in this video I'm going to show you guys three slightly harder slightly trickier integrals and
we still have to use the u substitution for them and did you guys notice that I'm going to be using
this black and red marker for all these integrals this is so cool the new markup I export the two in
one version okay I will have the links down below in the description so you can check them out as
well they have different color combinations and thanks to our math department here sherry for this
wonderful amazing surprising gift I'm so excited to use this because I've never seen anything like
this before and you guys will see how this marker work on the whiteboard in this video okay so let's
get to it first one the integral x over 1 plus x 2 fourth power DX what shall we let u be right if
we let u be the denominator 1 plus X to the fourth power D you will be 4x to the third power however
on the top we only have X to the first power right in that case we know they won't cancel each other
out well that's nothing by choice for you then right well we do have this X on the top hmm I
want to cancel that in a case will be great right so you should be what so that we can cancel this
X to the first power well this is X to the first power I want to let u equal to something with X
to the second power because once I differentiate that I get something X to the first power isn't it
okay I want to produce X to the second power somewhere in this integral and no place it's
going to be right here let me show you okay okay I'm so excited all right so this is going to be
the integral on the top I'm just going to let it to be X and on the bottom we have 1 plus in step
this X with fourth power let's look at this as x squared and a squared how's that so I'm going
to open a parenthesis you see I'm going to just switch the mark like this this is so cool huh
inside here I will X to the second power and the head this to the second power so that this is joe
de force power isn't it this is just so cool and now what well as our strategist earlier I'm going
to let u equals two X square right here so that when I differentiate it I get D u is equal to 2x
DX say my business let's divide 2x on both sides so we can isolate the X so we get DX equals 2t u
over 2 X and now we can take this integral from the X world to the you world and now we will see
this is going to be the integral of X and the top over 1 plus and notice that I only del u
equals 2x squared because inside here we only have x squared that's the inside function right
I cannot say 1 plus x squared okay so that's very different therefore we still have the one
class right here and then the inside you this is x squared actually x squared X W okay I'm going to
change that it was x squared and now it's going to be the U and then we still have that to the second
power and then for this DX is tu over 2x so let's put that down the U over 2 X very nice so that we
see this X and X can saw each other huh and then we are just going to be integrating 1 over 1 plus
u square tu and now let's see we're at focus on C 1/2 so that's for the 1/2 the front ok what's
the integral of this 1 over 1 plus u squared it's the inverse tangent u right so we still have this
1/2 right here in first tangent u at the end take this back to the ex world so U is x squared so we
have one half in first engine inside here is the x squared and then we are done so put plus C this
right here this it pretty cool isn't it okay for the second one we have the integral of tangent x
times Ln of cosine X as usual I like to tell you gasps okay maybe we'll a u is equal to the inside
function however in this case if u is equal to cosine X the derivative of cosine X is negative
sine X and here we don't have a sex this is what tangent X so they don't cancel each other right
away right maybe you'll work maybe not I don't know but here is the secret instead of dating u
equals two cosine X I'm going to let u equal to Ln of cosine X and let me show you why I may choose
I chose to do that so u equals to Ln cosine X I am going to differentiate this for you guys you
will see D u equal to and now remember how to differentiate with the Ln function derivative
of Ln of something is going to be one over the inside right and then that stays the same for now
so that's cosine X but don't forget the chain rule says we are going to do cutting thigh and then you
look for the roof to this the derivative of the inside the root of the cosine X is negative sign
so we multiply by this by negative sine X okay the chain rule and we still have the the X and now we
I'm going to just rewrite this again this is d u equals to negative we have sine x over cosine
X so that's what a tangent X so this is negative tangent X DX and you see I'm going to divide both
by negative tangent X so I can isolate the DX DX will be you over negative tangent X and you
see now this tangent X and that contain X will cancel each other out very nice huh all right
take this integral from the X row to your world this is going to be the integral tangent x times
this part the whole this right here it's the U so I will multiply by U and then the DX is that
D u over negative tangent X and now 10 genetics tangent X cancel each other out and we will have
fun don't forget the negative bring the negative to the front negative integral u te u and how to
integrate u in a new world well this is going to be you add a power 1 plus 1 is 2 and 8 divided by
2 which is the same as multiply 1/2 anyways this is going to give us 1/2 u square and we bring
