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

• 0:01 - 0:05
Sometimes integrals involving
trigonometric functions can be
• 0:05 - 0:09
evaluated by first of all using
trigonometric identities to
• 0:09 - 0:13
rewrite the integrand. That's
the quantity we're trying to
• 0:13 - 0:18
integrate an alternative form,
which is a bit more amenable to
• 0:18 - 0:22
integration. Sometimes a
trigonometric substitution is
• 0:22 - 0:26
appropriate. Both of these
techniques we look at in this
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unit. Before we start I want to
give you a couple of preliminary
• 0:31 - 0:35
results which will be using over
and over again and which will be
• 0:35 - 0:39
very important and the first one
is I want you to make sure that
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you know that the integral of
the cosine of a constant times
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X. With respect to X.
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Is equal to one over that
constant.
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Multiplied by the sign of KX
plus a constant of integration
• 0:55 - 1:00
as a very important result. If
you integrate the cosine, you
• 1:00 - 1:01
get a sign.
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And if there's a constant in
front of the X that appears down
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here will take that as read in
all the examples which follow.
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Another important results is the
integral of a sign. The integral
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of sine KX with respect to X
is minus one over K cosine KX.
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Plus a constant we're
integrating assign. The result
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is minus the cosine and the
constant factor. There appears
• 1:31 - 1:35
out down here as well, so those
• 1:35 - 1:37
two results. Very important.
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You should have them at your
fingertips and we can call upon
• 1:41 - 1:43
them whenever we want them in
the rest of the video.
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We also want to call appan
trigonometric identity's. I'm
• 1:48 - 1:52
going to assume that you've seen
a lot of trigonometric
• 1:52 - 1:55
identities before. We have a
table of trigonometric
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identities here, such as the
table that you might have seen
• 1:59 - 2:02
many times before. If you want
this specific table, you'll find
• 2:02 - 2:03
it in the printed notes
• 2:03 - 2:07
accompanying the video. Why
might we want to use
• 2:07 - 2:10
trigonometric identities?
Well, for example, we've just
• 2:10 - 2:14
seen that we already know how
to integrate the sign of a
• 2:14 - 2:18
quantity and the cosine of the
quantity. But suppose we want
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to integrate assign multiplied
by a cosine or cosine times
• 2:21 - 2:24
cosine or assigned times
assign. We don't actually know
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how to do those integrals.
Integrals at the moment, but
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if we use trigonometric
identities, we can rewrite
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these in terms of just single
sine and cosine terms, which
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we can then integrate.
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Also, the trigonometric
identities identities allow us
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to integrate powers of sines and
cosines. You'll see that using
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these identity's? We've got
powers of cosine powers of sign
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and the identity is allow us to
write into grams in terms of
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cosines and sines of double
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angles. We know how to integrate
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I've just reminded you of, so
I'm going to assume that you've
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got a table like this at your
fingertips, and we can call
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appan it whenever we need to.
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OK, let's have a look at the
first example and the example
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that I'm going to look at is a
definite integral. The integral
• 3:16 - 3:22
from X is not to X is π of the
sine squared of X DX. So note in
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particular, we've gotta power
here. We're looking at the sign
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squared of X. What I'm going to
do is go back to the table.
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And look for an identity that
will allow us to change the sign
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squared X into something else.
Let me just flip back to the
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table of trigonometric
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identities. The identity that
I'm going to use this one, the
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cosine of 2A.
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Is 1 minus twice
sign square day?
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If you inspect this carefully,
you'll see that this will enable
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us to change a sine squared into
the cosine of a double angle.
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Let me write that down again.
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Cosine of 2 A
is equal to 1
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minus twice sign squared
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A. First of all, I'm going
to rearrange this to get
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sine squared on its own.
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If we add two sine squared data
both sides, then I can get it on
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this side. And if I subtract
cosine 2A from both sides, are
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remove it from the left.
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Finally, if I divide both
sides by two, I'll be
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left with sine squared A.
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And this is the result that I
want to use to help me to
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evaluate this integral because
of what it will allow me to do.
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Is it will allow me to change a
quantity involving the square of
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a trig function into a quantity
involving double angles. So
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let's use it in this case.
