Sometimes integrals involving trigonometric functions can be
evaluated by first of all using trigonometric identities to
rewrite the integrand. That's the quantity we're trying to
integrate an alternative form, which is a bit more amenable to
integration. Sometimes a trigonometric substitution is
appropriate. Both of these techniques we look at in this
unit. Before we start I want to give you a couple of preliminary
results which will be using over and over again and which will be
very important and the first one is I want you to make sure that
you know that the integral of the cosine of a constant times
X. With respect to X.
Is equal to one over that constant.
Multiplied by the sign of KX plus a constant of integration
as a very important result. If you integrate the cosine, you
get a sign.
And if there's a constant in front of the X that appears down
here will take that as read in all the examples which follow.
Another important results is the integral of a sign. The integral
of sine KX with respect to X is minus one over K cosine KX.
Plus a constant we're integrating assign. The result
is minus the cosine and the constant factor. There appears
out down here as well, so those
two results. Very important.
You should have them at your fingertips and we can call upon
them whenever we want them in the rest of the video.
We also want to call appan trigonometric identity's. I'm
going to assume that you've seen a lot of trigonometric
identities before. We have a table of trigonometric
identities here, such as the table that you might have seen
many times before. If you want this specific table, you'll find
it in the printed notes
accompanying the video. Why might we want to use
trigonometric identities? Well, for example, we've just
seen that we already know how to integrate the sign of a
quantity and the cosine of the quantity. But suppose we want
to integrate assign multiplied by a cosine or cosine times
cosine or assigned times assign. We don't actually know
how to do those integrals. Integrals at the moment, but
if we use trigonometric identities, we can rewrite
these in terms of just single sine and cosine terms, which
we can then integrate.
Also, the trigonometric identities identities allow us
to integrate powers of sines and cosines. You'll see that using
these identity's? We've got powers of cosine powers of sign
and the identity is allow us to write into grams in terms of
cosines and sines of double
angles. We know how to integrate these already using the results.
I've just reminded you of, so I'm going to assume that you've
got a table like this at your fingertips, and we can call
appan it whenever we need to.
OK, let's have a look at the first example and the example
that I'm going to look at is a definite integral. The integral
from X is not to X is π of the sine squared of X DX. So note in
particular, we've gotta power here. We're looking at the sign
squared of X. What I'm going to do is go back to the table.
And look for an identity that will allow us to change the sign
squared X into something else. Let me just flip back to the
table of trigonometric
identities. The identity that I'm going to use this one, the
cosine of 2A.
Is 1 minus twice sign square day?
If you inspect this carefully, you'll see that this will enable
us to change a sine squared into the cosine of a double angle.
Let me write that down again.
Cosine of 2 A is equal to 1
minus twice sign squared
A. First of all, I'm going to rearrange this to get
sine squared on its own.
If we add two sine squared data both sides, then I can get it on
this side. And if I subtract cosine 2A from both sides, are
remove it from the left.
Finally, if I divide both sides by two, I'll be
left with sine squared A.
And this is the result that I want to use to help me to
evaluate this integral because of what it will allow me to do.
Is it will allow me to change a quantity involving the square of
a trig function into a quantity involving double angles. So
let's use it in this case.
The integral will become the integral from note to pie.
Sine squared X using this formula will be 1 minus cosine
twice X. All divided by
two. Integrated with respect to X.
I've taken out the fact that 1/2 here and I'm left with the
numerator 1 minus cosine 2X to be integrated with respect to X.
This is straightforward to
finish off. So definite integral. So I have square
brackets. The integral of one with respect to X is simply X.
And the integral of cosine 2 X we know from our preliminary
work is just going to be sine 2X divided by two with a minus
sign there and the limits are not and pie.
We finish this off by first of all, putting the upper limit in,
so we want X replaced by pie here and pie here.
The sign of 2π is 0.
So when we put the upper limit in will just get.
Pie by substituting for X here.
Let me put the lower limit in.
X being not will be 0 here.
And sign of note here, which is not so both of those terms will
become zero when we put the lower limit in and so we're just
left with simply 1/2 of Π or π
by 2. And that's our first example of how we've used a
trigonometric identity to rewrite an integrand involving
powers of a trig function in terms of double angles, which we
already know how to integrate.
Let's have a look at another example. Suppose we want
to integrate the sign of three X multiplied by the
cosine of 2 X.
