so this is the second technique of integration and this is called u substitution I will just
call this s that you stop so what it's used up well back in the days when we're doing derivative
we have the chain rule right and now I'm going to show you guys the connection between the use
up and the chain rule that's focused on this one right here we are going to first take the
derivative of tangent of X to the fourth power so how can we do that well we have to remember what
it's the derivative of tangent first derivative of tangent is secant squared right and then the
inside function right here is X to the fourth power with key B to be the same for now so we
have secant squared and then the inside stays the same so secant squared X to the fourth
power right and the chain rule tells us that we have to look at the inside function which is X
to the fourth power we have to multiply by the derivative of the inside function the relative
of explore fourth power is four X to the third power right so we have to multiply this by four
X to the third power that's the chain rule and this right here is yet that's the derivative
for tangent of X to the fourth power at the end we usually write down this part first for X to
the third power and then write down the secant squared X to the fourth power and this is it
we don't do policy this is just the derivative okay so how can we integrate for X to a third
power secant squared of X to the fourth power DX I know you guys see what I'm doing right here but
I want to show you guys all the steps okay to integrate something we have to ask ourselves do
we know right away derivative function give us this assume we do not do this question okay so
that way I can write down the whole steps for you guys and this way I can introduce you guys the way
the how the you sup works and then the steps of the solution right down okay so here is that all
process I don't technically know the roughly of what function this right away right because the
relative table doesn't have this as one of the entry hmm so this is how we are going to do it
you see that we have secant square and then the inside is X to the fourth power and imagine if
you have to take the relative this part here it's usually the part that requires the chain
rule right in that case that's the candidate for the you this right here extra force power what I'm
going to do first is I'm going to set this to be the U and this is the procedure this is the step
of the scenes that will write down we usually let u equal to the inside function in this case
is the X to the fourth power and once again U is usually the part that requires the chain rule if
you imagine this what's the derivative right okay U is equal to X to the fourth power and
what we have to do next is we have to talk about the derivative this so that's differentiate both
sides and we'll just talk about the differential because usually you can write this down as okay
differentiate we get D u DX it's equal to 4x to the third power right however I'm going to just
write down this for you guys so this we can save one step talk about a differential wise on the
left hand side when you differentiate we just get tu on the right hand side we do the usual
power rule we have 4x to the third power by then we have the DX okay so that's the differential
part anyways what I want you guys to do next is we would like to isolate the DX so we can look
at this equation and divide both sides by 4 X to the third power this way we can get DX equals 2t
you will be on the top over this 4x to the third power and this is the things that I like to tell
my student this original integral it's in the X world because the DX says so what we're going to
do is we will take this integral from the X world into the you world let me show you so this is
going to be integral this right here well is just 4x to a third power I would like to just write it
down as how D is but now secant square that's a function part so I will just write down secant
square however for the inside this is X to a fourth power and we said that to be you so we
just put this down as you right here and we also have the DX don't forget that yet this is crucial
when we do integration DX in the new world is EU over 4x to the third power so let's put it down
right here tu over 4x to the third power so what do we have to do next you see this right here we
have 4x to the third power and then we are also dividing by 4x to the third power of course they
cancel each other precisely right at the end what we'll have is integral of secant squared u EU
and you see originally the integral was in the X world everything was in terms of X right and now
this integral is in the you world we have secant square u only that we have to worry about right
and this is the idea after we have done the used substitution this integral right here is something
that we can recognize from the derivative table we have to ask yourself the relative what will
give us secant square u well the answer for that will be tangent u isn't it and at the end we are
done with the integration however we do have to go back to the X world because right here we know
you write is equal to X to the fourth power and originally you know fundamentally the question was
in the X work so let's go back at yep tangent of X to the fourth power and we are done usually I just
like to tell my student after we are completely done with the integral we put a plus C at the
very yet ok you don't world too much about putting a plus C right here it doesn't really matter that
much so what it's the connection between the chain rule and the user the chain rule is that when we
are taking the derivative we have to multiply by the derivative of the exact function the use up
is that we have to divide by the derivative of the inside function right and you see when we divided
by the derivative of the inside function this is something that we can recognize and I do want to
make a remark sometimes there are integrals that we just can now work out and usually when we are
doing the questions from the textbook those other questions that definitely work right so
once again this is how the use up will work and I'm going to show you guys the next example okay
let's once again look at integral and another root equation side-by-side but this time we'll look at
the integral first so we have to integrate 1 over 5x minus 2 DX and do we know the roughly of what
function will give us exactly 1 over Phi X minus 2 right away maybe it's not too obvious right and
this is where the u sub K help us out and you see right here we have 5x minus 2 in the denominator
if I differentiate Phi X minus 2 I get just a 5 which is just a number right and that would be a
good choice for you let me show you how it works so let u equal to 5x minus 2 just the denominator
and then when you differentiate both sides differential wise okay on the left hand side we
have the U on the right hand side the derivative of 5x minus 2 is just the 5 this is zero and don't
forget to put down the DX and I like to tell my students isolate the DX so that's divided by Phi
on both sides then we can get DX equals to D U on the top over 5 if you isolate a DX like this every
time it's easier to see do you need to multiply what you into D by over cancel what seems like
that ok so I highly recommend you guys to isolate that yet like this anyways we are ready to take
this integral from the X world into the you world so we have the Inca people right here one on the
top over Phi X minus two is the U so we put on a u in the denominator D X is D u over five so put
that down d u over five alright so a few things that we can do here and there first you see we do
have a factor 1 over 5 right here isn't it we can take a constant multiple outside of integration
so I can look at this as 1 over v integral of 1 over u and then we have the T U and now we see
what's the integral of 1 over u 2-u the ref D of what will give us 1 over u the answer to that is
natural log of U right and we are done with the integration policy don't forget to write down the
constant multiple 1 over 5 and this give us Ln U and remember whenever we go for the integral of
1 over u EU to this we have to attach an absolute value around the U and why do we have to do that
you can check out my previous video on that but anyways attach the absolute value and then we're
kind of done with the integration part at the very end take the you back to the X so we have 1 over 5
Ln absolute value the U is Phi X minus 2 so let's put that down right here for you close it at the
end we are done so we put a plus C at the end this is it this is the answer for that integral and now
I will have to ask you what if we differentiate this right here and that's exactly this part of
the question so let's go ahead and differentiate 1 over 5 Ln of 5 X minus 2 and notice that for the
rift equation the absolute value is none needed we will just focus on the parentheses but when
you integrate be sure you have the absolute value all right so to take the derivative this this is
just a constant multiple so we can write that down right away 1 over 5 and we will just focus on the
derivative of this part right here this is the function part the derivative of Ln of something
is what this will give us 1 over this thing which is 5x minus 2 and then you see the inside is 5x
minus 2 isn't it chain rule says we have to take the derivative this which will give us 5 and then
multiply right here so we multiply by 5 and you see this is why I chose you to be 5x minus 2 use
up and chain rule they are related use up is just pretty much the backwards version of the chain
rule so this part right here is the chain rule this part here what's the use up see at the end
you have 1 over 5 times this times 5 what happens 1 over 5 & 5 cancels each other out well we have
just 1 over 5x minus 2 we are done so you see they confirm with each other you stop and change and
you check out my next video I'll work on more examples and the trickier ones for you guys