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The integral will become the
integral from note to pie.
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Sine squared X using this
formula will be 1 minus cosine
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twice X. All divided by
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two. Integrated with
respect to X.
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I've taken out the fact that 1/2
here and I'm left with the
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numerator 1 minus cosine 2X to
be integrated with respect to X.
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This is straightforward to
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finish off. So definite
integral. So I have square
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brackets. The integral of one
with respect to X is simply X.
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And the integral of cosine 2 X
we know from our preliminary
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work is just going to be sine
2X divided by two with a minus
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sign there and the limits are
not and pie.
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We finish this off by first of
all, putting the upper limit in,
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so we want X replaced by pie
here and pie here.
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The sign of 2π is 0.
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So when we put the upper limit
in will just get.
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Pie by substituting for X here.
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Let me put the lower limit in.
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X being not will be 0 here.
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And sign of note here, which is
not so both of those terms will
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become zero when we put the
lower limit in and so we're just
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left with simply 1/2 of Π or π
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by 2. And that's our first
example of how we've used a
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trigonometric identity to
rewrite an integrand involving
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powers of a trig function in
terms of double angles, which we
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• 7:01 - 7:09
Let's have a look at
another example. Suppose we want
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to integrate the sign of
three X multiplied by the
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cosine of 2 X.
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With respect to X.
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Now we already know how to
integrate signs. We know how to
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integrate cosines, but we have a
problem here because there's a
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product. These two terms are
multiplied together and we don't
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know how to proceed.
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What we do is look in our table
of trigonometric identities for
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an example where we've gotta
sign multiplied by a cosine.
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Let's go back to the table.
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The first entry in our
table involves assign multiplied
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by a cosine.
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Let me write this formula down
again. 2 sign a cosine be.
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Is
equal
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to. The
sign of the sum of A and be
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added to the sign of the
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difference A-B. And this is
the identity that I
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will use in order
to rewrite this integrand
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as two separate integrals.
We identify the A's
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3X. The B is 2 X.
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The factor of 2 here isn't a
problem. We can divide
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everything through by two.
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So we lose it from this side.
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So our integral? What will it
become? Well, the integral of
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sign 3X cosine 2X DX will
become. We want the integral of
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the sign of the sum of A&B.
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Well, there's some of A&B will
be 3X plus 2X, which is 5X. So
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we want the sign of 5X.
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Added to the sign of the
difference of amb. Well a
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being 3X B being 2X A-B
will be 3X subtract 2 X
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which is just One X. So we
want the sign of X all
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divided by two and we want
to integrate that with
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respect to X.
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So what have we done? We've used
the trig identity to change the
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product of a signing cosine into
the sum of two separate sign
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terms, which we can integrate
straight away. We can integrate
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that taking the factor of 1/2
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out. The integral of sign 5X
will be minus the cosine of 5X
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divided by 5.
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And the integral of sine X will
be just minus cosine X, and
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they'll be a constant of
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integration. And just to tidy it
up, at the end we're going to
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have minus the half with the
five at the bottom. There will
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give you minus cosine 5X all
divided by 10.
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And there's a half with this
term here, so it's minus cosine
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X divided by two.
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Plus a constant of integration.
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And that's the solution of this
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problem. Let's explore the
integral of products of sines
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and cosines a little bit
further, and what I want to look
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at now is integrals of the form
the integral of sign to the
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power MX multiplied by cosine to
the power NX DX.
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Well, look at a whole family of
integrals like this, but in
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particular for the first example
I'm going to look at the case of
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what happens when M is an odd
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number. Whenever you have an
integral like this, when M is
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odd, the following process will
work. Let's look at a specific
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case, supposing I want to
integrate sine cubed X.
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Multiplied by cosine
squared XDX.
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Notice that M.
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Is an odd number and is 3.
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There's a little trick here that
we're going to do now, and it's
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the sort of trick that comes
with practice and seeing lots of
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examples. What we're going to do
is we're going to rewrite the
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sign cubed X in a slightly
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different form. We're going to
recognize that sign cubed can be
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written as sine squared X
multiplied by Sign X.
