Video Transcript
In this video, we’ll learn how to
find indefinite integrals of trigonometric functions. We’ll begin by recalling what we
mean by an antiderivative before looking at how this helps us to integrate a number
of trigonometric functions.
Let’s start by recalling what the
antiderivative of a function is. Uppercase 𝐹 is the antiderivative
of lowercase 𝑓 if uppercase 𝐹 prime of 𝑥 is equal to lowercase 𝑓 of 𝑥. And that’s where uppercase 𝐹 prime
of 𝑥 is the derivative with respect to 𝑥 of uppercase 𝐹 of 𝑥. In fact, we can say this is true
for any function 𝐺 of 𝑥, where 𝐺 of 𝑥 is equal to uppercase 𝐹 of 𝑥 plus 𝑐,
for any constant 𝑐. Now this is very useful as we’ll be
using it to define an indefinite integral.
We can say that the indefinite
integral of lowercase 𝑓 of 𝑥 with respect to 𝑥 is equal to uppercase 𝐹 of 𝑥
plus 𝑐, where uppercase 𝐹 is the antiderivative of lowercase 𝑓. And it’s very important to remember
our constant of integration 𝑐 when performing an indefinite integral. And note that it’s called an
indefinite integral as there are no limits of integration specified. That is, we’re not integrating over
a specific interval of 𝑥-values as we would be in a definite integral. And this is a little beyond the
scope of this video since it touches on the fundamental theorem of calculus.
However, we can see why the
constant 𝑐 is necessary if we perform the reverse operation on our indefinite
integral equation. So that’s differentiating with
respect to 𝑥. And since we’re simply performing
the inverse operation on the left, the derivative with respect to 𝑥 of the integral
of lowercase 𝑓 of 𝑥 is just lowercase 𝑓 of 𝑥. And on the right-hand side, when we
differentiate uppercase 𝐹 of 𝑥 with respect to 𝑥, we simply get 𝐹 prime of
𝑥. And our constant 𝑐 will simply
vanish because differentiating a constant gives zero. And now when we go back the other
way to our integral, the constant will appear again. But since we don’t know the value
of this constant, we simply label it 𝑐. It’s just an unknown constant.
We can use this idea of the
integral as antiderivative together with our knowledge of the derivatives of various
trigonometric functions to find some specific integrals. So let’s begin by looking at the
function 𝑓 of 𝑥 is equal to the sin of 𝑎𝑥, for real constants 𝑎 and where for
this to be valid 𝑥 must be in radian measure. And we want to find the indefinite
integral of 𝑓 of 𝑥 evaluated with respect to 𝑥. We recall that the derivative of
the cos of 𝑎𝑥 with respect to 𝑥 is equal to negative 𝑎 times the sin of
𝑎𝑥. We can therefore say that the
indefinite integral of negative 𝑎 sin of 𝑎𝑥 evaluated with respect to 𝑥 must be
the cos of 𝑎𝑥. Don’t forget though that since
we’re working with an indefinite integral, we need to add a constant of integration,
which we’ll call lowercase 𝑐.
Okay, so this is good start. But we’re actually looking to find
the indefinite integral of the sin of 𝑎𝑥, not negative 𝑎 times the sin of
𝑎𝑥. We’re allowed though to take the
constant, negative 𝑎, outside of the integral. And so we see that negative 𝑎
times the indefinite integral of sin 𝑎𝑥 is equal to the cos of 𝑎𝑥 plus the
constant 𝑐. And since negative 𝑎 is a
constant, we can divide both sides by negative 𝑎. And we obtain the indefinite
integral of sin 𝑎𝑥 to be negative one over 𝑎 times the cos of 𝑎𝑥 plus uppercase
𝐶. And notice that we’ve changed from
lowercase 𝑐 to uppercase 𝐶 because we’ve divided our original constant by another
constant, negative 𝑎, and we need to represent this change in value.