With respect to X.
Now we already know how to integrate signs. We know how to
integrate cosines, but we have a problem here because there's a
product. These two terms are multiplied together and we don't
know how to proceed.
What we do is look in our table of trigonometric identities for
an example where we've gotta sign multiplied by a cosine.
Let's go back to the table.
The first entry in our table involves assign multiplied
by a cosine.
Let me write this formula down again. 2 sign a cosine be.
Is equal
to. The sign of the sum of A and be
added to the sign of the
difference A-B. And this is the identity that I
will use in order to rewrite this integrand
as two separate integrals. We identify the A's
3X. The B is 2 X.
The factor of 2 here isn't a problem. We can divide
everything through by two.
So we lose it from this side.
So our integral? What will it become? Well, the integral of
sign 3X cosine 2X DX will become. We want the integral of
the sign of the sum of A&B.
Well, there's some of A&B will be 3X plus 2X, which is 5X. So
we want the sign of 5X.
Added to the sign of the difference of amb. Well a
being 3X B being 2X A-B will be 3X subtract 2 X
which is just One X. So we want the sign of X all
divided by two and we want to integrate that with
respect to X.
So what have we done? We've used the trig identity to change the
product of a signing cosine into the sum of two separate sign
terms, which we can integrate straight away. We can integrate
that taking the factor of 1/2
out. The integral of sign 5X will be minus the cosine of 5X
divided by 5.
And the integral of sine X will be just minus cosine X, and
they'll be a constant of
integration. And just to tidy it up, at the end we're going to
have minus the half with the five at the bottom. There will
give you minus cosine 5X all divided by 10.
And there's a half with this term here, so it's minus cosine
X divided by two.
Plus a constant of integration.
And that's the solution of this
problem. Let's explore the integral of products of sines
and cosines a little bit further, and what I want to look
at now is integrals of the form the integral of sign to the
power MX multiplied by cosine to the power NX DX.
Well, look at a whole family of integrals like this, but in
particular for the first example I'm going to look at the case of
what happens when M is an odd
number. Whenever you have an integral like this, when M is
odd, the following process will work. Let's look at a specific
case, supposing I want to integrate sine cubed X.
Multiplied by cosine squared XDX.
Notice that M.
Is an odd number and is 3.
There's a little trick here that we're going to do now, and it's
the sort of trick that comes with practice and seeing lots of
examples. What we're going to do is we're going to rewrite the
sign cubed X in a slightly
different form. We're going to recognize that sign cubed can be
written as sine squared X multiplied by Sign X.
That's a little trick. The sign cubed can be written as sine
squared times sign. So our integral can be
written as sine squared X times sign X
multiplied by cosine squared X DX.
And then I'm going to pick a trigonometric identity involving
sine squared to write it in terms of cosine squared. Let's
find that identity.
With an identity here, which says that sine squared of an
angle plus cost squared of an
angle is one. If we rearrange this, we can write that sine
squared of an angle is 1 minus the cosine squared of an angle
will use that.
Sine squared of any
angle. Is equal to 1 minus the cosine squared over any
angle. Will use that in here to change the sign squared X into
terms involving cosine squared X. Let's see what happens. This
integral will become the integral of or sign squared X.
Will become one minus cosine
squared X. There's still the terms cynex.
And at the end we still got
cosine squared X. Now this is looking a bit complicated, but
as we'll see it's all going to come out in the Wash. Let's
remove the brackets here and see
what we've got. There's a one multiplied by all this sign X
times cosine squared X.
So that's just sign X times cosine squared X
will want to integrate that with respect to X.
There's also cosine squared X multiplied by all this.
Now the cosine squared X with this cosine squared X will give
us a cosine, so the power 4X.
There's also the sign X.
And we want to integrate that.
Also, with respect to X and there was a minus sign in front,
so that's going to go in there.
So we've expanded the brackets here and written.
This is 2 separate integrals.
Now, each of these integrals can be evaluated by making a
substitution. If we make a substitution and let you equals
cosine X. The differential du is du DX.
DX Do you DX if we differentiate cosine, X will get
minus the sign X.
So we've got du is minus sign X
DX. Now look at what we've got when we make this substitution.
The cosine squared X will become simply you squared and sign X DX
altogether can be written as a minus du, so this will become.
Minus the integral.
Of you squared.
Do you?