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That's a little trick. The sign
cubed can be written as sine
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squared times sign. So
our integral can be
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written as sine squared
X times sign X
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multiplied by cosine
squared X DX.
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And then I'm going to pick a
trigonometric identity involving
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sine squared to write it in
terms of cosine squared. Let's
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find that identity.
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With an identity here, which
says that sine squared of an
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angle plus cost squared of an
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angle is one. If we rearrange
this, we can write that sine
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squared of an angle is 1 minus
the cosine squared of an angle
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will use that.
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Sine squared of any
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angle. Is equal to 1 minus
the cosine squared over any
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angle. Will use that in here to
change the sign squared X into
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terms involving cosine squared
X. Let's see what happens. This
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integral will become the
integral of or sign squared X.
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Will become one minus cosine
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squared X. There's still
the terms cynex.
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And at the end we still got
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cosine squared X. Now this is
looking a bit complicated, but
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as we'll see it's all going to
come out in the Wash. Let's
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remove the brackets here and see
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what we've got. There's a one
multiplied by all this sign X
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times cosine squared X.
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So that's just sign X
times cosine squared X
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will want to integrate
that with respect to X.
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There's also cosine squared X
multiplied by all this.
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Now the cosine squared X with
this cosine squared X will give
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us a cosine, so the power 4X.
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There's also the sign X.
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And we want to integrate that.
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Also, with respect to X and
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so that's going to go in there.
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So we've expanded the
brackets here and written.
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This is 2 separate integrals.
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Now, each of these integrals can
be evaluated by making a
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substitution. If we make a
substitution and let you equals
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cosine X. The differential du
is du DX.
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DX Do you DX if we
differentiate cosine, X will get
• 14:31 - 14:33
minus the sign X.
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So we've got du is minus sign X
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DX. Now look at what we've got
when we make this substitution.
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The cosine squared X will become
simply you squared and sign X DX
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altogether can be written as a
minus du, so this will become.
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Minus the integral.
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Of you squared.
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Do you?
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cosine to the power four cosine
• 15:06 - 15:10
to the power 4X will be you to
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the powerful. And sign X
DX sign X DX is minus DU.
• 15:15 - 15:18
There's another minus
sign here, so overall
• 15:18 - 15:23
will have plus the
integral of you to the
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four, do you?
• 15:25 - 15:30
Now these are very very simple
integrals to finish the integral
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of you squared is you cubed over
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3? The integral of you to the
four is due to the five over 5
• 15:39 - 15:40
plus a constant of integration.
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All we need to do to finish off
• 15:48 - 15:53
variables. Remember, you was
cosine of X, so we finish off by
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writing minus 1/3.
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You being cosine X means that
we've got cosine cubed X.
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Plus 1/5. You to
the five will be Co sign
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to the power 5X.
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Plus a constant of integration.
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And that's the solution to the
problem that we started with.
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Let's stick with the same sort
of family of integrals, so we're
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still sticking with the integral
of sign to the power MX cosine
• 16:30 - 16:33
to the power NX DX.
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And now I'm going to have a look
at what happens in the case when
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M is an even number.
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And N is an odd number.
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This method will always work
when M is even. An is odd.
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Let's look at a specific case.
Suppose we want to integrate the
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sign to the power 4X.
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Cosine cubed X.
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DX
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Notice that M the power of sign
is now even em is full.
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And N which is the power of
cosine, is odd an IS3.
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What I'm going to do is I'm
going to use the identity that
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cosine squared of an angle is 1
minus sign squared of an angle
• 17:21 - 17:25
and you'll be able to lift that
directly from the table we had
• 17:25 - 17:28
at the beginning, which stated
the very important and well
• 17:28 - 17:31
known results that cosine
squared of an angle plus the
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sine squared of an angle is
always equal to 1.