Making a note of our first result
for the indefinite integral of sin 𝑎𝑥, we can repeat this process for the function
𝑓 of 𝑥 is equal to cos 𝑎𝑥, where again 𝑎 is a real constant and 𝑥 is measured
in radians. We’re going to use the fact that
the derivative with respect to 𝑥 of sin 𝑎𝑥 is equal to 𝑎 times the cos of
𝑎𝑥. And we can therefore say that the
indefinite integral of 𝑎 cos 𝑎𝑥 evaluated with respect to 𝑥 is equal to sin 𝑎𝑥
plus the constant of integration lowercase 𝑐. And again, we’re allowed to take
the constant 𝑎 outside the integral, and the right-hand side remains unchanged. And finally, we divide through by
𝑎 and obtain the indefinite integral of cos 𝑎𝑥 evaluated with respect to 𝑥 to be
one over 𝑎 times the sin of 𝑎𝑥 plus the new constant of integration uppercase
𝐶.
Now you might recall that the
process for differentiating sine and cosine functions forms a cycle. That is, the derivative of sin 𝑥
is equal to cos 𝑥, where here our constant 𝑎 is equal to one. Differentiating again gives us
negative sin 𝑥. And differentiating again, we get
negative cos 𝑥. And differentiating one more time,
we go back to our original function, sin 𝑥. We reverse this cycle as shown for
integration. Let’s now look at some examples
illustrating the integration of sine and cosine functions.
Determine the indefinite integral
of negative sin 𝑥 minus nine times the cos of 𝑥 evaluated with respect to 𝑥.
Before trying to evaluate this, it
can be useful to recall some of the properties of integrals. Firstly, the integral of the sum of
two or more functions is 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 𝑥 itself. These properties mean we can
rewrite our integral as negative the integral of sin 𝑥 evaluated with respect to 𝑥
minus nine times the integral of cos of 𝑥 evaluated with respect to 𝑥.
And now we recall the general
results for the integrals of the sine and cosine functions. The indefinite integral of sin of
𝑎𝑥 is equal to negative one over 𝑎 times cos 𝑎𝑥 plus the constant of
integration 𝑐. And the integral of cos of 𝑎𝑥
evaluated with respect to 𝑥 is equal to one over 𝑎 times sin 𝑎𝑥 plus the
constant 𝑐. So in our case, the constant 𝑎 is
equal to one, and our integral is negative negative cos 𝑥 plus the constant 𝐴
minus nine times sin 𝑥 plus the constant 𝐵. And we’ve chosen 𝐴 and 𝐵 to show
that these are different constants of integration.
Now distributing the parentheses
and combining the two constants 𝐴 and 𝐵 into a single constant 𝐶, we have the
indefinite integral of negative sin 𝑥 minus nine cos 𝑥 evaluated with respect to
𝑥 is equal to cos 𝑥 minus nine times sin 𝑥 plus the constant 𝐶.
Determine the indefinite integral
with respect to 𝑥 of negative eight times the sin of eight 𝑥 minus seven times the
cos of five 𝑥.
In this question, we’re looking to
integrate the sum of two functions of 𝑥. We begin by recalling that the
integral of the sum of two or more functions is equal to the sum of the integrals of
those respective functions. And so we can write our integral as
the integral of negative eight sin of eight 𝑥 with respect to 𝑥 plus the integral
of negative seven cos of five 𝑥 d𝑥. We also know that we can take any
constant factors outside of the integrals and focus on integrating each expression
in 𝑥. And this means we can rewrite our
integrals as negative eight times the integral with respect to 𝑥 of sin eight 𝑥
minus seven times the integral of cos five 𝑥 d𝑥.
Next, we recall the general results
for the integrals of sin 𝑎𝑥 and cos 𝑎𝑥. The indefinite integral with
respect to 𝑥 of sin of 𝑎𝑥 is equal to negative one over 𝑎 times cos 𝑎𝑥 plus
the constant of integration 𝑐. And the indefinite integral of cos
𝑎𝑥 with respect to 𝑥 is one over 𝑎 sin 𝑎𝑥 plus the constant 𝑐. In our case, the constant 𝑎 is
eight in the first integral and five in the second integral. And applying these results to our
integrals, we see that the integral of sin of eight 𝑥 is negative one over eight
cos eight 𝑥 plus the constant 𝐴. And the integral with respect to 𝑥
of cos of five 𝑥 is one over five times the sin of five 𝑥 plus 𝐵. Note that we’ve chosen 𝐴 and 𝐵 as
the constant of integration, not just a single value of 𝑐, to show that these are
actually different constants.
