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How To Find The Integral OF inverse Trigonometric Functions
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Feb 19, 2025
Today we learned how to integrate inverse trigonometric functions. This is a really tricky section because of the many wrinkles that it involves
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so we are talking about listen up
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integrating inverse Tri function all
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right that's important you better write
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down the formulas right so the first one
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is this when you have
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uh an integral in the form d a s- U2
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this give you r u a plus c right so you
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have these three forms that I gave you
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is it on our it's on your stuff okay the
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question is this you need to understand
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that U is a function of X okay and a is
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just an integer that is greater than
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zero but now how do you apply these
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formulas you have to figure out how to
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identify them right so if you look at
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the first problem there the very first
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problem is already hard you know when
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I'm looking at it so you have 1 over xun
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of 4 - 1 DX right if I look here if you
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look at
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three separate forms you the last one
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the last one but is how you use it is
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the question it's not I'm going to use
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the last one yes it is but okay so a
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would be one a would be okay that's fine
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a would be one U would be 4X that's not
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right oh U is X sorry it's X U is 2x U
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is U is 2x right d but problem is this
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there's way the things you can do here
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right U supposed to be 2x right that
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mean that me because what if you look
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here if you is just uh x what should be
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u s x² but here what do we have 4X right
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4X so that means there is a problem
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somewhere so this has to be what 2x so
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that means I'm going to rewrite this
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integral right I can change it I can
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make my life easier and if I turn this
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into a two what do I have to do here
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turn that into a what why would it even
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be two all right two it can't be just
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two look again brother what do I have
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here I have U and this has to be what u
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s right if U is just uh 4X that's not
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going to happen because I need to
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transform this let me rewrite this
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expression for you this is X over what
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2x the whole thing what Square - 1 s
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right because 1 is 1 squ correct so now
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U is now what 2x because 4x² is 2x the
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whole thing squ do you see what I'm
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talking about
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you see it yeah but isn't the square
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connected to X why is connect do
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what so you're just rewriting for I have
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to rewrite it so that way I can match it
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up with what I have here okay right now
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the problem is this even here I'm still
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not done right because this is U this
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has to be u s but if you look here this
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is X right this is X so I need to turn
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this into what two 2x so now what I do
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is this if I turn this into two I have
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to add I have to multiply this by what
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two as well right so I'm going to
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rewrite this so that gives me 2 over
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2x < TK 2x 2 - 1 2 which is now what we
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wanted now we have exactly what we
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needed here okay so now I can apply my
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formula on so U is 2X and a is 1
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so now we can just go ahead and then
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apply the formula right so that gives
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me 1 / a which is 1/ 1 because a is 1
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Arc secant of u u is 2x over one right
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plus
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C what what is under U = 2x a
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a = 1 that's a right AAL 1 so U = 2x = 1
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and then you just apply the formula you
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you this is really tricky this is where
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we have to be careful because we have to
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be to recognize a pattern if you don't
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recognize a pattern then we have a big
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problem
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good all right so the next problem is
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going to involve some more
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transformation again we have to be
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careful here so let me write that down
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can I what is this I'm sorry to
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interrupt you those of you in the AP
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class could you communicate with the
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rest of your classmates I opened it up
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for one more
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[Music]
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day you want group
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chat
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that like my bad
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that's there's no
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way all right so now
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look now we have a second one right so
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if you look here which one do you think
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this matches up to the second one the
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first one right oh yeah the first one
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the first one but now we have to do some
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transformation right we can't just apply
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regular can we apply it straight up no
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because if I look here right if I if I
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let you equals t to the 4 what's du TQ
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right so what can I do here to make my
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life easy I'm going to I'm going to
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change this again I'm going to rewrite
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this so I'm going to have t over what 1
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- t^ 2 s because I want to make sure I
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get to this point so that way I can
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apply my formula DT all right and this
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is like 1 s because a sare - U2 now do
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you see how it works if U Is t² Right du
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is what 2T DT right so if you let
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U = t²
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du is 2 T DT right and DT T DT by itself
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is what 12 du right so if I was to do
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that so because now it matches up with
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what I have here right you see what I'm
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talking about so now let's not forget
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that we have a two there if you're going
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to apply this formula because that's
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that's extremely careful we have to be
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extremely careful right we have to be
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able to do that so now we can go ahead
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and uh apply our formula right so I'm
