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JOEL LEWIS: Hi, welcome
back to recitation.

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In lecture, Professor Miller
did a bunch of examples of

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integrals involving
trigonometric functions.

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So I thought I would give you
a couple more, well I guess

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three more examples, some of
them are a little different

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flavor than the ones he did, but
some nice examples of some

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integrals you can compute now.

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One thing you might need for
them is you're going to have

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to remember some of your
trig identities.

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Just like you had to remember
some of them last time.

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So in particular, one identity
that you might need today that

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you didn't need in lecture,
was the angle sum

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identity for cosine.

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So let me just remind you what
that is, it says the cosine of

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a plus b is equal to cosine a
cosine b minus sine a sine b.

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So you're going to need that
formula to compute one of

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these three integrals.

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So why don't you pause the
video, take some time

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to work these out.

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Come back you can check your
answers against my work.

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Hopefully you had some
fun and some luck

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working on these integrals.

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Let's have a go at them.

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The first integral is sine
cubed x secant squared x.

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So I don't think that Professor
Miller did any

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integrals that involved the
secant function, but one thing

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to remember is that the secant
function is very closely

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related to the cosine
function.

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It's just cosine
to the minus 1.

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Sorry, by that I mean 1 over
cosine, not the arc cosine.

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So in order to, so we can view
this first integral here as an

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integral of a power of sine
times a power of cosine, just

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like the ones you
had in lecture.

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So we can write this as sine
cubed of x times--

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I'm going to write--

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cosine x to the minus 2 dx.

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And here we see this is a nice
example where the sine occurs

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with an odd power.

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So when the sine occurs with an
odd power, or when one of

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them at least occurs with
an odd power, life

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is relatively simple.

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And what we do is just what we
did in lecture, which is we

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break it up so that we just have
one instance of the one

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that occurs to an odd power and
then a bunch of powers of

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the other one.

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So in this case, we have
sine to an odd power.

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So we want to pull out all the
even, you know, all the extra

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multiples, so in this case we
have sine times sine squared.

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So we take all those sine
squared's and we convert them

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into 1 minus cosine squared's.

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This is going to be equal to the
integral of sine x times,

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well, times sine squared x,
which is 1 minus cosine

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squared x times cosine x to
the minus second power dx.

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And now you take all these
cosines and you multiply them

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together and you see
what you've got.

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So in this case that's equal
to the integral--

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OK, so 1 times cosine to the
minus 2, is just cosine

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to the minus 2.

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So this first term is cosine x
to the minus 2 sine x and then

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minus, OK we have cosine squared
times cosine to the

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minus 2, so that one's even
simpler, that's just

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minus sine x dx.

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So at this point, if you like,
you could do a substitution of

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a trigonometric function.

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You could set u equal
to cosine x.

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All right?

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What's going to happen if you
set u equal to cosine x, well

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du is going to be
minus sine x dx.

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Each of these have exactly one
appearance of sine x in them,

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so this would become--

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so if we set u equal cosine x.

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So, as I said du equals
minus sine x dx.

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Because each of these have a
sine x dx in them, that's

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going to get clumped into the
du and the cosines will just

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get replaced with
u everywhere.

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So I'm going to take this
over here so I have

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a little more space.

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So this becomes the integral of
u to the minus 2, I guess

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it's minus u to the
minus 2 plus 1 du.

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That minus sine x is just
exactly equal to du.

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OK.

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And now this is easy to
finish from here.

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This is equal to, well OK, so
minus 2, so we have to go up

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one to minus 1, so that's just
u to the minus 1 plus 1 gives

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us a u plus a constant, and
then of course we have to

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substitute back in.

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So this is u is cosine x, so u
to the minus 1 is secant x

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plus cosine x plus a constant.

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So that's the first one.

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For the second one, sine
x cosine of 2x.

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We actually have
a slightly more

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complicated situation here.

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Because the cosine 2x is no
longer, it's not just a

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trigonometric function of x
alone, it's a trigonometric

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function of 2x.