Our final step is to distribute the
parentheses. Negative eight times negative one
over eight. cos of eight 𝑥 is just the cos of
eight 𝑥. And negative seven times one over
five sin five 𝑥 is negative seven over five sin five 𝑥. And finally, we multiply negative
eight by 𝐴 and negative seven by 𝐵. And since we don’t know the values
of 𝐴 and 𝐵, we choose to represent this as a single constant 𝑐. We found then that the integral we
required is the cos of eight 𝑥 minus seven over five times the sin of five 𝑥 plus
the constant 𝑐.
We’re now going to consider some
alternative derivatives. We recall that the derivative of
tan 𝑎𝑥 is 𝑎 sec squared 𝑎𝑥. Now, you might like to pause this
video for a moment to consider what this tells us about the integral of sec squared
𝑎𝑥. Well, let’s have a look. We recall that an integral can be
considered as the antiderivative. That is, integration is the reverse
process for differentiation. And we can see then that the
indefinite integral of 𝑎 sec squared 𝑎𝑥 with respect to 𝑥 must be tan of 𝑎𝑥
plus the constant of integration 𝑐. We take out the constant factor of
𝑎. And then we divide through by
𝑎. Then we can see that the integral
of sec squared 𝑎𝑥 with respect to 𝑥 is one over 𝑎 tan of 𝑎𝑥 plus uppercase
𝐶.
Using much the same method, we
obtain the following integrals relating to the derivatives of the reciprocal
trigonometric functions. The integral of csc 𝑎𝑥 cot 𝑎𝑥
with respect to 𝑥 is negative one over 𝑎 times csc 𝑎𝑥 plus 𝑐. The integral of sec 𝑎𝑥 tan 𝑎𝑥
with respect to 𝑥 is one over 𝑎 sec 𝑎𝑥 plus 𝑐. And the integral of csc squared
𝑎𝑥 with respect to 𝑥 negative one over 𝑎 times cot of 𝑎𝑥 plus 𝑐. We’re now going to look at some
examples of these results and how often we use trigonometric identities to help us
evaluate these integral.
Determine the indefinite integral
of negative sec squared six 𝑥 with respect to 𝑥.
To answer this question, it’s
almost simply enough to quote the general result for the integral of sec squared
𝑎𝑥, where in our case the constant 𝑎 is equal to six. The result tells us that the
integral of sec squared 𝑎𝑥 with respect to 𝑥 is one over 𝑎 times the tan of 𝑎𝑥
plus constant 𝑐. It’s sensible, however, before
using this result to take the factor of negative one outside the integral as
shown. So when we do perform our
integration, we obtain the solution to be negative one times one-sixth of tan of six
𝑥 plus 𝑐, recalling that in our case the constant 𝑎 is equal to six.
And now all that’s left is to
distribute the parentheses. Negative one times one-sixth of tan
six 𝑥 is negative one-sixth tan of six 𝑥. And negative one times the constant
𝑐 gives us this new constant uppercase 𝐶. And so we find our indefinite
integral to be negative one-sixth of tan of six 𝑥 plus 𝐶.
Determine the indefinite integral
of two cos cubed of three 𝑥 plus one over nine cos squared of three 𝑥 with respect
to 𝑥.
Now at first glance, this integral
looks quite complicated. We should spot, however, that we
can actually simplify the quotient. How we do this is to essentially
reverse the process we follow when adding two fractions. And we see that we can write the
quotient as two cos cubed three 𝑥 over nine cos squared three 𝑥 plus one over nine
cos squared three 𝑥. By dividing through in the first
fraction by cos squared three 𝑥, our first term simplifies to two over nine cos
three 𝑥. And then to help us to spot what to
do next, let’s rewrite the second fraction as a ninth times one over cos squared of
three 𝑥.
