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JOEL LEWIS: Hi, welcome
back to recitation.

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In lecture, Professor
Miller did a bunch

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of examples of
integrals involving

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trigonometric functions.

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So I thought I would
give you a couple more,

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well I guess three more
examples, some of them

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are a little different
flavor than the ones he did,

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but some nice examples of some
integrals you can compute now.

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One thing you might
need for them is you're

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going to have 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

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

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for cosine.

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So let me just
remind you what that

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is, it says the
cosine of a plus b

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

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So you're going to
need that formula

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to compute one of
these three integrals.

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So why don't you
pause the video,

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take some time 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

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did any integrals that
involved the secant function,

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but one thing to remember is
that the secant function is

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very closely 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 arccosine.

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So, in order to--

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so we can view this
first integral here

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as an integral of a power of
sine times a power of cosine,

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just 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 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 them

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at least occurs
with an odd power,

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life is relatively simple.

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And what we do is
just what we did

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in lecture, which is we
break it up so that we just

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have one instance of the one
that occurs to an odd power

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and then a bunch of
powers of 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,

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you know, all the extra
multiples, so in this case

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we have sine times sine squared.

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So we take all
those sine squareds

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and we convert them into
1 minus cosine squareds.

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

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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

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and you multiply them together
and you see what you've got.

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

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so 1 times cosine to the minus
2 is just cosine to the minus 2.

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So this first term is
cosine x to the minus 2

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sine x and then minus--
OK, we have cosine

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squared times cosine
to the minus 2,

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so that one's even simpler,
that's just minus sine x dx.

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So at this point,
if you like, you

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could do a substitution of
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

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u equal to cosine x, well 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-- 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,

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that's going to get
clumped into the du.

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And the cosines will just get
replaced with u everywhere.

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So I'm going to
take this over here

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so I have a little more space.

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

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I guess 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,

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so we have to go
up one to minus 1,

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

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a u, plus a constant,
and then of course we

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have to substitute back in.

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

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is secant x 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 complicated situation here.

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Because the cosine
2x is no longer,

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it's not just a trigonometric
function of x alone,

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it's a trigonometric
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,

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you're going to have to expand
this out in terms of just sine

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x and cosine x, rather
than cosine of 2x.

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So in order to do that
you just need to remember

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your double angle formula.

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

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one way 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 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,

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there are 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

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we have an odd power of
sine, so it would be nice

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if everything else
could be written

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in terms of just cosine.

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

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is that 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 you,

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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 then

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the integral of minus
sine x is cosine x

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plus a 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,

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

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So again, when you want
to integrate something

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where you have-- so here we have
both trigonometric functions

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occur in multiples.

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Now, if it was
the same multiple,

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if it were 2x and 2x,
say, that would be fine.

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You know, you could
just make a substitution

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like u equals 2x or something
and then proceed as usual.

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But because they're
different multiples,

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we need, when we do
these trig integrals,

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we need to apply the methods
that we 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

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of the same thing.

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So here the thing
to do, I think,

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is to try and put
everything in terms

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of just sine x and cosine x.

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So sine 2x is one you should
remember, 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

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learned it once upon a time.

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The other thing you
can do for cosine 3x,

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is you can use this
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 formula,

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so you get this is equal to
cosine 2x cosine x minus sine

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2x sine x.

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OK and now you can use
the double angle formulas

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here and here.

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

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you get is you're
going to get 4 cosine

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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 sine squared x

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and I want to write
everything in terms of cosine.

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So I've replaced the-- I've
done an extra step here

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that I'm not showing,
where I replaced

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the sine squared x with a
1 minus cosine squared x.

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So you make this
substitution, and you also

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make the substitution
for sine x-- sorry,

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for sine 2x that you have.

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

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you again have an expression
that's got powers of cosine

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with a single power of sine.

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So doing these simplifications,
you again get this nice form.

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One of them's odd,
the other one-- one

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of the trig functions,
sine or cosine

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appears to just the
power of 1, so, OK,

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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 angle

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complication.

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

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but I did write down the answer
so you can check your answer.

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When you-- the third
integral that we had,

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which is integral sine 2x
cosine 3x dx, when you work it

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all out you should get,
let's see what I got, OK,

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you should get minus 8/5
cosine to the fifth x

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plus 2 cosine cubed x plus
a constant of integration.

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So if you work out
all the details,

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this is what you should get.

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I'll stop there.