What about this term? We've got cosine to the power four cosine
to the power 4X will be you to
the powerful. And sign X DX sign X DX is minus DU.
There's another minus sign here, so overall
will have plus the integral of you to the
four, do you?
Now these are very very simple integrals to finish the integral
of you squared is you cubed over
3? The integral of you to the four is due to the five over 5
plus a constant of integration.
All we need to do to finish off is return to our original
variables. Remember, you was cosine of X, so we finish off by
writing minus 1/3.
You being cosine X means that we've got cosine cubed X.
Plus 1/5. You to the five will be Co sign
to the power 5X.
Plus a constant of integration.
And that's the solution to the problem that we started with.
Let's stick with the same sort of family of integrals, so we're
still sticking with the integral of sign to the power MX cosine
to the power NX DX.
And now I'm going to have a look at what happens in the case when
M is an even number.
And N is an odd number.
This method will always work when M is even. An is odd.
Let's look at a specific case. Suppose we want to integrate the
sign to the power 4X.
Cosine cubed X.
DX
Notice that M the power of sign is now even em is full.
And N which is the power of cosine, is odd an IS3.
What I'm going to do is I'm going to use the identity that
cosine squared of an angle is 1 minus sign squared of an angle
and you'll be able to lift that directly from the table we had
at the beginning, which stated the very important and well
known results that cosine squared of an angle plus the
sine squared of an angle is always equal to 1.
What I'm going to do is I'm going to use this to rewrite the
cosine term. In here, in terms
of signs. First of all, I'm going to apply the little trick
we had before. And split the cosine turn up like this cosine
cubed. I'm going to write this
cosine squared. Multiplied by
cosine. So I've changed the cosine cubed to these two terms
here. Now I can use the identity to change cosine
squared X into terms involving
sine squared. So the integral will become the integral of
sign. To the power 4X.
Cosine squared X. We can write as one minus sign, squared X.
And there's still this term cosine X here as well.
And all that has to be integrated with respect to X.
Let me remove the brackets here. When we remove the brackets,
there will be signed to the 4th X Times one all multiplied by
cosine X. That'll be signed to the 4th X
multiplied by sign squared
X. Which is signed to the 6X or multiplied by cosine X.
And there's a minus sign in the middle, and we want to integrate
all that. With respect to X.
Again, a simple substitution will allow us to finish this
off. If we let you.
Be sign X.
So do you.
Is cosine X DX.
This will become immediately the integral of well signed to the
4th X sign to the 4th X will be you to the four.
The cosine X times the DX cosine X DX becomes du.
Subtract. Sign to the six, X will become you to
the six. And the cosine X DX is du.
So what we've achieved are two very simple integrals that we
can complete to finish the
problem. The integral of you to the four is due to the five over
5. The integral of you to the six is due to the 7 over 7.
Plus a constant.
And then just to finish off, we return to the original variables
and replace EU with sign X, which will give us 1/5.
Sign next to the five or sign to the power 5X.
Minus.
One 7th. You to the Seven will be signed to the 7X.
Plus a constant of integration.
So that's how we deal with integrals of this family. In the
case when M is an even number and when N is an odd number. Now
in the case when both M&N are even, you should try using the
double angle formulas, and I'm not going to do an example of
that because there isn't time in this video to do that. But there
are examples in the exercises accompanying the video and you
should try those for yourself.
I'm not going to look at some integrals for which
a trigonometric substitution is
appropriate. Suppose we want to evaluate this integral.
The integral of 1 / 1
plus X squared.
With respect to X.
Now the trigonometric substitution that I want to use
is this one. I want to let X be the tangent of a new variable, X
equals 10 theater.
While I picked this particular substitution well, all will
become clear in time, but I want to just look ahead a little bit
by letting X equal 10 theater.
What will have at the denominator down here is
1 + 10 squared theater.
One plus X squared will become 1 + 10 squared and we have an
identity already which says that 1 + 10 squared of an
angle is equal to the sequence squared of the angle. That's
an identity that we had on the table right at the beginning,
so the idea is that by making this substitution, 1 + 10
squared can be replaced by a single term sequence squared,
as we'll see, so let's progress with that
substitution.
If we let X be tongue theater, the integrals going to become 1
/ 1 plus X squared will become 1 + 10 squared.
Theater. And we have to take care of the DX in an appropriate
way. Now remember that DX is going to be given by the XD
theater multiplied by D theater.
