BAM! Mr. Tarrou. In this video, we are going to look at working out definite integrals
or evaluating definite integrals involving integration by parts and my full length lesson
where I introduced this concept…the link to that lesson is in the description of this
video, everything was indefinite integrals. So if you happen upon your homework, the definite
integral from a to b of u dv is going to be equal to, well we are going to evaluate the
product of uv using those lower limits and upper limits of a and b, minus the definite
integral from a to b of v du. And if you did not see my full length lesson on integration
by parts, we want to identify the factor of the integrand which, when we take the derivative
of it, we get a simpler expression and that's going to be u and the rest of the integrand
or the more complicated factor that will still be, where you can still integrate it using
the basic integration rule, is going to be set equal to dv. Okay, so our example that
we are going to look at is the definite integral from 0 to 1 of 3x times the inverse cosine
or the arc cosine of x^2 dx. Now what do you let equal u and what do you equal dv.
Well, in our full length lesson, we were introduced to an acronym which would help us, in a slight
variation, you can find this on YouTube or the internet or maybe even in your textbook
if it gives you one, to give you an aid to help identify what you want to set equal to
u. And you go down in sequence and first you look for logarithms, then inverse trig functions,
then algebraic function, trigonometric function or factor and finally an exponential factor
to your integrand. And whatever you hit first as you go down the acronym is a suggestion
to what maybe something good to set equal to u. So do we have any logarithms? No. Do
we have any inverse trig functions? Why…yes we do. We have the inverse cosine or arc cosine
of x^2 and so let’s set that equal to u and see what we get as far as uv and du that
we can plug into our integration by parts rule here and finish up the problem. Okay,
so we are going to let u equal the inverse cosine of x^2. You put the dx portion to rest
of the, we use the dx part of this integrand as you set up your dv equation and if you
let u equal the inverse cosine of x^2, then du is going to be equal to, okay, the derivative
of inverse cosine that's negative u prime, so this is my u or negative du and so that's
going to be -2x over the square root of 1- u^2. Okay so the square root of 1- my my inside
function u squared dx. Okay and that’s going to be equal to -4.. Excuse me…where did
that come from.. -2x over the square root of 1-x^4, for du. Okay, so if my u is going
to be the inverse cosine of x^2, then my dv is going to be equal to the rest of that and
I can do this now or later, I am going to go ahead and do this now, I am going to take
this constant that's in front of the x and move it out in front of my definite integral
and have 3 times the definite integral from 0 to 1 of x times the inverse cosine of x^2
dx and now my dv is going to be equal to x dx and to figure out what my v is going to
be, to go onto this rule of course, I just integrate this and so v is going to be equal
to the indefinite integral..whoops..of x dx. Which is going to give, well I am going to
take that power and increase it by one so it's going to be x^2. Excuse me. And I am
trying to write the symbol and I am also doing the integration at the same time of course
that's not I really mean to do. So let's try that again. v is equal to the indefinite integral
of xdx. Now to actually integrate that, we get v is equal to we are going to raise that
power by 1, we get x^2, we will divide by that raised power so we going to have 1/2
x^2. Let's try that again. How about 1/2 x^2. Okay, now that we have our u, our du and integrating
our equation that we set equal to dv, our expression for v, let's plug into this integration
by parts rule and see where we stand. nananananana So we have our 3 times uv, being analyzed
with that lower limit of 0 and upper limit of 1 minus the definite integral from 0 to
1 of v du. And we are letting the 2's cancel out. I am also, a lot of times, when this
numerator is negative you could bring that constant, if you will, of -1 out front, let
it cancel with that minus sign there and I could write + the definite integral from 0
to 1 of x^3 over the square root of 1-x^4. We'll see in a second why I left the negative
there though. I don't like to integrate with a radical sign in the expression. So I am
going to write that, that square root of course is a power of 1/2 and I am writing that as
a power of - 1/2 because I am letting it float out of the denominator. So as it stands now,
we have 3 times 1/2 x^2 times the inverse cosine of x^2 with that lower limit of 0 and1
that's going to be evaluated, I am going to evaluate that with. Minus the definite integral
from 0 to 1 of -x^3 times (1-x^4) to the negative 1/2 power dx.
Now this definite integral that's left in the problem, are we going to need integration
by parts again to evaluate and do that integration or can we just use some kind of basic integration
rule that we have studied already in the class and we have got a lot of integration rules-
basic integration rules at our disposal. There's some kind of, something going on over here
with a power attached to it. So can we integrate using the power rule? Well if I let this inside
function be my u, I have already used u, so let's say w. So let w= 1-x^4. Well then w
is equal to that inside function 1-x^4, then dw, taking the derivative, we are going to
get, is going to be -4 x^3 dx. And then I can see that, well I have a negative x^3,
negative x^3 dx. So I can do a couple of things. In my early introduction and many of my lessons
involving integration, I like to do the u substitution completely but as we have moved
more and more into our book.... if you can just complete the chain rule and recognize
that in your pattern that you are trying to use some kind of basic integration rule with,
you can do that as well, something which I mean is I can take this and I can divide both
sides by 4. And I can now, with this w is equal to 1-x^4. And with this equation here,
1/4 dw = -x^3 dx, (it is not really -x^3 times dx). But at any rate, I can rewrite this as
expression now, removing all references to x and rewrite it in terms of w. But then I
would have to change the lower and upper limits of my definite integration or expression here.
If I am going to go into…and remove all references of x then I can’t have my lower
and upper limits in terms of x either. I will need to have it in terms of w. So instead
of going through all of that, I can just complete the chain rule...the chain rule meaning that
if I could introduce a -4 here, then I would have my u to the power of -1/2, du out front
and my chain rule would be completed and I can finish with the integration rule, the
power rule of integration. So I am going to rewrite this as 3 times 1/2 x^2 the inverse
cosine of x^2, lower and upper limits of 0 and 1 again. Now I need a 4 to be there, so
I am going to have, minus the definite integral from 0 to 1 of -4 x^3 times (1-x^4) raised
to -1/2 power dx. So now, inside my power, I have a u and my u prime is outside of that
power, so I have the chain rule complete. But if I want to introduce a multiple of 4,
I need to cancel that out with a division of 4. And now, I am going to step out and
reveal this solution one step at a time, so you don’t have to waste your time watching
me write, but with that u and that u prime, now with that chain rule complete, I can just
integrate using the power rule and get to a final answer for this definite integral.
Okay, so again, just one more time remembering that we have a u and a u prime, the chain
rule pattern completed, I can just use the power rule of integration. So we are going
to take that power of - 1/2, we are going to increase it by 1, we are going to divide
by that increased power, but dividing by 1/2 of course is the same as multiplying by 2.
I can leave the 2 in these brackets or pull that constant, multiple out front, so 2 times
1/4 is equal to 1/2. And now start plugging in our lower and upper limits in each of these
expressions. So we have 3 times 1/2 times 1^2 times the inverse cosine of 1 minus 1/2
times 0 squared times the inverse cosine of 0 and so on and so on just continuing that
process through the rest of this line and the inverse cosine of 1 of course...what angle
measure gives you a cosine value of 1. Zero. So this expression comes out to be 0, because
the inverse cosine of 1 is 0. And 0 times anything is 0. We have a factor of 0 here,
we have -1/2… 1-1 is zero. So that’s going to be zero again. Minus the square root of
1 which is going to be 1. And cleaning this up, we have a final answer of 3/2. And that
is how you do a definite integral or evaluate a definite integral when you are going to
need to use integration by parts. I am Mr. Tarrou. BAMM! Go do your homework!