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What I'm going to do is I'm
going to use this to rewrite the
• 17:40 - 17:43
cosine term. In here, in terms
• 17:43 - 17:47
of signs. First of all, I'm
going to apply the little trick
• 17:47 - 17:53
we had before. And split the
cosine turn up like this cosine
• 17:53 - 17:56
cubed. I'm going to write this
• 17:56 - 17:59
cosine squared. Multiplied by
• 17:59 - 18:05
cosine. So I've changed the
cosine cubed to these two terms
• 18:05 - 18:12
here. Now I can use
the identity to change cosine
• 18:12 - 18:15
squared X into terms involving
• 18:15 - 18:21
sine squared. So the integral
will become the integral of
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sign. To the power 4X.
• 18:24 - 18:30
Cosine squared X. We can write
as one minus sign, squared X.
• 18:31 - 18:36
And there's still this term
cosine X here as well.
• 18:38 - 18:41
And all that has to be
integrated with respect to X.
• 18:44 - 18:49
Let me remove the brackets here.
When we remove the brackets,
• 18:49 - 18:55
there will be signed to the 4th
X Times one all multiplied by
• 18:55 - 19:02
cosine X. That'll be signed
to the 4th X
• 19:02 - 19:05
multiplied by sign squared
• 19:05 - 19:11
X. Which is signed to the 6X
or multiplied by cosine X.
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middle, and we want to integrate
• 19:19 - 19:23
all that. With
respect to X.
• 19:25 - 19:29
Again, a simple substitution
will allow us to finish this
• 19:29 - 19:31
off. If we let you.
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Be sign X.
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So do you.
• 19:36 - 19:39
Is cosine X DX.
• 19:40 - 19:44
This will become immediately the
integral of well signed to the
• 19:44 - 19:48
4th X sign to the 4th X will be
you to the four.
• 19:49 - 19:54
The cosine X times the DX cosine
X DX becomes du.
• 19:56 - 20:03
Subtract. Sign
to the six, X will become you to
• 20:03 - 20:09
the six. And the cosine
X DX is du.
• 20:09 - 20:13
So what we've achieved are two
very simple integrals that we
• 20:13 - 20:15
can complete to finish the
• 20:15 - 20:21
problem. The integral of you to
the four is due to the five over
• 20:21 - 20:26
5. The integral of you to the
six is due to the 7 over 7.
• 20:26 - 20:28
Plus a constant.
• 20:29 - 20:34
And then just to finish off, we
• 20:34 - 20:38
and replace EU with sign X,
which will give us 1/5.
• 20:39 - 20:42
Sign next to the five or sign to
the power 5X.
• 20:44 - 20:45
Minus.
• 20:46 - 20:52
One 7th. You to the
Seven will be signed to the 7X.
• 20:53 - 20:56
Plus a constant of integration.
• 20:58 - 21:02
So that's how we deal with
integrals of this family. In the
• 21:02 - 21:06
case when M is an even number
and when N is an odd number. Now
• 21:06 - 21:10
in the case when both M&N are
even, you should try using the
• 21:10 - 21:13
double angle formulas, and I'm
not going to do an example of
• 21:13 - 21:17
that because there isn't time in
this video to do that. But there
• 21:17 - 21:20
are examples in the exercises
accompanying the video and you
• 21:20 - 21:21
should try those for yourself.
• 21:22 - 21:29
I'm not going to look
at some integrals for which
• 21:29 - 21:31
a trigonometric substitution is
• 21:31 - 21:37
appropriate. Suppose we want to
evaluate this integral.
• 21:37 - 21:43
The integral of
1 / 1
• 21:43 - 21:46
plus X squared.
• 21:47 - 21:48
With respect to X.
• 21:50 - 21:53
Now the trigonometric
substitution that I want to use
• 21:53 - 21:59
is this one. I want to let X be
the tangent of a new variable, X
• 21:59 - 22:00
equals 10 theater.
• 22:01 - 22:04
While I picked this particular
substitution well, all will
• 22:04 - 22:09
become clear in time, but I want
to just look ahead a little bit
• 22:09 - 22:11
by letting X equal 10 theater.
• 22:12 - 22:15
What will have at the
denominator down here is
• 22:15 - 22:17
1 + 10 squared theater.
