>> Welcome to the Cypress College math review
on integration by parts.
Integration by parts is a method that allows us
to integrate more types of integrals.
Especially those consisting of a product of two functions.
Let u and v be differentiable functions
of x. I can write the derivative of u either way.
Also, the derivative of v. The differential
of u can be written this way.
And here's the differential of v.
The product rule
for a derivative gives us u prime times v plus v prime times
u. Integrating both sides of this equation with respect to x.
Gives us the antiderivative of v. And then we have u prime dx.
Which, of course, from what we have above, is du.
And then we're integrating u. And then v prime dx gives us dv.
If we solve for this integral.
We have the formula for integration by parts.
Once you've determined that you're going to use integration
by parts to evaluate your integral.
You need to determine which factor will be u.
Which you will be differentiating.
And which factor will be dv, which you will be integrating.
For dv, choose the most complicated part
of the integrand that fits a basic integration rule.
When choosing the factor that will be u,
use the acronym L.I.A.T.E. Which stands for logarithmic,
inverse trig, algebraic, trig, or exponential -- in that order.
In other words, if you have a factor that's logarithmic.
For sure, that factor is u.
If you have a factor that's inverse trig, make that u.
Recall once again what our formula
for integration by parts is.
Now, a lot of instructors will allow you
to use a tabular method.
With a tabular method, we set it up like this.
I've seen other ways to do it.
But something like this.
Which part do you differentiate?
You differentiate the factor u. Which part do you integrate?
You integrate the factor dv, yes?
When you differentiate u, you get du.
When you integrate dv, you get v. And then, of course,
because of the formula that we have here.
We have to tack on the plus and the minus
to get the appropriate answers.
So this original integral.
Is equal to this product, minus this integral.
Now, did that really save you time?
No. What's nice about the tabular method,
is you can continue to do this process numerous times.
Numerous steps over and over again.
And often, integration by parts has to be done numerous steps.
U substitution does not work on this problem.
And since we have a logarithmic factor, u would be natural log
of x. So dv is x squared dx.
We differentiate to get 1 over x dx.
We integrate to get x cubed over 3.
So our original integral is equal to u times v.
Minus the integral of v du.
Pull out the 1/3.
We cancel, and now we just have x squared to integrate.
So to integrate x squared using the power rule.
We would get x cubed over 3 times the other 3
in the denominator, gives me 9.
We have our answer.
I'm going to show the same problem done
over using the tabular method.
So in this column, we're going to differentiate.
And in this column, we're going to integrate.
So looking at natural log and x squared.
We're definitely going to differentiate natural log.
Since we don't know how to integrate natural log.
So x squared will be the factor that we integrate.
The derivative of natural log is 1 over x. The antiderivative
of x squared is x cubed over 3.
Then, because of the formula, you put on your signs.
So plus, minus.
And then, if this process continued on
and you did integration by parts through multiple steps.
You would continue this process down with the signs.
And you'd keep differentiating, keep integrating.
So the original integral.
Is equal to -- this is the u times v part.
So on the diagonal is u times v. So that is a product --
-- Minus. And horizontal is the integral.
And you can see, that's the same problem I just did.
And here we go.
On this problem you have a choice.
You can let u be the exponential function,
or the trigonometric function.
It really doesn't matter.
I'll let u be the trigonometric function.
So we'll let u be cosine of 3x.
Which means that du will be negative 3 times sine of 3xdx.
Dv then, will be e to the 4x dx.
And v then, will be 1/4 e to the 4x.
So our original integral.
Is equal to u times v. Which is 1/4 e to the 4x cosine of 3x.
Minus the integral of v du.
I have a double negative.
So I have plus and then 3/4, e to the 4x sine of 3xdx.
But once again, I have another integral that I'm going to have
to use integration by parts to do.
So we're starting over.
This time, u is equal to sine of 3x.
Du is equal to 3 cosine of 3xdx.
Dv, e to the 4x dx.
