WEBVTT
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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.