Next, we recall that the integral
of the sum of two or more functions is the sum of their integrals. And we also know that we can take
any constant factors outside of the integral and focus on integrating the expression
in 𝑥. And applying these properties gives
us two over nine times the integral of the cos of three 𝑥 with respect to 𝑥 plus
one over nine times the integral of one over the cos squared of three 𝑥 again with
respect to 𝑥. We can quote the result for the
integral of cos of 𝑎𝑥. It’s one over 𝑎 times the sin of
𝑎𝑥 plus the constant of integration 𝑐. And this means that the integral of
cos of three 𝑥 is a third sin of three 𝑥 plus some constant of integration which
we’ll call 𝐴.
But what do we do about the second
integral? Well, we know the trigonometric
identity one over cos of 𝑥 is equal to sec 𝑥. And so we see that we can rewrite
one over cos squared of three 𝑥 as sec squared of three 𝑥. So now we can use the general
result for the integral of sec squared of 𝑎𝑥 with respect to 𝑥. And that’s one over 𝑎 times the
tan of 𝑎𝑥 plus some constant 𝑐. And this means we can write the
integral of sec squared three 𝑥 with respect to 𝑥 as one over three tan three 𝑥
plus a constant of integration which we’ll call 𝐵.
Finally, we distribute our
parentheses and we see that two over nine times one-third of sin three 𝑥 is equal
to two over 27 times sin three 𝑥. And one over nine times one over
three tan three 𝑥 is equal to one over 27 tan three 𝑥. We multiply each of our constants
by two over nine and one over nine, respectively. And we end up with a new constant
𝐶. We find then that our integral is
two over 27 times the sin of three 𝑥 plus one over 27 times tan of three 𝑥 plus
the constant of integration 𝐶.
Let’s consider one more example
that requires the integrals we’ve looked at together with some trigonometric
identities.
Determine the indefinite integral
of negative three tan squared of eight 𝑥 times csc squared eight 𝑥 evaluated with
respect to 𝑥.
This does at first look quite
tricky. However, if we recall some of our
trigonometric identities, we can change this into something more manageable. We know that tan 𝑥 is equal to sin
𝑥 over cos 𝑥 and that csc 𝑥 is equal to one over six 𝑥. We can therefore rewrite our
integrand, that’s the function we want to integrate, as negative three times sin
squared of eight 𝑥 over cos squared of eight 𝑥 times one over sin squared of eight
𝑥. And then we see that we can cancel
the sin squared of eight 𝑥. We can take the factor of negative
three outside our integral to make the next step easier. And we have negative three times
the integral of one over cos squared eight 𝑥 evaluated with respect to 𝑥.
Now we know that one over cos 𝑥 is
equal to sec 𝑥. So our integral becomes negative
three times the integral of sec squared of eight 𝑥 with respect to 𝑥. But of course the integral of sec
squared 𝑎𝑥 with respect to 𝑥 is one over 𝑎 tan of 𝑎𝑥 plus constant 𝑐. And so in our case where the
constant 𝑎 is equal to eight, the integral of sec squared eight 𝑥 with respect to
𝑥 is one over eight times the tan of eight 𝑥 plus the constant 𝑐. We distribute our parentheses and
we see that our answer is negative three-eighths times tan of eight 𝑥 plus a new
constant, since we multiplied our original one by negative three. So let’s call that uppercase
𝐶.
Let’s now complete this video by
recalling some of the key points we’ve covered. In this video, we’ve seen that we
can use the fact that integration is essentially the reverse process of
differentiation to evaluate the indefinite integrals of sin of 𝑎𝑥, cos of 𝑎𝑥,
and sec squared of 𝑎𝑥. We’ve also seen that recalling
certain trigonometric identities such as tan equals sin 𝑥 over cos 𝑥 or one over
sin 𝑥 is equal to cos of 𝑥 can make integrals easier to evaluate.