### Video Transcript

In this video, weβll learn how to
find indefinite integrals of trigonometric functions. Weβll begin by recalling the first
part of the fundamental theorem of calculus before looking at how this helps us to
integrate a number of trigonometric functions and the applications of these
integrals. We begin by stating the first part
of the fundamental theorem of calculus. In this theorem, we let π be a
continuous real valued function defined on a closed interval π to π. Then capital πΉ is the function
defined for all π₯ in this closed interval by capital πΉ of π₯ equals the integral
of π of π‘ with respect to π‘ evaluated between π and π₯. Then capital πΉ must be uniformly
continuous on this closed interval and differentiable on the open interval π to π
such that capital πΉ prime of π₯ is equal to π of π₯ for all π₯ in the closed
interval π to π.

This last part says πΉ is the
antiderivative of the function π. The function whose derivative is
equal to the original function. And essentially, what this tells us
is that integration is the reverse process, the differentiation. So letβs begin by looking at the
function π of π₯ equals sin of ππ₯ for real constants π. And of course, π₯ must be a radiant
measure. We recall that the derivative of
cos of ππ₯ is negative π sin of ππ₯. We can, therefore, say that the
indefinite integral of negative π sin of ππ₯ evaluated with respect to π₯ must be
cos of π₯. And donβt forget that since weβre
working with an indefinite integral, we need to add this constant of integration;
letβs call that π.

Okay, this is a great start. But weβre actually looking to find
the indefinite integral of sin of ππ₯ not negative π sin of ππ₯. Weβre allowed, though, to take the
constant negative π outside of the integral. And we see that negative π times
the indefinite integral of sin of ππ₯ is equal to cos of ππ₯ plus π. And since negative π is just a
constant, we divide both sides by negative π. And we obtain the indefinite
integral of sin of ππ₯ to be negative one over π times cos of ππ₯ plus capital
πΆ. And you might notice that Iβve
changed from a lowercase π to an uppercase πΆ. And this is simply because weβve
divided our original constant by another constant. So I want to represent that this
number has actually changed in value.

We can repeat this process for the
function π of π₯ equals cos π of π₯. Once again, π is a real constant
and π₯ is measured in radians. Weβre going to use the fact that
the derivative of sin of ππ₯ is equal to π cos of ππ₯. And we can, therefore, say that the
indefinite integral of π cos of ππ₯ evaluated with respect to π₯ must be sin ππ₯
plus the constant of integration π. We take the constant π outside of
the integral, and the right-hand side of our equation remains unchanged. Finally, we divide through by
π. And we obtain the integral of cos
of ππ₯ evaluated with respect to π₯, to be one over π times sin of ππ₯ plus
capital πΆ.

Now you might recall that the
process for differentiating sine and cosine functions forms a cycle. That is, the derivative of sin π₯
is cos π₯. And if we differentiate again, we
get negative sin π₯. Differentiating again, we get
negative cos of π₯. And differentiating one more time,
we go back to the start and get sin of π₯. We reverse this cycle for
integration as shown. Letβs now look at some examples
demonstrating the integration of the sin and cos functions.

Determine the indefinite integral
of negative sin of π₯ minus nine cos of π₯ evaluated with respect to π₯.

It can be useful to recall the
properties of integrals before evaluating this. Firstly, we know that the integral
of the sum of two or more functions is actually equal to the sum of the integrals of
those respective functions. And we also know that we can take
any constant factors outside of the integral and focus on integrating the expression
in π₯ itself. This means we can rewrite our
integral as negative the integral of sin π₯ evaluated with respect to π₯ minus nine
times the integral of cos of π₯ evaluated with respect to π₯. This means we can rewrite our
integral as shown. And this means we can recall the
general results for the integral of the sine and cosine functions.

