WEBVTT

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>> Welcome to the Cypress
College math review

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on integration by parts.

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Integration by parts is
a method that allows us

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to integrate more
types of integrals.

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Especially those consisting
of a product of two functions.

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Let u and v be differentiable
functions

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of x. I can write the
derivative of u either way.

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Also, the derivative
of v. The differential

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of u can be written this way.

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And here's the differential
of v.

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The product rule

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for a derivative gives us u
prime times v plus v prime times

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u. Integrating both sides of
this equation with respect to x.

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Gives us the antiderivative of
v. And then we have u prime dx.

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Which, of course, from
what we have above, is du.

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And then we're integrating u.
And then v prime dx gives us dv.

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If we solve for this integral.

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We have the formula for
integration by parts.

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Once you've determined that
you're going to use integration

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by parts to evaluate
your integral.

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You need to determine
which factor will be u.

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Which you will be
differentiating.

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And which factor will be dv,
which you will be integrating.

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For dv, choose the
most complicated part

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of the integrand that fits
a basic integration rule.

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When choosing the
factor that will be u,

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use the acronym L.I.A.T.E.
Which stands for logarithmic,

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inverse trig, algebraic, trig,
or exponential -- in that order.

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In other words, if you have
a factor that's logarithmic.

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For sure, that factor is u.

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If you have a factor that's
inverse trig, make that u.

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Recall once again
what our formula

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for integration by parts is.

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Now, a lot of instructors
will allow you

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to use a tabular method.

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With a tabular method,
we set it up like this.

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I've seen other ways to do it.

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But something like this.

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Which part do you differentiate?

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You differentiate the factor
u. Which part do you integrate?

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You integrate the
factor dv, yes?

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When you differentiate
u, you get du.

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When you integrate dv, you
get v. And then, of course,

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because of the formula
that we have here.

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We have to tack on
the plus and the minus

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to get the appropriate answers.

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So this original integral.

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Is equal to this product,
minus this integral.

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Now, did that really
save you time?

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No. What's nice about
the tabular method,

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is you can continue to do
this process numerous times.

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Numerous steps over
and over again.

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And often, integration by parts
has to be done numerous steps.

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U substitution does not
work on this problem.

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And since we have a logarithmic
factor, u would be natural log

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of x. So dv is x squared dx.

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We differentiate
to get 1 over x dx.

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We integrate to get
x cubed over 3.

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So our original integral
is equal to u times v.

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Minus the integral of v du.

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Pull out the 1/3.

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We cancel, and now we just
have x squared to integrate.

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So to integrate x squared
using the power rule.

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We would get x cubed
over 3 times the other 3

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in the denominator, gives me 9.

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We have our answer.

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I'm going to show
the same problem done

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over using the tabular method.

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So in this column, we're
going to differentiate.

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And in this column,
we're going to integrate.

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So looking at natural
log and x squared.

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We're definitely going to
differentiate natural log.

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Since we don't know how
to integrate natural log.

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So x squared will be the
factor that we integrate.

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The derivative of natural log
is 1 over x. The antiderivative

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of x squared is x cubed over 3.

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Then, because of the formula,
you put on your signs.

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So plus, minus.

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And then, if this
process continued on

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and you did integration by
parts through multiple steps.

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You would continue this
process down with the signs.

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And you'd keep differentiating,
keep integrating.

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So the original integral.

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Is equal to -- this
is the u times v part.

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So on the diagonal is u times
v. So that is a product --

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-- Minus. And horizontal
is the integral.

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And you can see, that's the
same problem I just did.

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And here we go.

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On this problem you
have a choice.

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You can let u be the
exponential function,

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or the trigonometric function.

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It really doesn't matter.

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I'll let u be the
trigonometric function.

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So we'll let u be cosine of 3x.

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Which means that du will be
negative 3 times sine of 3xdx.

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Dv then, will be e to the 4x dx.

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And v then, will
be 1/4 e to the 4x.

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So our original integral.

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Is equal to u times v. Which is
1/4 e to the 4x cosine of 3x.

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Minus the integral of v du.

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I have a double negative.

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So I have plus and then 3/4,
e to the 4x sine of 3xdx.

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But once again, I have another
integral that I'm going to have

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to use integration
by parts to do.

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So we're starting over.

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This time, u is equal
to sine of 3x.

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Du is equal to 3 cosine of 3xdx.

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Dv, e to the 4x dx.

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Which gives us v is
equal to 1/4 e to the 4x.

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So our original integral.

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Is equal to our original
u times v.

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Plus 3/4 times our new integral.

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Our new integral has the new
u times v, which is 1/4 e

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to the 4x sine of 3x,
minus the integral of v du.

