in this video we introduce u-substitution which is primarily a tool for working the
chain rule backwards when we're getting an anti-derivative of a composite function
and the goal with this video is to give a quick by-example overview of the key points
rather than a formal theoretical discussion so we start with an example and i want to demonstrate
this example with three different levels of formality and in the first level of formality
i'm just going to say okay x cosine x squared i'm trying to get the antiderivative of this thing
i think it's pretty close to saying sine x squared and the reason why is because the chain
rule should produce a factor of 2x out of that and that gives me basically what i started with
so this is kind of a guess and check and adjust kind of method and when i differentiate this
guess i take the derivative with respect to the argument and that gives me a cosine x squared
and then the chain rule tax on the derivative of x squared which is 2x
and i basically have the integrand i started with it's just there's an annoying factor of
two that appeared because of the chain rule so what i'm going to do is adjust for that by
throwing a one half into this function one half sine x squared and now when i differentiate that
that 2 from the 2x is going to cancel the one half out in front and i'll get an x cosine x squared so
we're good don't forget your plus c so another way of saying what we want here is that i'd like
to get the derivative of the interior function sitting right next to the composite function
so this is a way of highlighting the chain rule backwards so in my next method and this is pretty
close to how most people handle integrals and differential equations physics and engineering
i'm going to just build that derivative right next to the cosine function and of course you
can't get away with just throwing twos around but i could compensate with a one-half out in front
just keep in mind that constant multiples can move in and out of an integral but variables can't and
there i have it i have a composite function with the derivative of the interior function sitting
right next to it showing me exactly what the chain rule has done and i can just guess that this came
from differentiating the sine of x squared finally we introduced the formal u substitution and this
is just a way of symbolically transforming the integral to account for the chain rule what we
want to do is express the interior function of this composite function as a u and then express
the integral with a differential of d u instead of dx and finally we'll compute the u integral
and then transform the result back in terms of x so in this case we're going to let u be x squared
and then i have to get at the differential dx here and so i'm going to take the derivative du dx
and that's 2x and that allows me to relate the differentials of u and x so i multiply by dx
on both sides and get a 2x dx now there's two basic paths that i could take from here
one way to finish this thing is to build that differential 2x dx inside the integrand and the
only reason this is possible is because i happen to have the correct variable piece there that's
the x that's out in front and this seems familiar because it's basically the same thing that we were
just doing in that second integral so i multiply by 2 on the interior multiplied by one half out
in front to compensate for it and i've built my 2x dx and i have a name for that that is d u so
i replace the x squared with a u and i get cosine u i replace 2x dx with d u and i have a cosine u d
u and now i can simply guess the antiderivative of the cosine which is the sine function
and i'm not done yet because it was an x integral and i better have an
x answer and i get one half sine x squared plus c so same answer as before now the other way i could
handle the transformation of differentials is to actually explicitly solve for dx
and that would give me a du over 2x so taking my original integral and replacing dx with du over
2x and replacing the x squared with u inside that cosine function i get here so x cosine u d u over
2x and again the only reason this works out is because we had the correct functional piece in
here to give me the derivative of the interior function if i didn't have that x tacked on
this wouldn't work if it was an x squared this wouldn't work so the reason this is working is
basically because i was set up to do the chain rule backwards so that 2 in the denominator comes
out in front and i have a one-half integral cosine of udu and we get to the same place
half sine u plus c and replace the u with an x squared and i get one half sine x squared plus c
and we're done now before we move on i want to point out that it's easy to get lost in
the formalism when we're doing an explicit u substitution so that third integral the
way i do it i would call that an explicit u substitution ultimately it's better to
solve the integral like the second approach so we're trying to see the chain rule backwards
by seeing the derivative of the interior function tacked on to the integrand so even if you elect
to use an explicit u substitution it's not going to work unless you essentially have
the derivative of the interior function sitting there in the integrand because we're doing the
chain rule backwards so that's usually what we're looking for before we launch into a u substitution
now let's take a look at how to do a definite integral that involves the chain rule backwards