• 22:18 - 22:23
One plus X squared will become
1 + 10 squared and we have an
• 22:23 - 22:27
that 1 + 10 squared of an
• 22:27 - 22:31
angle is equal to the sequence
squared of the angle. That's
• 22:31 - 22:36
an identity that we had on the
table right at the beginning,
• 22:36 - 22:41
so the idea is that by making
this substitution, 1 + 10
• 22:41 - 22:44
squared can be replaced by a
single term sequence squared,
• 22:44 - 22:48
as we'll see, so let's
progress with that
• 22:48 - 22:48
substitution.
• 22:49 - 22:55
If we let X be tongue theater,
the integrals going to become 1
• 22:55 - 22:59
/ 1 plus X squared will become 1
+ 10 squared.
• 23:00 - 23:05
Theater. And we have to take
care of the DX in an appropriate
• 23:05 - 23:12
way. Now remember that DX is
going to be given by the XD
• 23:12 - 23:14
theater multiplied by D theater.
• 23:14 - 23:18
DXD theater we want to
differentiate X is 10 theater
• 23:18 - 23:20
with respect to theater.
• 23:20 - 23:25
Now the derivative of tongue
theater is the secant squared,
• 23:25 - 23:28
so we get secret squared Theta D
• 23:28 - 23:33
theater. So this will allow us
to change the DX in here.
• 23:34 - 23:40
Two, secant squared, Theta D
Theta over on the right.
• 23:40 - 23:45
At this stage I'm going to use
the trigonometric identity,
• 23:45 - 23:50
which says that 1 + 10 squared
of an angle is equal to the
• 23:50 - 23:55
sequence squared of the angle.
So In other words, all this
• 23:55 - 23:59
quantity down here is just the
sequence squared of Theta.
• 23:59 - 24:05
And this is very nice now
because this term here will
• 24:05 - 24:11
cancel out with this term down
in the denominator down there,
• 24:11 - 24:17
and we're left purely with the
integral of one with respect to
• 24:17 - 24:20
theater. Very simple to finish.
• 24:21 - 24:25
The integral of one with respect
to theater is just theater.
• 24:25 - 24:26
Plus a constant of integration.
• 24:28 - 24:33
original variables and if X was
• 24:33 - 24:38
10 theater than theater is the
angle whose tangent, his ex. So
• 24:38 - 24:41
theater is 10 to the minus one
• 24:41 - 24:44
of X. Plus a constant.
• 24:46 - 24:48
And that's the problem finished.
• 24:48 - 24:51
This is a very important
standard result that the
• 24:51 - 24:55
integral of one over 1 plus
X squared DX is equal to the
• 24:55 - 24:58
inverse tan 10 to the minus
one of X plus a constant.
• 24:58 - 25:01
That's a result that you'll
see in all the standard
• 25:01 - 25:04
tables of integrals, and
it's a result that you'll
• 25:04 - 25:07
need to call appan very
frequently, and if you can't
• 25:07 - 25:10
remember it, then at least
you'll need to know that
• 25:10 - 25:14
there is such a formula that
exists and you want to be
• 25:14 - 25:15
able to look it up.
• 25:17 - 25:20
I want to generalize this a
little bit to look at the case
• 25:20 - 25:25
when we deal with not just a one
here, but a more general case of
• 25:25 - 25:28
an arbitrary constant in there.
So let's look at what happens if
• 25:28 - 25:30
we have a situation like this.
• 25:31 - 25:37
Suppose we want to integrate one
over a squared plus X squared
• 25:37 - 25:39
with respect to X.
• 25:39 - 25:43
Where a is a
• 25:43 - 25:50
constant. This time I'm going to
make this substitution let X be
• 25:50 - 25:56
a town theater, and we'll see
• 25:56 - 25:58
in just a little while.
• 25:59 - 26:04
With this substitution, X is
a Tan Theta. The differential
• 26:04 - 26:09
DX becomes a secant squared
Theta D Theta.
• 26:12 - 26:14
Let's put all this into this
• 26:14 - 26:20
integral here. Will have the
integral of one over a squared.
• 26:21 - 26:27
Plus And X squared
will become a squared 10.
• 26:27 - 26:28
Squared feet are.