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going to have a uh du by this is tdt is
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du/ 2 so that's going to give
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me uh 12 right uh
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arc S of U which is uh T t² right over a
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which which is 1 plus C so that's going
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to be our solution here okay so get how
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did you get the2 1/2 because du is 2 tdt
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uhhuh so D tdt is 1 12 du so I didn't
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make sure I put a 1/2 next to it okay if
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I don't do that it's going to screw my
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my my my numbers up I don't understand
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that well look what do I have here t d
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DT that tdt is equal to what what du
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over what
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two okay right so that's why I have to
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put a 1/2 there right that's why I have
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a 1/2 again this is a really tricky it's
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just a regular problem you put the one
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half in front of it one half in front of
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it okay it's a little bit tricky because
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you have to recognize the P that's the
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only difference here that's why a lot of
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people don't like the trigonometry im an
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because it's a lot of patterns that we
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have to recognize okay now the very next
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one can I raise this are you still
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writing this down no I'm good all right
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but the next one's a little bit
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easier just U is 2 I don't know we find
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that out so I'm not going to I'm just
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going to do it like regularly I'm just
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going to have one 1+ 4x² right DX all
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right so obviously which one do you
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think that matches up to the second one
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the second one right but now again we
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have to rewrite this so that we make
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sense to us right so it can be 2 and
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yeah so it's going to be great that's
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great 2x what square right and then one
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square here DX but now if you look if
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you that U = 2x right du should be
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what 2 2 DX right so you have to divide
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it again by
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what2 you see it yeah so now we're going
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to have U is 2x right a is one so this
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is going to give us2 again RC
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wait wait wait there's there's a three
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there's three app to zero yeah don't
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worry about it I'm not really worried
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about those I just want to treat this
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ass said regular test we going have to
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worry about yeah yeah yeah we will do
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that don't worry about it but I just I'm
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more worried about getting this first
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and then we're going to apply that later
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on right so let's do that this is 2x
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right over what over one right now plus
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C now suppose you have your what do you
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have here again wait why did you why'
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you put the square to both what do you
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mean why is one square here yeah because
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I'm trying to make it look like this
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it's the same as just one one square is
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one but I want to make sure it matches
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up the so you see what's going on right
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so now now if you have let's let's
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assume that you have the same problem
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but now you have wait hold up you put
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the 1/2 arc tangent but isn't it going
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to be 12 times oh wait it's one over one
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so it doesn't matter right I me at least
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now you're getting it though you know
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do you have any idea what he's doing I
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haven't been locked in at all maybe
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should all right now watch this for a
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second right so once we let's say I
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added theun of 32 and then zero
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right my answer will be the same one 12
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AR
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T 2x right now we have between what 32
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and zero now you just got to go and do
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the one R 12 R 10 U uh 2 3 /2 right - 12
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R
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10 0 * 2 2 * 0 right so put that in your
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calculator and whatever it gives you is
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the answer
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okay so that's how you will solve that
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all right does that make
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sense yeah this is a really I I know
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this is a very difficult so you can put
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arc tangent in your calculator yeah AR
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Tang your calculator looks like this
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this one R 10 is
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101 okay that's the key and R sign is
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S1 and then arant is see1 that's what
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the key is your on your ti 83 all right
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that make sense yeah all right so now
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there's another problem Kevin I want you
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to help me solve this one all right all
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right the next
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one so I'll write it on the board and
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see you can recognize the par so we have
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sin x right DX
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over uh 1 + cosine s x so what do you
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see here which one do you think you're
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going to use the second one the second
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one all right second one so U is going
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to be what so let's make
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uine X right and du is what
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if sin x so that means when you write it
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you going to have to put what negative
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there you go that's it you see how easy
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it is when you recognize the P so now
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what's the answer it's going to
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be one over or just one over one uh AR
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tangent s or cosine x 1+ c plus c right
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that's it so once you recognize because
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you realize that root of that is sin x x
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because it's negative you have to make
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sure negative and that's it all right so
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that's how this this is going to work so
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what I'm going to do is I want to stop
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here and then tomorrow we're going to
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pick up and do the completing the square
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because that's a different one so I want
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to I want to give you time to like work
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on several problems like this today and
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then tomorrow we're going to do the
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completing the square which is a
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different it's the same thing but a
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different section