DXD theater we want to differentiate X is 10 theater
with respect to theater.
Now the derivative of tongue theater is the secant squared,
so we get secret squared Theta D
theater. So this will allow us to change the DX in here.
Two, secant squared, Theta D Theta over on the right.
At this stage I'm going to use the trigonometric identity,
which says that 1 + 10 squared of an angle is equal to the
sequence squared of the angle. So In other words, all this
quantity down here is just the sequence squared of Theta.
And this is very nice now because this term here will
cancel out with this term down in the denominator down there,
and we're left purely with the integral of one with respect to
theater. Very simple to finish.
The integral of one with respect to theater is just theater.
Plus a constant of integration.
We want to return to our original variables and if X was
10 theater than theater is the angle whose tangent, his ex. So
theater is 10 to the minus one
of X. Plus a constant.
And that's the problem finished.
This is a very important standard result that the
integral of one over 1 plus X squared DX is equal to the
inverse tan 10 to the minus one of X plus a constant.
That's a result that you'll see in all the standard
tables of integrals, and it's a result that you'll
need to call appan very frequently, and if you can't
remember it, then at least you'll need to know that
there is such a formula that exists and you want to be
able to look it up.
I want to generalize this a little bit to look at the case
when we deal with not just a one here, but a more general case of
an arbitrary constant in there. So let's look at what happens if
we have a situation like this.
Suppose we want to integrate one over a squared plus X squared
with respect to X.
Where a is a
constant. This time I'm going to make this substitution let X be
a town theater, and we'll see why we've made that substitution
in just a little while.
With this substitution, X is a Tan Theta. The differential
DX becomes a secant squared Theta D Theta.
Let's put all this into this
integral here. Will have the integral of one over a squared.
Plus And X squared will become a squared 10.
Squared feet are.
The
DX Will become a sex squared Theta D
Theta. Now what I can do now is I can take out a common
factor of A squared from the
denominator. Taking an A squared out from this term will leave me
one taking a squared out from this term will leave me tan
squared theater. And it's still on the top. I've got a sex
squared Theta D Theta.
We have the trig identity that 1 + 10 squared of any angle is sex
squared of the angle.
So I can use that identity in here to write the denominator as
one over a squared and the 1 + 10 squared becomes simply
sequence squared theater.
We still gotten a secant squared theater in the numerator, and a
lot of this is going to simplify and cancel now.
The secant squared will go the
top and the bottom. The one of these at the bottom will go with
the others at the top, and we're left with the integral of one
over A with respect to theater.
Again, this is straightforward to finish. The integral of one
over a one over as a constant with respect to Theta is just
going to give me one over a.
Theater. Plus the constant of
integration. To return to the original variables, we've got to
go back to our original substitution. If X is a tan
Theta, then we can write that X over A is 10 theater.
And In other words, that theater is the angle whose
tangent is 10 to the minus one of all this X over a.
That will enable me to write our final results as one over a town
to the minus one.
X over a.
Plus a constant of integration.
And this is another very important standard result that
the integral of one over a squared plus X squared with
respect to X is one over a 10 to the minus one of X over a plus a
constant, and as before, that's a standard result that you'll
see frequently in all the tables of integrals, and you'll need to
call a pawn that in lots of situations when you're required
to do integration.
OK, so now we've got the standard result that the
integral of one over a squared plus X squared DX is equal
to one over a town to the minus one of X of
A. As a constant of integration.
Let's see how we might use this formula in a slightly
different case. Suppose we have the integral of 1 / 4 +
9 X squared DX.
Now this looks very similar to the standard formula we have
here. Except there's a slight
problem. And the problem is that instead of One X squared, which
we have in the standard result, I've got nine X squared.
What I'm going to do is I'm going to divide everything at
the bottom by 9, take a factor of nine out so that we end up
with just a One X squared here. So what I'm going to do is I'm
going to write the denominator
like this. So I've taken a factor of nine out. You'll see
if we multiply the brackets again here, there's 9 * 4 over
9, which is just four and the nine times the X squared, so I
haven't changed anything. I've just taken a factor of nine out
the point of doing that is that now I have a single. I have a
One X squared here, which will match the formula I have there.