Which gives us v is equal to 1/4 e to the 4x.
So our original integral.
Is equal to our original u times v.
Plus 3/4 times our new integral.
Our new integral has the new u times v, which is 1/4 e
to the 4x sine of 3x, minus the integral of v du.
Which, in this case, would be 3/4 times the integral of e
to the 4x cosine of 3xdx.
Please note that this integral and the integral
on the left-hand side.
Which was my original integral, are multiples of each other.
Those integrals are the -- it's the same integral.
But we have a different constant in front of them.
Alright, let's take a look at those constants.
So let's deal with the coefficients in front of these.
So here I have 3/16.
And here I have negative 9/16.
So on the left-hand side, I have 1 times this integral.
On the right-hand side,
I have negative 9/16 times this integral.
So I'm going to add 9/16 times this same integral
to both sides.
Yes? When I do that, I get 25/16 times that integral is equal
to these two terms here.
Then I simply multiply both sides
by the reciprocal of 25/16.
So once again, I'm multiplying by 16 over 25.
And I would get 4 over 25.
Once again, over here, I'm multiplying by 16 over 25.
Both sides of the equation by that.
Yes. So distribute it to there and you distribute it to here.
So I have 4 over 25 here.
And here I get 3 over 25.
And then, of course, plus c. And we have our answer.
We're going to do this problem again using the tabular method.
We also want to talk about the number of steps of integration
by parts it takes to do a problem.
So we differentiate what's on the left
and we integrate what's on the right.
So we have our steps that we're --
our functions that we're differentiating on the left.
Integrating on the right.
We put on our signs.
So we're going down through.
And how do we know when to stop?
The original integral is equal to this product.
Plus this product, plus this product, and so on and so forth.
Plus this integral.
So either you can integrate this.
Or, it is a multiple --
-- Of the original integral.
Like the last problem.
That's how many steps it takes to do integration by parts.
That's when you know you're done.
So let's differentiate cosine 3x and integrate e to the 4x.
So the derivative is negative 3 sine of 3x.
The antiderivative is 1/4 e to the 4x.
Are we done?
Can we integrate this bottom row?
No. Is this bottom row a multiple
of the original integral?
No. Keep going.
Differentiate again.
Alright, let's see if we're done.
Can we integrate this bottom row?
No. Is this bottom row a multiple
of my original integral?
Yes. We're done.
Tack on the signs, let's do this.
So the original integral --
-- Is equal to our -- this product.
Plus this product.
So 3/16 u to the 4x sine of 3x, plus this integral.
So that's minus 9/16 times the integral of e
to the 4x cosine of 3x.
So I'm going to add 9/16 times that integral to both sides.
So we get 25/16 times the integral.
Just like we did before.
So we have the same answer.
Skip a step since we've already done this problem.
There we go.
For multiple step problems, there's nothing
like the tabular method.
So, since our algebraic function will differentiate basically
into nothing.
We will do that.
Our trigonometric function will just alternate
between sines and cosines.
We'll let it do that.
How far -- how do I, how do I know how far to go?
Until it's disappeared into nothing.
I didn't really have to go that far.
How far did I have to go?
I had to go until I could integrate.
So I could've done one less step.
I could've integrated 6 cosine x. But why not go ahead and go
down to just integrating zero.
Yeah? Nothing wrong with that.
So our answer is.
This product, negative x cubed cosine x. Plus this product,
3x squared sine of x. Plus this product,
6x cosine x. Plus this product, negative 6 sine
of x. Plus this integral.
Now what are you integrating?
You're integrating zero.
What's the antiderivative of zero?
C, yeah? It's the constant of integration, yeah?
When you integrate zero with respect to x,
you simply get constant of integration.
Don't you?
It's just c. There's nothing like the tabular method
for integration by parts when you have
to do it multiple steps.
Here are some more problems that you might want to do.
Pause the video.
Try them on your own.
Then restart the video and check your answers.
Here are the answers to the practice problems.