The indefinite integral of sin of
ππ₯ is negative one over π cos ππ₯ plus that constant π. And the indefinite integral of cos
of ππ₯ is one over π times sin of ππ₯ plus π. So our integral is negative
negative cos of π₯ plus π΄ minus nine times sin of π₯ plus π΅. Then Iβve chosen π΄ and π΅ here to
show that these are different constants of integration. Distributing the parentheses and
combining our constants in to one constant, we find the integral of negative sin π₯
minus nine cos of π₯ evaluated with respect to π₯ to be cos of π₯ minus nine sin of
π₯ plus π.

Determine the indefinite integral
of negative eight sin of eight π₯ minus seven cos of five π₯ evaluated with respect
to π₯.

In this question, weβre looking to
integrate the sum of two functions in π₯. We begin by recalling the fact that
the integral of the sum of two or more functions is actually equal to the sum of the
integrals of those respective functions. So we can write our problem as the
integral of negative eight sin of eight π₯ with respect to π₯ plus the integral of
negative seven cos of five π₯ dπ₯. We also know that we can take any
constant factors outside of the integral and focus on integrating each expression in
π₯. So we can rewrite our problem
further as negative eight times the integral of sin of eight π₯ with respect to π₯
minus seven times the integral of cos of five π₯ with respect to π₯.

Next, we recall the general results
for the integral of sine and cosine. The integral of sin of ππ₯ is
negative one over π cos of ππ₯ plus π. And the indefinite integral of cos
of ππ₯ is one over π sin of ππ₯ plus π. We integrate each function,
respectively, and we see that the integral of sin of eight π₯ is negative
one-eighths cos of eight π₯ plus π΄. And the indefinite integral of cos
of five π₯ is a fifth sin of five π₯ plus π΅. And Iβve chosen π΄ and π΅, as
opposed to just one value of π, to show that these are actually different
constants.

Our final step is to distribute the
parentheses. Negative eight times negative
one-eighths cos of eight π₯ is just cos of eight π₯. Negative seven times one-fifth of
sin of five π₯ is a negative seven-fifths sin of five π₯. Finally, we multiply negative eight
by π΄ and negative seven by π΅. And we end up with this new
constant πΆ. And we found that the integral we
required is cos of eight π₯ minus seven-fifths of sin of five π₯ plus πΆ.

Weβre now going to consider some
alternative derivatives. We recall that the derivative of
tan ππ₯ is π sec squared ππ₯. Now, you might like to pause the
video for a moment and consider what this tells us about the integral of sec squared
ππ₯. Well, letβs have a look. We recall that the first part of
the fundamental theorem of calculus essentially tells us that integration is the
reverse process for differentiation. And we can see then that the
integral of π sec squared ππ₯ evaluated with respect to π₯ must be tan of ππ₯ and
since this is an indefinite integral plus π. We take out this constant factor of
π and then we divide. And we see that the integral of sec
squared of ππ₯ evaluated with respect to π₯ is one over π tan of ππ₯ plus capital
πΆ.

It is outside of the scope of this
video to spend too long looking at the other integrals. But using much the same method, we
obtain the following integrals relating to the derivatives of the reciprocal
trigonometric functions. The integral of csc ππ₯ cot ππ₯
is negative one over π csc ππ₯. The integral of sec ππ₯ tan ππ₯
is one over π sec ππ₯. And the derivative of csc squared
ππ₯ is negative one over π times cot of ππ₯. Weβre now going to have a look at
some of the examples of these results and how often times the trigonometric identity
will be required to help us evaluate these integrals.

Determine the indefinite integral
of negative sec squared six π₯ evaluated with respect to π₯.

To answer this question, itβs
almost simply enough to quote the general result for the integral of sec squared of
ππ₯. Itβs one over π times the tan of
ππ₯. It is sensible, though, to take a
factor of negative one outside of the integral as shown. And when we do, we obtain the
solution to be negative one times a sixth of tan six π₯ plus π. And all thatβs left is to
distribute the parentheses. Negative one times a sixth of tan
six π₯ is negative one-sixth tan of six π₯. And negative one times π gives us
this new constant, capital πΆ. And so we find our indefinite
integral to be negative one sixth of tan of six π₯ plus πΆ.