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So in order to do the sort of
the things we did in lecture,

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what you're going to have to do
is, you're going to have to

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expand this out in terms of
just sine x and cosine x,

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rather than cosine of 2x.

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So in order to do that you just
need to remember your

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double angle formula.

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Now if you don't remember your
double angle formulas, one way

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you can remember them is
remember the angle sum formula

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and apply it with a and
b both equal to x.

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If you like.

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But, or you could just
remember your

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double angle formulas.

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So let's go over here.

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And so in this case, we have the
integral of sine x times,

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well cosine 2x-- so there are
actually, right, there are

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several different ways you
can write cosine 2x.

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So one way you can write it is
cosine squared x minus sine

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squared x, but we want
everything-- again we have an

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odd power of sine, so it would
be nice if everything else

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could be written in terms
of just cosine.

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So an equivalent form for cosine
2x is to write, is that

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it's equal to 2 cosine
squared x minus 1 dx.

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And here, this is very, very
similar to the last question.

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Again, you can make the
substitution u equals cosine,

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if you like.

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And so I'm not going to do out
the whole substitution for

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you, but this is going to work
out to 2 cosine squared x sine

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x, the integral of that is minus
2/3 cosine cubed x and

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then the integral of minus
sine x is cosine x plus a

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constant of integration.

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So that's the second one.

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The third one now is of a
similar flavor, except you

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have here, you have sine
2x and cosine 3x.

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So again, when you
want to integrate

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something where you have--

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so here we have both
trigonometric functions occur

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in multiples.

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No if it was the same multiple,
if it were 2x and

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2x, say, that would be fine.

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You know, you could just make
a substitution like u equals

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2x or something and then
proceed as usual.

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But because they're different
multiples, we need, when we do

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these trig integrals, we need
to apply the methods that we

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used in lecture.

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You want the arguments
to agree.

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So you want it all to be sine of
something and cosine of the

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same thing.

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So here the thing to do, I
think, is to try and put

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everything in terms of just
sine x and cosine x.

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So sine 2x is one you
should remember,

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double angle formula.

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Cosine 3x, well in order to
figure out what cosine 3x is,

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you could just know it,
maybe you learned

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it once upon a time.

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The other thing you can do for
cosine 3x, is you can use this

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angle sum formula.

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So let me just work that
out for you quickly.

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So cosine of 3x, well how do you
use the angle sum formula?

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You can write 3x as 2x plus x.

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So this is equal to cosine
of 2x plus x.

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And now this is a sum, so you
can use the cosine sum

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formula, so you get this is
equal to cosine 2x cosine x

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minus sine 2x sine x.

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OK and now you can use
the double angle

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formulas here and here.

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so in the end, if you do this
all out, what you get is

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you're going to get 4 cosine
cubed x minus 3 cosine x.

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You might say, oh what happened
to the sine x's?

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We're going to get a sign
squared x and I want to write

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everything in terms of cosine.

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So I've replaced the, I've done
an extra step here that

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I'm not showing, where I
replaced the sine squared x

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with a 1 minus cosine
squared x.

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So you make this substitution,
and you also make the

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substitution for sign x--
sorry-- for sine 2x that you

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have. OK, sine 2x is equal
to-- you know,

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double angle formula--

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2 sine x cosine x.

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And so when you multiply sine 2x
by cosine 3x you again have

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an expression that's got powers
of cosine with a single

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power of sine.

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So doing these simplifications,
you again get

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this nice form.

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One of them's odd, the other
one, so, one of the trig

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functions, sine or cosine
appears to just the power of

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1, so, OK, so you can do this
nice simple substitution.

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You don't need to do any of
the double angle, or half

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angle complication.

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I'm not going to finish this
one out for you, but I did

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write down the answer so you
can check your answer.

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When you, the third integral
that we had, which is integral

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sine 2x cosine 3x dx, when you
work it all out you should

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get, let's see what I got, OK,
you should get minus 8/5

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cosine to the fifth x plus 2
cosine cubed x plus a constant

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of integration.

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So if you work out all the
details, this is what you

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should get.

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I'll stop there.

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