down this negative ok we're done at the end I'm going to rewrite this nicely we have the negative
1/2 right here and then for the new let me open a parenthesis to make everything looks clear inside
here for the U which is Ln of Cossacks let's put that down here Ln cosine X and then we are going
to raise this to the second power and we are done so co+ see this right here is it okay and one more
example for you guys okay for this one we have the integral 1 over 1 plus square root of x we don't
have a lot to work with isn't it because it seems like our only choice for the use the denominator
1 plus square root of x and the reason that I want to attach the 1 plus right here is that
when I differentiate the whole denominator the 1 goes away anyways so that's a good choice and
I'm not sure what will happen and let's just try our best to make things happen let's go u equals
to the denominator 1 plus square root of x and I'm going to differentiate both sides so we get D you
equals 2 this is zero the relative of square root bags is what 1 over 2 square root of x and then we
are down at the riveted part so attached at the X this time we are going to multiply both sides by
2 square root of x so that we can get DX equals to 2 square root of x 2 you write 2 square root X do
you and let's just take this into the new world to see what we get so this is going to be integral 1
on the top over the whole thing right here is U so put on the new right here in the denominator
and then for the DX we know it's this but this somebody that's not inviting the new world namely
the square root of x but a causing right here is that if you refer back to this equation here you
see you it's equal to 1 plus square root of x why don't we just subtract 1 on both sides this
way we can put this way we can see that square root of x it's equal to u minus 1 right so you
snow this right here it's a same as u minus 1 and we can rewrite this again for you guys this is
saying DX it's equal to 2 the square of X is that u minus 1 so how to use the parenthesis so we have
to do minus 1 here and then D U and bam plugging this into the DX let's see what we happen right
so that's right down to parenthesis u minus 1 D and D you okay and now let's continue a few thing
that we can do first let's bring the two to the front and then we have the integral you - one goes
on the top right over you and this is bu and now but well I cannot just integrate this you because
this U is affected by this u so we are going to split the fraction okay this is going to be two
integral u over u let me write down all the work for you guys and then minus one over u we split
the fraction and we know this is just two integral u over u is 1 and then we have the minus one over
u the U and we're ready to integrate so this is going to give us two and then open the parentheses
for the result integration one in the you world the integral for that is going to give us u and
then minus the integral of 1 over u in the new world that's Ln absolute value of U and then we
are done and now I have to take this back to the eggs world we were done with the integration part
right so I'm going to do two steps right here I'm going to take the two and it is shipped inside in
the meantime we know Yuman is equal to 1 plus guru bags right that's what we said to be to
start off with therefore we will have 2 times the parenthesis U is 1 plus square root of x so that's
ready down right here and then minus 2 times that so we have 2 right here and then Ln absolute value
the U is 1 plus guru bags let's put that down here 1 plus square root of x and we close this
up to value and technically we are all done right here but a few things I'm going to mention in this
particular example first as you can see of course we can distribute the two in this parentheses
right in front of these parentheses and we'll end up with two plus two square root of x minus
2 Ln absolute value of 1 plus square root of x like oh this isn't it all right and the moment
we are done you can call this 3 plus C but you may find the answer to be differing the back
of the textbook if you have this question for the homework first you see you have a constant
right here which is the two and then when you have 2 plus C this is a constant this is also a
constant two constants when you add them together it is still a constant so sometimes they may merge
into just one big constant C so right here I want to label this as c1 and what you can do
is you can ignore the two but then just count label this s and you'll see just right on the
function part that's all so that's right down to and then we have the square root of x minus 2 and
then we have the Ln and now this is another thing I want to mention if you look at this power here
we know Spooner bags the output is always but the output of school bags is always positive and then
we're adding and this is the past one Part D plus a positive inside here we are always guaranteed to
have something positive in our case the absolute value is now needed therefore I can change the
absolute value into just a regular parentheses and then putting 1 plus square root of x all right and
as I said this and that together I'm just going to write D as another constant so plus C and
technically I would applaud us C 2 however this is I'm going to present for the answer
okay in that case I will just write this as the constant that we are going to use and this right
here is it let me know what you guys think about this black red marker it's really cool isn't it