In these next videos I'm going to start talking about some
U substitution
And U substitution is typically the first integration technique, you learn
And the idea with U substitution is you're simply trying to turn an integral into a rather
Somewhat complicated integral into one that's a little more
recognizable
So suppose we have the following problem
suppose we have
[2x] and then we have x squared
Plus 4. Let's make it raised to the 100 power
Now if you didn't know anything about U substitution, and you've seen a little bit of integration
You could simply multiply this thing out you'd have [a] whole bunch of terms, and then you could integrate those piece-by-piece
But probably you wouldn't want to do that at least I wouldn't want to
The idea is what we're going to do is we're going to do kind of a relabeling scheme
so I'm going to call the stuff on the inside the x squared plus 4 I'm going to re blable that as
you so, I'm going to make U equal to the stuff inside the parentheses and
The trick with [U] substitution is look for things that are derivatives of each other
Notice if you take the derivative of this inside stuff you'll actually get the 2x term
It doesn't have to be
Exactly the other stuff as as we'll see but that's usually a clue that U substitution works
so again
I'm calling you the stuff on the inside and basically taking the
Derivative of U which means I have to take the derivative of the right side as well
The derivative [of] x squared is 2x 4 becomes 0 and then we tack on a DX
what you're actually doing here is calculating a differential and
Notice the stuff inside the parentheses now if I go to rewrite this
Well, we relabeled x squared plus 4 that's what we called you. So now I have [u] to the
100 Power and
Notice my 2x dx that's left over because again. I've replaced the [U] squared
Plus 4 raised to the 100 power the only thing that's left over is the 2x dx term that's exactly
what d u is equal to and
I'm just going to put my d u out to the right and
Now the idea is this is just a basic integration problem. You've got a variable raised to the 100th power
It's just like having x raised to the 100 well, you'll get u to the 101 we have to add 1 to the power
divide by that new power
Plus C. And then the idea is since you started with x's you want to stop with x's right now
We have use but simply just use your original substitution and plug that back in
So again u is the same thing as x squared plus 4 raised to the 101 power
divided by 101
plus C, and
That will be your antiderivative
so again use substitution is just kind of a trick for relabeling things to make it look like a
more basic problem
Let's see let's do let's do some more of these
Suppose I have cosine
to the fourth of theta times sine of theta
d-Theta
So again the thing I recognize here that kind of makes this stand out to me as well sine and cosine are
Derivatives of each other yeah one of them you pick up a negative
But that's okay as long as you're only you're getting the correct derivative what they say up to a constant [you]
substitution will work
remember that cosine of theta to the fourth you can rewrite that as
The following and I usually [like] [to] see these a little bit better
[I've] gotten used to realizing in my head that they're equivalent, but I used to always rewrite them just to make things a little clearer
usually whatever is inside the
Parentheses being raised to a power a lot of times that's going to be a good guess [that] that's going to be what you should
Equal so I'm going to let U equal cosine of theta
well the derivative or the differential is going to be negative sine of theta d-Theta [and]
Notice though okay?
So I'm calling
Cosine of theta U so that'll take care of my u to the fourth and then I have a sine of theta D-Theta left over
The differential though says I get negative sine theta D theta
well, I can just multiply both sides by [a] negative and get negative d u
equals sine Theta D theta, [and] that's now [what] I'll replace the sine theta d theta stuff with
so
it
says that instead of cosine [of] theta that's what I'm calling you so I'll have you to the fourth and
Now the sine theta d theta that's left hanging out
That's what I'm going to plug in for the I'll plug negative d [U] in for that
okay, so you can think about the negative is simply being a negative one and you can pull that all the way out front if
You want to?
and
Again now all you have [to] do is integrate [U] to the fourth becomes U to the fifth over five plus C
And then again, we'll simply plug back in what our original U-substitution was back in for you
So it says you was cosine of theta
raised to the Fifth [Power] divided by Five plus
C and that will be your antiderivative for this original problem, [and]
You can always check if you take the derivative of this thing
You know you should get what you are integrating back at the very beginning
See if we can't find one other
One other good problem here. Let's do a
student maybe one Tricky one
Okay, so in that problem. I had to manipulate the you let's do one other one here
Suppose [I] integrating x over the fourth root of x plus two now
[usually] when I see things off by a power like an x squared and an x or maybe x to the ninth and x to
The eighth it really stands out to me that it's used substitution
But in fact, this is [also] a U-substitution problem
Again if you want to rewrite this we can make this x over x plus 2 to the 1/4 power
So I'm going to use the same logic in the last one where whatever is being raised to the power
That's what will make you equal to
so U is equal to x plus 2
the differential d U is simply going to be 1
Dx and
if I go to real able things
I'm going to leave myself a little space here [because] I'm going to talk about a little bit [more]
The x plus 2 is equivalent to you, so I've got you to the 1/4
Dx is equivalent to d. You all right there that is
But I still have an x left hanging out that I need to get rid of here
well
I can I've got a relationship involving U and x I can simply subtract 2 and have that U minus 2 is
equal [to] x and
That's what I'll plug in for my x on. Top is U minus 2 and
Now we'll simply break this up
U over u to the 1/4 that's going to become u to the 3/4
minus 2 over U to the 1/4 we'll bring that up and make it to the negative 1/4 and
Now we'll simply integrate
So you when we add 4 over 4 I'll get to the 7/4
Dividing by 7/4 is the same thing as multiplying by 4/7
minus 2 U to the negative 1/4 if I add 4 over 4 I'm going to get you to the
3/4 I'll have to divide by 3 [for] [werth's] which is the same as multiplying by
4/3
+ [C] and again the last thing I need to do here is replace my use
with my x plus 2 that, I had found a written from the original substitution [and]
if there's any simplification, I'll do that as well, [so]
I'll have 4/7 [U] which is x plus 2 to the 7 fourths minus
8/3 you again which is x plus 2 to the 3/4 plus C and?
there is your antiderivative, so
Again U substitution already becomes a little trickier unfortunately, there's much longer and harder integration techniques than [U] substitution
[but] you'll see those soon enough unfortunately, [I] guess so
if you want to see some more examples of U substitution
I've only done indefinite integrals here. I've also got a lot of videos with
Definite integrals involving new substitution along with pretty much every other
Integration technique as well on my website so feel free to hop over there and shoot me an email you got any questions