• 26:29 - 26:32
The
• 26:32 - 26:39
DX Will
become a sex squared Theta D
• 26:39 - 26:47
Theta. Now what I can do
now is I can take out a common
• 26:47 - 26:50
factor of A squared from the
• 26:50 - 26:57
denominator. Taking an A squared
out from this term will leave me
• 26:57 - 27:04
one taking a squared out from
this term will leave me tan
• 27:04 - 27:09
squared theater. And it's still
on the top. I've got a sex
• 27:09 - 27:11
squared Theta D Theta.
• 27:13 - 27:20
We have the trig identity that 1
+ 10 squared of any angle is sex
• 27:20 - 27:22
squared of the angle.
• 27:23 - 27:29
So I can use that identity in
here to write the denominator as
• 27:29 - 27:35
one over a squared and the 1 +
10 squared becomes simply
• 27:35 - 27:36
sequence squared theater.
• 27:37 - 27:41
We still gotten a secant squared
theater in the numerator, and a
• 27:41 - 27:45
lot of this is going to simplify
and cancel now.
• 27:46 - 27:48
The secant squared will go the
• 27:48 - 27:52
top and the bottom. The one of
these at the bottom will go with
• 27:52 - 27:56
the others at the top, and we're
left with the integral of one
• 27:56 - 27:58
over A with respect to theater.
• 28:00 - 28:03
Again, this is straightforward
to finish. The integral of one
• 28:03 - 28:06
over a one over as a constant
with respect to Theta is just
• 28:06 - 28:08
going to give me one over a.
• 28:09 - 28:12
Theater. Plus the constant of
• 28:12 - 28:17
original variables, we've got to
• 28:17 - 28:22
go back to our original
substitution. If X is a tan
• 28:22 - 28:27
Theta, then we can write that X
over A is 10 theater.
• 28:27 - 28:31
And In other words, that
theater is the angle whose
• 28:31 - 28:35
tangent is 10 to the minus
one of all this X over a.
• 28:37 - 28:41
That will enable me to write our
final results as one over a town
• 28:41 - 28:43
to the minus one.
• 28:43 - 28:46
X over a.
• 28:46 - 28:48
Plus a constant of integration.
• 28:49 - 28:53
And this is another very
important standard result that
• 28:53 - 28:57
the integral of one over a
squared plus X squared with
• 28:57 - 29:04
respect to X is one over a 10 to
the minus one of X over a plus a
• 29:04 - 29:08
constant, and as before, that's
a standard result that you'll
• 29:08 - 29:12
see frequently in all the tables
of integrals, and you'll need to
• 29:12 - 29:17
call a pawn that in lots of
situations when you're required
• 29:17 - 29:18
to do integration.
• 29:18 - 29:24
OK, so now we've got the
standard result that the
• 29:24 - 29:31
integral of one over a squared
plus X squared DX is equal
• 29:31 - 29:38
to one over a town to
the minus one of X of
• 29:38 - 29:40
A. As a constant of integration.
• 29:41 - 29:46
Let's see how we might use
this formula in a slightly
• 29:46 - 29:52
different case. Suppose we
have the integral of 1 / 4 +
• 29:52 - 29:54
9 X squared DX.
• 29:55 - 29:59
Now this looks very similar to
the standard formula we have
• 29:59 - 30:01
here. Except there's a slight
• 30:01 - 30:05
problem. And the problem is that
instead of One X squared, which
• 30:05 - 30:08
we have in the standard result,
I've got nine X squared.
• 30:09 - 30:12
What I'm going to do is I'm
going to divide everything at
• 30:12 - 30:16
the bottom by 9, take a factor
of nine out so that we end up
• 30:16 - 30:19
with just a One X squared here.
So what I'm going to do is I'm
• 30:19 - 30:21
going to write the denominator
• 30:21 - 30:25
like this. So I've taken a
factor of nine out. You'll see
• 30:25 - 30:30
if we multiply the brackets
again here, there's 9 * 4 over
• 30:30 - 30:35
9, which is just four and the
nine times the X squared, so I
• 30:35 - 30:40
haven't changed anything. I've
just taken a factor of nine out
• 30:40 - 30:45
the point of doing that is that
now I have a single. I have a
• 30:45 - 30:50
One X squared here, which will
match the formula I have there.
