβre now going to look at an
example that requires just a little more manipulation.

Evaluate the indefinite integral of
one over the square root of four π₯ squared minus 16 with respect to π₯.

At first glance, it might appear
that this has a simple result of an inverse trigonometric function. However, once we know that the
derivative of the inverse sec of π₯ over π is π over π₯ times the square root of
π₯ squared minus π squared, when the inverse secant of π₯ over π is greater than
zero, less than π, but not equal to π by two. Our integrand isnβt quite of this
form. We noticed particularly that we
have four π₯ squared instead of just π₯ squared. Weβre, therefore, going to need to
perform some manipulation. Weβre going to begin by multiplying
both the numerator and denominator of our fraction by two. Now remember, this doesnβt actually
change the integrand because itβs the equivalent of multiplying by one.

So weβre looking to evaluate the
indefinite integral of two over two π₯ times the square of four π₯ squared minus
16. Now, we notice that four π₯ squared
and 16 are both square numbers. This means we can write four π₯
squared minus 16 as two π₯ all squared minus four squared. And now, we notice that we can make
a substitution. If we let π’ be equal to two π₯
squared, then inside our square root sign, weβll have π’ squared minus four
squared. Notice that thatβs a lot closer to
the form weβre looking for. If π’ was equal to two π₯, then we
know that dπ’ by dπ₯ must be equal to two. And we can say equivalently that
dπ’ must be equal to two π₯.

Well notice, we can now replace two
dπ₯ with dπ’. We can replace two π₯ and two π₯
with π’, and our function is now in the form weβre looking for. We need to integrate one over π’
times the square root of π’ squared minus four squared with respect to π’. Looking back to our original
derivative, we noticed that the inverse sec of π₯ over π is the antiderivative of
π over π₯ times the square root of π₯ squared minus π squared. In our example, π must be equal to
four. Now, the numerator, of course, of
our fraction is one not four. So our integral will be a quarter
of the inverse sec of π’ over four plus π. Our final step is a look back at
our substitution, and we replace π’ with two π₯. And we see that our indefinite
integral is a quarter of the inverse sec of π₯ over two plus the constant of
integration π.

In our final example, weβre going
to look at an integral which involves manipulation by completing the square and a
clever little substitution.

Evaluate the indefinite integral of
one over π₯ squared minus π₯ plus one with respect to π₯.

Now, this isnβt a nice function to
integrate at all. So weβre going to need to do
something a little bit clever. Itβs certainly not the product of
two functions. So weβre not going to use
integration by parts. But if we do something special to
the denominator, we can actually use integration by substitution. Weβre going to complete the square
of the denominator of the expression π₯ squared minus π₯ plus one. Remember, we halve the coefficient
of π₯. Here, thatβs negative one, so half
of that is negative one-half. We, therefore, have π₯ minus a half
all squared in the brackets. Negative a half squared is
one-quarter, so we subtract that one quarter. And we see that our expression is
equivalent to π₯ minus a half all squared plus three quarters. And now, this is the integral that
weβre looking to evaluate.

Next, we need to spot that we know
that indefinite integral of π over π squared plus π₯ squared. Itβs the inverse tan of π₯ over
π. So to ensure that our function
looks a little like this, weβre going to perform a substitution. Weβre going to let π₯ minus a half
be equal to π’. Then this part will be π’
squared. The derivative of π₯ minus one-half
is one. So dπ’ by dπ₯ equals one, which
means that dπ’ is equal to dπ₯. So we can replace dπ₯ with dπ’ and
π₯ minus a half with π’. And we see that weβre actually
looking to find the indefinite integral of one over π’ squared plus
three-quarters.

Now, this still doesnβt quite look
like what weβre after. We need it to be π squared on the
denominator. Well, three-quarters is the same as
the square root of three quarters squared. So π here is equal to the square
root of three-quarters. And of course, since the numerates
of our fraction is one and not the square root of three-quarters, the integral is
one divided by the square root of three-quarters times the inverse tan of π’ over
the square root of three quarters plus π. One divided by the square root of
three quarters is two root three over three. And then we go back to our
substitution π’ equals π₯ minus a half. And we replace that in our
result. And finally, we distribute our
parentheses. The indefinite integral of one over
π₯ squared minus π₯ plus one with respect to π₯ is two root three over three times
the inverse tan of root three over three times two π₯ minus one plus the constant of
integration π.

In this video, weβve seen that we
can use the concept of antiderivatives to integrate functions which have inverse
trigonometric results. The integral of one over the square
root of π squared minus π₯ squared is the inverse sin of π₯ over π. The integral of π over π squared
plus π₯ squared is the inverse tan of π₯ over π. And the integral of π over π₯
times the square root of π₯ squared minus π squared is the inverse sec of π₯ over
π. We also saw that sometimes we need
to perform some manipulation and a clever substitution to achieve our results.