Determine the indefinite integral
of two cos cubed three π₯ plus one over nine cos square three π₯ evaluated with
respect to π₯.

This question does, at first
glance, look quite tricky. However, we should spot that we can
actually simplify this quotient somewhat. We essentially reversed the process
we would take when adding two fractions. And we see that we can write the
quotient as two cos cubed three π₯ over nine cos squared three π₯ plus one over nine
cos squared three π₯. The first fraction simplifies to
two-ninths of cos of three π₯. And then to help us spot what to do
next, letβs rewrite the second fraction as a ninth times one over cos squared three
π₯. Next, we recall that the integral
of the sum of two or more functions is actually equal to the sum of the integrals of
those respective functions. And we write this as the integral
of two-ninths of cos of three π₯ evaluated with respect to π₯ plus the integral of a
ninth times one over cos squared three π₯, again, evaluated with respect to π₯.

We also know that we can take any
constant factors outside of the integral and focus on integrating the expression in
π₯. So we write this further as
two-ninths times the integral of cos of three π₯ dπ₯ plus a ninth times the integral
of one over cos squared three π₯ dπ₯. We can quote the general result for
the integral of cos of ππ₯. Itβs one over π sin of ππ₯. And this means the integral of cos
of three π₯ is a third sin of three π₯ plus some constant of integration. Letβs call that π΄. But what do we do about the second
integral? Well, we know the trigonometric
identity one over cos of π₯ equals sec of π₯. And we see that we can rewrite one
over cos squared of three π₯ as sec squared three π₯. And then we have the general result
for the integral of sec squared ππ₯ dπ₯. Itβs one over π tan of ππ₯ plus
some constant. And this means we can write the
integral of sec squared three π₯ as a third tan of three π₯ plus another constant of
integration π΅.

We distribute our parentheses and
we see that two ninths times a third sin of three π₯ is two twenty-sevenths sin of
three π₯. Similarly, we obtain a ninth times
a third of tan three π₯ to be one twenty-seventh of tan of three π₯. And finally, when we multiply each
of our constants by two-ninths and one-ninth, respectively, we end up with a new
constant πΆ. And we find that our integral is
equal to two twenty-sevenths of sin of three π₯ plus one twenty-seventh of tan of
three π₯ plus πΆ.

Weβll consider one more example
that requires the integrals that weβve looked at plus some trigonometric
identities.

Determine the indefinite integral
of negative three tan squared eight π₯ times csc squared eight π₯ evaluated with
respect to π₯. These does at first look quite
tricky. However, If we recall some of our
trigonometry identities, it does get a little nicer. We know that tan π₯ is equal to sin
π₯ over cos of π₯. And we also know that csc π₯ is
equal to one over sin π₯. We can, therefore, rewrite our
entire integrant as negative three times sin squared eight π₯ over cos squared eight
π₯ times one over sin squared eight π₯. And then we noticed that the sin
squared eight π₯ cancels. We can take the factor of negative
three outside of the integral sin to make the next step easier. And we have a negative three times
the integral of one over cos squared eight π₯ dπ₯.

But we know that one over cos of π₯
is equal to sec of π₯. So our integral becomes negative
three times the integral of sec squared eight π₯ evaluated with respect to π₯. But of course, the integral of sec
squared ππ₯, evaluated with respect to π₯, is one over π tan of ππ₯ plus some
constant of integration π. And so we see that the integral of
sec squared eight π₯ is an eighth tan of eight π₯ plus π. We distribute our parentheses. And we see that weβre left with
negative three-eights of tan of eight π₯ plus a new constant since we multiplied our
original one by negative three. Letβs call that capital πΆ.

In this video, weβve seen that we
can use the fact that integration is essentially the reverse process of
differentiation to evaluate the indefinite integrals of sin of ππ₯, cos of ππ₯,
and sec squared of ππ₯. Weβve also seen that recalling
certain trigonometric identity, such as tan π₯ equals sin π₯ over cos π₯ or one over
sin π₯ is equal to csc of π₯, can make integrals easier to evaluate.