and in this one the interior function is a polynomial and that's being raised to a power of 4
and then the derivative of the interior function is essentially provided to us in the integral
that's the x out in front now the derivative of that interior function is actually negative 2x
but we can adjust for constant multiples and again we take three approaches to the integral so in the
first one i'm going to take the informal approach of constructing the derivative of the interior
function in the integrand and the derivative of that interior function is negative two x
so i multiply by negative one half out in front and negative two inside the integral to compensate
and now i have a function raised to a power so 3 minus x squared raised to the 4th sitting right
next to the derivative of that function which makes it clear how the chain rule is operating
and i'm able to guess the antiderivative so i have this function raised to a fourth
sitting next to its derivative and i use the power rule on this and i get three minus x squared
to the fifth divided by five and we still need to evaluate that as x goes from zero to one now
before i do the evaluation because this is an informal method i'm going to check that
that antiderivative actually made sense so if i differentiate that thing move the one-fifth out in
front there it would give me a negative one-half times one-fifth and then differentiating three
minus x squared to the fifth with respect to its interior i get five times three minus x squared to
the fourth that's the power rule for derivatives multiplied by negative two x that's the chain rule
my fives cancel my negative one half and my negative two cancel and i'm left with an x
times the quantity three minus x squared to the fourth so we've done everything correctly now
what about evaluating this thing across the limits of integration i have a negative one tenth after
i gather the constants and then when i sub in a one into this thing i end up with a three minus
one or a two to the fifth and when i sub n is zero i end up with a three to the fifth so i have a two
to the fifth minus three to the fifth and this works out to 211 tenths or 211 over 10. in the
next approach to the integral i'm going to use a formal u substitution and as usual we're going to
say u is the interior function so we're going to let u be 3 minus x squared and then we take the
derivative of that to get d u dx that's negative 2x but we multiply both sides by dx and we get the
du as a negative 2x dx and this time i'm going to opt for building this differential negative 2x dx
inside the integral again keep in mind this is only possible because we had the right function
piece in there we had that x in there making it so we could build the derivative of the interior
and of course i can't just multiply things by negative 2 whenever i want i have to compensate
by putting a negative one half out in front and now the negative 2x dx that's d u and
the 3 minus x squared to the fourth that's u to the fourth so i end up with a negative one half
integral of u to the fourth d u and then i have to be careful about my limits of integration
because those are x limits of integration and i'm going to label them that way i'm integrating as x
goes from zero to one now the whole argument for doing explicit u substitution is that it makes it
easier to guess the anti-derivative when i get to that point in the problem so u to the fourth
has a very easy anti-derivative it's one-fifth
u to the fifth and then i'm evaluating as x goes from 0 to 1 and this means i better transform
my antiderivative back in terms of x so u was actually 3 minus x squared and now i'll just
write that my limits of integration are 0 to 1 because i'm back in terms of x here
and this is the same expression i had before when i sub in my upper limit i get a 2 to the fifth
my lower limit gives me a 3 to the fifth and when i clean things up i get 211 over 10.
now the reason i wrote down a third version of this thing is because we actually have an
option to transform the limits of integration into u space and then we can forget about x
so all the substitution is the same here i'm going to go ahead and bounce to this step from the
previous version of the solution i have a negative one-half and i have the integral as x goes from
zero to one of u to the fourth d u but instead of guessing an anti-derivative and transforming it
back in terms of x i'm just going to ask myself when x is 0 what's u and i go back to where i
originally defined u and i plug in x equals 0 and i find that u is 3 minus 0 or 3. so my lower
limit of the integral then becomes a u limit of three then i ask myself when x is equal to one
what's u and u is going to be three minus a one squared or three minus one or two so now i have
negative one half integral from three to two of u to the fourth d u and i guess the antiderivative
giving me a one fifth u to the fifth and i'm going to evaluate this as u goes from three to two
so the cool thing about this approach is that it allows us to forget about x forever so we just
completely transformed the problem into u space and we're going to get
our final number out of u space so i have a negative 1 10 out in front when i plug in
the upper limit i get a 2 to the fifth i plug in the lower limit and i get a 3 to the fifth
and that simplifies to 211 over 10. if you enjoyed this video or at least found it useful
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