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Integrating exponential functions using the substitution method
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subscribe to hey so we're going to let U
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equal what let you equal x right yeah so
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what's the derivative of Ro X without
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thinking one
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what 12 what are you saying X DX it's 12
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right of x d if you do you can right now
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we know that now we can split this right
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so so we know that the uh
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du 12 two I
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mean 2 du will be equal to what 1 x DX
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correct because I'm trying to get 1 s of
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X DX by itself wait couldn't do you also
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equal DX over two yeah yeah you can do
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that too right and now we just go back
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and just substitute that in here so that
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gives us now what two outside and then e
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to what E3 U d right because 1 / X DX is
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2D and then of X is U it's just e to the
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U
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now are we good all right so the
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anti-derivative will be what two e to
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the U plus C and then we replace U by by
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the of X so that we get 2 Eun x + C all
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right now let's try and work on the next
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one it's a little more complex what
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again did you do the the next one yet
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three no all right let's try that so we
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got e to thex over 1 + what e tox DX
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right what would be the easiest way to
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solve this
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problem what would you do let
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= you should go let you equals the whole
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thing because if you do this here it'll
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be harder for you to like find the DU
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right letal
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e because the of one is what zero right
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so let U we going to let u = 1 + e x e x
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d x e to X I mean and what's
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du what's the dtive of 1 0 what's the do
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with e to the
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x 1 orative e to the X DX right and now
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we just go back and substitute that in
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here so we have A1 so we get Negative
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uh
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duu over U right now what's the an of
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one U du we did it what is
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it y forgot what's one over U one
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no natural log of absolute value natural
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log of absolute value got negative
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natural log of What U plus C right and
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we know what U is U is what 1 + e x so
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this is negative natural log of 1 + e x
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+ C all right that make
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sense yeah now the next one will be a
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little more complicated cuz now we have
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like different things there so let's try
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and work on that too let me raise this
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here so we have um 5 - e to
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X over
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e 2x DX
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right so what do you
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think the easiest thing to do here is I
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think you should make U let = 5us no no
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no let U equal e to the power right we
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can do that
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2 and remember we can split these two we
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can split them if you want right because
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if you don't split them it be hard for
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us to solve it so it's better to split
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them to begin with right so I'll start
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by saying this is like 5 e 2x right
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minus e x e 2x right DX now we know we
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can take this what up can we yeah yeah
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that that'll be um e to all right so
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this gives us what 5 e to the what -2X
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to right
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minus E to
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thex DX and now we can use two U
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substitution to solve this here right we
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can use two U substitution to solve this
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now we can go let U here first let U
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equals e
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-2X right are we good oh you just cancel
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out top of this yeah I just took it up
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and then now becomes e x and let = e -2X
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du gives you -2 right e to -2X DX right
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and
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then oh no no no I'm I'm doing too much
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here hold on that was a mistake I
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apologize so let U equal let U = -2X not
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E2 what and then
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D2 DX right so now we can do and then
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here we're going to do the same thing
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we're going to let u = x and then du
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here will be1 DX and now we can just go
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ahead and substitute that in there right
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so now here we going to have this is
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five outside we 5 over 5 2 e to U du U
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and then minus here since you have a
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negative and a negative becomes positive
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or we just put ne1 here to make it easy
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e to the U du and now we can find the
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antiderivative of that right that'll be5
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/ 2 e to the
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U minus
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or plus because we have two negative
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plus e to U plus C and then substitute U
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by -2X here and then sub U byx on this
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one we don't have to put two C's just
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one C is enough so that give this wait
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why you put the U on the outside of the
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which one on the first five five five
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over wait shouldn't it be why I put the
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the five on the outside yeah why did you
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put the outside well cuz I I don't want
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to I mean you can do that too you can
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keep it in here it's not going to change
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anything you can put it outside or
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inside shouldn't the the in the
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parentheses right there be -2 inste 1
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here yeah no because it's 1 DX yeah but
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it should be2 though right no that = X
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for the second one I separated this into
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two right for the for the left one we
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let U = -2X and for the second one we
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let u = x so we have two separate we
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doing the same thing but twice right so
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now this is e to -2X X Plus e
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tox and then plus C and I should take
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care of that all right are we
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good all
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right I also want to try one of them
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with the co
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cosine and then for the cosine let's not
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worry about the corner point for now
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okay are you good yeah we can't see bro
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all right let me
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move
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I feels like so prop right now you feel
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like what it's a proper like this that's
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how you're supposed to see
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anyway no yes you are yes you are it
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good for your back you have to be
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propped up like that because then you
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don't Str your
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leg all right y trying to work with this
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one here so we have P sin pix
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cosine PX DX you just make up a problem
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no I didn't make it up it's on your
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no
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I she has one yeah yeah but for us let's
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not worry about those let's just do like
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this I I'll just put this on because I'm
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not really concerned because for those
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you just have to plug it in I'm just
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worried about really understanding what
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to do here first right so let's try and
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find
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this so why would you let U equal
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here um you let U equal sin pi x good
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let U equal sin Pi that's tough and
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what's
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du first PI right Pi cosine p x DX don't
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forget the pi because you have the
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inside right and now cosine XX DX is
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what du over
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P
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right don't forget that okay don't
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forget to take the pie out to divide it
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by pi U so now you going to have 1 over
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Pi e to the U du and then the rest is
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just a piece of cake right you just find
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the in
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of
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tough and then replace You by Co sin pix
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so one over pi
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i c all
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right e to the pi x and that should take
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care of
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the yeah so that this should be good so
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I told you this was going to be
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