www.mathcentre.ac.uk/.../9.7%20Integration%20requiring%20use%20of%20trigonometric%20identities.mp4

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.