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Which, in this case, would be
3/4 times the integral of e

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to the 4x cosine of 3xdx.

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Please note that this
integral and the integral

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on the left-hand side.

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Which was my original integral,
are multiples of each other.

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Those integrals are the
-- it's the same integral.

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But we have a different
constant in front of them.

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Alright, let's take a
look at those constants.

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So let's deal with the
coefficients in front of these.

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So here I have 3/16.

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And here I have negative 9/16.

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So on the left-hand side, I
have 1 times this integral.

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On the right-hand side,

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I have negative 9/16
times this integral.

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So I'm going to add 9/16
times this same integral

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to both sides.

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Yes? When I do that, I get 25/16
times that integral is equal

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to these two terms here.

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Then I simply multiply
both sides

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by the reciprocal of 25/16.

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So once again, I'm
multiplying by 16 over 25.

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And I would get 4 over 25.

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Once again, over here, I'm
multiplying by 16 over 25.

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Both sides of the
equation by that.

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Yes. So distribute it to there
and you distribute it to here.

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So I have 4 over 25 here.

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And here I get 3 over 25.

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And then, of course, plus
c. And we have our answer.

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We're going to do this problem
again using the tabular method.

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We also want to talk about the
number of steps of integration

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by parts it takes
to do a problem.

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So we differentiate
what's on the left

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and we integrate
what's on the right.

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So we have our steps
that we're --

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our functions that we're
differentiating on the left.

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Integrating on the right.

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We put on our signs.

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So we're going down through.

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And how do we know when to stop?

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The original integral is
equal to this product.

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Plus this product, plus this
product, and so on and so forth.

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Plus this integral.

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So either you can
integrate this.

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Or, it is a multiple --

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-- Of the original integral.

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Like the last problem.

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That's how many steps it takes
to do integration by parts.

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That's when you know
you're done.

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So let's differentiate cosine
3x and integrate e to the 4x.

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So the derivative is
negative 3 sine of 3x.

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The antiderivative
is 1/4 e to the 4x.

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Are we done?

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Can we integrate
this bottom row?

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No. Is this bottom
row a multiple

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of the original integral?

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No. Keep going.

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Differentiate again.

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Alright, let's see
if we're done.

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Can we integrate
this bottom row?

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No. Is this bottom
row a multiple

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of my original integral?

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Yes. We're done.

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Tack on the signs,
let's do this.

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So the original integral --

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-- Is equal to our
-- this product.

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Plus this product.

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So 3/16 u to the 4x sine
of 3x, plus this integral.

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So that's minus 9/16
times the integral of e

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to the 4x cosine of 3x.

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So I'm going to add 9/16 times
that integral to both sides.

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So we get 25/16 times
the integral.

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Just like we did before.

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So we have the same answer.

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Skip a step since we've
already done this problem.

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There we go.

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For multiple step
problems, there's nothing

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like the tabular method.

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So, since our algebraic function
will differentiate basically

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into nothing.

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We will do that.

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Our trigonometric function
will just alternate

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between sines and cosines.

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We'll let it do that.

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How far -- how do I, how
do I know how far to go?

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Until it's disappeared
into nothing.

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I didn't really have
to go that far.

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How far did I have to go?

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I had to go until
I could integrate.

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So I could've done
one less step.

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I could've integrated 6 cosine
x. But why not go ahead and go

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down to just integrating zero.

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Yeah? Nothing wrong with that.

00:19:59.386 --> 00:20:00.946 A:middle
So our answer is.

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This product, negative x cubed
cosine x. Plus this product,

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3x squared sine of
x. Plus this product,

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6x cosine x. Plus this
product, negative 6 sine

00:20:32.596 --> 00:20:35.936 A:middle
of x. Plus this integral.

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Now what are you integrating?

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You're integrating zero.

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What's the antiderivative
of zero?

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C, yeah? It's the constant
of integration, yeah?

00:20:51.686 --> 00:20:54.646 A:middle
When you integrate
zero with respect to x,

00:20:54.646 --> 00:20:57.706 A:middle
you simply get constant
of integration.

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Don't you?

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It's just c. There's nothing
like the tabular method

00:21:06.446 --> 00:21:08.706 A:middle
for integration by
parts when you have

00:21:08.706 --> 00:21:10.886 A:middle
to do it multiple steps.

00:21:13.886 --> 00:21:18.396 A:middle
Here are some more problems
that you might want to do.

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Pause the video.

00:21:20.476 --> 00:21:21.906 A:middle
Try them on your own.

00:21:23.136 --> 00:21:26.646 A:middle
Then restart the video
and check your answers.

00:21:31.816 --> 00:21:33.976 A:middle
Here are the answers to
the practice problems.