• 30:50 - 30:54
If I take the 9 outside the
• 30:54 - 30:59
integral. I'm left with 1 /, 4
ninths plus X squared integrated
• 30:59 - 31:06
with respect to X and I hope you
can see that this is exactly one
• 31:06 - 31:11
of the standard forms. Now when
we let A squared B4 over nine
• 31:11 - 31:17
with a squared is 4 over 9. We
have the standard form. If A
• 31:17 - 31:23
squared is 4 over 9A will be 2
over 3 and we can complete this
• 31:23 - 31:28
integration. Using the standard
result that one over 9 stays
• 31:28 - 31:30
there, we want one over A.
• 31:31 - 31:35
Or A is 2/3. So
we want 1 / 2/3.
• 31:36 - 31:38
10 to the minus one.
• 31:38 - 31:40
Of X over a.
• 31:41 - 31:45
X divided by a is X divided by
• 31:45 - 31:48
2/3. Plus a constant of
• 31:48 - 31:53
integration. Just to tide to
these fractions up, three will
• 31:53 - 31:58
divide into 9 three times, so
we'll have 326 in the
• 31:58 - 32:04
denominator. 10 to the minus one
and dividing by 2/3 is like
• 32:04 - 32:10
multiplying by three over 2, so
I'll have 10 to the minus one of
• 32:10 - 32:12
three X over 2 plus the constant
• 32:12 - 32:17
of integration. So the point
here is you might have to do a
• 32:17 - 32:19
bit of work on the integrand
in order to be able to write
• 32:19 - 32:22
it in the form of one of the
standard results.
• 32:23 - 32:29
OK, let's have a look at another
case where another integral to
• 32:29 - 32:33
look at where a trigonometric
substitution is appropriate.
• 32:33 - 32:39
Suppose we want to find the
integral of one over the square
• 32:39 - 32:42
root of A squared minus X
• 32:42 - 32:47
squared DX. Again,
A is a constant.
• 32:50 - 32:56
The substitution that I'm
going to make is this one.
• 32:56 - 33:01
I'm going to write X equals
a sign theater.
• 33:02 - 33:09
If I do that, what will happen
to my integral, let's see.
• 33:09 - 33:11
And have the integral of one
• 33:11 - 33:18
over. The square root. The A
squared will stay the same, but
• 33:18 - 33:23
the X squared will become a
squared sine squared.
• 33:23 - 33:26
I squared sine squared Theta.
• 33:27 - 33:31
Now the reason I've done that
is because in a minute I'm
• 33:31 - 33:35
going to take out a factor of a
squared, which will leave me
• 33:35 - 33:40
one 1 minus sign squared, and I
do have an identity involving 1
• 33:40 - 33:43
minus sign squared as we'll
see, but just before we do
• 33:43 - 33:47
that, let's substitute for the
differential as well. If X is a
• 33:47 - 33:51
sign theater, then DX will be a
cosine, Theta, D, Theta.
• 33:52 - 33:59
So we have a cosine Theta D
Theta for the differential DX.
• 34:01 - 34:07
Let me take out the factor of a
squared in the denominator.
• 34:08 - 34:14
first term will leave me one
• 34:14 - 34:20
and a squared from the second
term will leave me one minus
• 34:20 - 34:22
sign squared Theta.
• 34:23 - 34:27
I have still gotten a costly to
the theater at the top.
• 34:28 - 34:32
Now let me remind you there's a
trig identity which says that
• 34:32 - 34:35
the cosine squared of an angle
plus the sine squared of an
• 34:35 - 34:37
angle is always one.
• 34:37 - 34:41
So if we have one minus the sine
squared of an angle, we can
• 34:41 - 34:42
replace it with cosine squared.
• 34:43 - 34:50
So 1 minus sign squared Theta
we can replace with simply
• 34:50 - 34:52
cosine squared Theta.
• 34:52 - 34:54
Is the A squared out the
frontier and we want the square
• 34:54 - 34:56
root of the whole lot.