If I take the 9 outside the
integral. I'm left with 1 /, 4 ninths plus X squared integrated
with respect to X and I hope you can see that this is exactly one
of the standard forms. Now when we let A squared B4 over nine
with a squared is 4 over 9. We have the standard form. If A
squared is 4 over 9A will be 2 over 3 and we can complete this
integration. Using the standard result that one over 9 stays
there, we want one over A.
Or A is 2/3. So we want 1 / 2/3.
10 to the minus one.
Of X over a.
X divided by a is X divided by
2/3. Plus a constant of
integration. Just to tide to these fractions up, three will
divide into 9 three times, so we'll have 326 in the
denominator. 10 to the minus one and dividing by 2/3 is like
multiplying by three over 2, so I'll have 10 to the minus one of
three X over 2 plus the constant
of integration. So the point here is you might have to do a
bit of work on the integrand in order to be able to write
it in the form of one of the standard results.
OK, let's have a look at another case where another integral to
look at where a trigonometric substitution is appropriate.
Suppose we want to find the integral of one over the square
root of A squared minus X
squared DX. Again, A is a constant.
The substitution that I'm going to make is this one.
I'm going to write X equals a sign theater.
If I do that, what will happen to my integral, let's see.
And have the integral of one
over. The square root. The A squared will stay the same, but
the X squared will become a squared sine squared.
I squared sine squared Theta.
Now the reason I've done that is because in a minute I'm
going to take out a factor of a squared, which will leave me
one 1 minus sign squared, and I do have an identity involving 1
minus sign squared as we'll see, but just before we do
that, let's substitute for the differential as well. If X is a
sign theater, then DX will be a cosine, Theta, D, Theta.
So we have a cosine Theta D Theta for the differential DX.
Let me take out the factor of a squared in the denominator.
Taking a squad from this first term will leave me one
and a squared from the second term will leave me one minus
sign squared Theta.
I have still gotten a costly to the theater at the top.
Now let me remind you there's a trig identity which says that
the cosine squared of an angle plus the sine squared of an
angle is always one.
So if we have one minus the sine squared of an angle, we can
replace it with cosine squared.
So 1 minus sign squared Theta we can replace with simply
cosine squared Theta.
Is the A squared out the frontier and we want the square
root of the whole lot.
Now this is very simple. We want the square root of A squared
cosine squared Theta. We square root. These squared terms will
be just left with.
A cosine Theta.
In the denominator and within a cosine Theta in the numerator.
And these were clearly cancel out.
And we're left with the integral of one with respect to theater,
which is just theater plus a constant of integration.
Just to return to the original variables, given that X was a
sign theater, then clearly X over A is sign theater.
So theater is the angle who sign is or sign to the minus one of X
over a, so replacing the theater with sign to the minus one of X
over a will get this result.
And this is a very important standard result that if you want
to integrate 1 divided by the square root of A squared minus X
squared, the result is the inverse sine or the sign to the
minus one of X over a.
Plus a constant of integration.
Will have a look one final example which is a variant on
the previous example. Suppose we want to integrate 1 divided by
the square root of 4 - 9 X squared DX.
Now that's very similar to the one we just looked at. Remember
that we had the results that the integral of one over the square
root of A squared minus X
squared DX. Was the inverse sine of X over a plus a constant?
That's keep that in mind. That's the standard result we've
already proved. We're almost there. In this case. The problem
is that instead of a single X squared, we've got nine X
squared. So like we did in the other example, I'm going to take
the factor of nine out to leave us just a single X squared in
there, and I do that like this.
Taking a nine out from these terms here, I'll have
four ninths minus X squared.
Again, the nine times the four ninths leaves the four which we
had originally, and then we've got the nine X squared, which we
have there. The whole point of doing that is that then I'm
going to extract the Route 9, which is 3 and bring it right
outside. And inside under the integral sign, I'll be left with
one over the square root of 4 ninths minus X squared DX.
Now in this form, I hope you can spot that we can use the
standard result immediately with the standard results, with a
being with a squared being equal to four ninths.
In other words, a being equal to
2/3. Putting all that together will have a third. That's the
third and the integral will become the inverse sine.
X. Divided by AA
was 2/3. Plus a constant of integration.
And just to tidy that up will be left with the third inverse sine
dividing by 2/3 is the same as multiplying by three over 2, so
will have 3X over 2 plus a constant of integration.
And that's our final result. So we've seen a lot
of examples that have integration using
trigonometric identities and integration using trig
substitutions. You need a lot of practice, and there
are a lot of exercises in the accompanying text.