• 34:56 - 35:03
Now this is very simple. We want
the square root of A squared
• 35:03 - 35:09
cosine squared Theta. We square
root. These squared terms will
• 35:09 - 35:11
be just left with.
• 35:11 - 35:13
A cosine Theta.
• 35:13 - 35:19
In the denominator and within a
cosine Theta in the numerator.
• 35:20 - 35:22
And these were clearly
cancel out.
• 35:23 - 35:28
And we're left with the integral
of one with respect to theater,
• 35:28 - 35:31
which is just theater plus a
constant of integration.
• 35:34 - 35:39
variables, given that X was a
• 35:39 - 35:44
sign theater, then clearly X
over A is sign theater.
• 35:44 - 35:50
So theater is the angle who sign
is or sign to the minus one of X
• 35:50 - 35:55
over a, so replacing the theater
with sign to the minus one of X
• 35:55 - 35:57
over a will get this result.
• 35:58 - 36:02
And this is a very important
standard result that if you want
• 36:02 - 36:08
to integrate 1 divided by the
square root of A squared minus X
• 36:08 - 36:12
squared, the result is the
inverse sine or the sign to the
• 36:12 - 36:15
minus one of X over a.
• 36:15 - 36:17
Plus a constant of integration.
• 36:18 - 36:25
Will have a look one final
example which is a variant on
• 36:25 - 36:32
the previous example. Suppose we
want to integrate 1 divided by
• 36:32 - 36:38
the square root of 4 -
9 X squared DX.
• 36:39 - 36:43
Now that's very similar to the
one we just looked at. Remember
• 36:43 - 36:48
that we had the results that the
integral of one over the square
• 36:48 - 36:50
root of A squared minus X
• 36:50 - 36:55
squared DX. Was the inverse sine
of X over a plus a constant?
• 36:55 - 36:59
That's keep that in mind. That's
the standard result we've
• 36:59 - 37:04
there. In this case. The problem
• 37:04 - 37:08
is that instead of a single X
squared, we've got nine X
• 37:08 - 37:12
squared. So like we did in the
other example, I'm going to take
• 37:12 - 37:16
the factor of nine out to leave
us just a single X squared in
• 37:16 - 37:18
there, and I do that like this.
• 37:19 - 37:26
Taking a nine out from
these terms here, I'll have
• 37:26 - 37:30
four ninths minus X squared.
• 37:30 - 37:34
Again, the nine times the four
ninths leaves the four which we
• 37:34 - 37:37
got the nine X squared, which we
• 37:37 - 37:42
have there. The whole point of
doing that is that then I'm
• 37:42 - 37:46
going to extract the Route 9,
which is 3 and bring it right
• 37:46 - 37:52
outside. And inside under the
integral sign, I'll be left with
• 37:52 - 37:58
one over the square root of 4
ninths minus X squared DX.
• 38:00 - 38:05
Now in this form, I hope you can
spot that we can use the
• 38:05 - 38:09
standard result immediately with
the standard results, with a
• 38:09 - 38:12
being with a squared being equal
to four ninths.
• 38:13 - 38:17
In other words, a being equal to
• 38:17 - 38:21
2/3. Putting all that together
will have a third. That's the
• 38:21 - 38:25
third and the integral will
become the inverse sine.
• 38:26 - 38:30
X. Divided by AA
• 38:30 - 38:35
was 2/3. Plus a
constant of integration.
• 38:37 - 38:43
And just to tidy that up will be
left with the third inverse sine
• 38:43 - 38:48
dividing by 2/3 is the same as
multiplying by three over 2, so
• 38:48 - 38:53
will have 3X over 2 plus a
constant of integration.
• 38:53 - 38:56
And that's our final
result. So we've seen a lot
• 38:56 - 38:58
of examples that have
integration using
• 38:58 - 39:00
trigonometric identities
and integration using trig
• 39:00 - 39:03
substitutions. You need a
lot of practice, and there
• 39:03 - 39:06
are a lot of exercises in
the accompanying text.
Title:
www.mathcentre.ac.uk/.../9.7%20Integration%20requiring%20use%20of%20trigonometric%20identities.mp4
Video Language:
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