how to integrate using u substitution

as we all know the hardest part using goose up for intercourse that what should we choose to pay you right therefore in this video I'm going to show you guys three examples and how to use the u substitution to do an integral and now each example I'm going to show you guys two very similar integrals because this way I can point out the little details that we have to focus on much better and hopefully after this video we can understand and like the U substitution much more anyways let's look at this one the integral of

secant of 2x times tangent of 2x DX and then we also have the integral of secant square x times tangent square X DX let's focus on this first off so this is the first tip of selecting the you as goes up what's an inside function right here we have secant of two x times tangent of 2x right and the good thing is that the two XL networks are the same so that's a good choice for you because it's an inside function right so I'm going to write down the steps first write down the selection of

the you you I'm selecting does 2x and once you have the you differentiate both sides on the left hand side we get in you and on the right hand side the derivative 2 X is 2 DX and then the next step is we are going to isolate DX I will highly recommend you guys to isolate the DX because sometimes you may have to defy you how to multiply well if you always isolate the X it's more clear and consistent anyways in this case we have 2 D PI by 2 on both sides so we can

say DX is equal to VU for 2 and with all this we can take this integral into the new world let me show you so this is going to be the integral secant of two works so that's right on see you can but then for the two it we set that to be you so we have secant of U and then multiplied by tangent of 2x so that's the same as saying tension of you right so you can say 10gen the to access to you at the end the DX is 5 d you over to

so I can put that down you over to and you see we don't have X anymore in this integral so this integral is completely in a new world that means we are on the right track and we see that we have a 1/2 right here and this is a constant multiple with this expression and what we can do is we can take the 1/2 to the front of integration so that's write down as 1/2 integral secant u times tangent u U and now how can we integrate this in an ad work do we know the

ripped apart function will give us seeking you tangent u this is the moment that we have to know our the row of the table really really well the answer to that is C can you eat the root key of C can you or keep us exactly C can you teach you therefore I can finish this by saying this is 1/2 and olders is just secant u and then we are done with the integration part okay and some people will tell you you can add a plus C right here but I would like to tell you

you can just put down a plus C at the very end because we are not done yet we have to take this expression from the new world back to the X world because we start off with X so the last touches that will say this is 1/2 secant u is d 2 X so let's put that right here and we are all done so I'll put a plus C let me very end and holbox the answer right here that's it for this one and now let's look at this integral secant square x times tangent squared

X DX first of all do we have any inside function well to figure that out we have to do caps secant squared X as the following I can look at this as parenthesis with secant X inside and then raised to the second power so secant X is one of the inside function right and the reason it's a one off because we can do the same thing for the tangent square X I can look at tangent square X as tangent X and then raised to the second power therefore we have two choices right because we are

two inside functions the first one is secant X and then the second one is tangent X which one should we choose to be the you well you can just write down U is equal to secant X and then once you write that down differentiate that you'll get D u it's equal to what's at the root position X secant x times tangent X DX right however if we do that unfortunately right here we have a tangent squared X right the real key of secant X is only secant X tangent to the first power X so the

tangent X cannot cancel each other completely that means secant X it's not a good choice for you on the other hand if we choose you to be tangent X the relative tangent X is what if secant square X isn't it we happen to have exactly secant squared X right here they will cancel each other now let's put up all the steps right here let u equal to tangent X and let me just make a remote before continue sometimes I may make a wrong choice for you I put it down and then unfortunately even our work

by it's okay just go back to the equation and try something else okay don't be afraid to write down the steps all right so I put down u it's equal to tangent X let's go ahead say D u is equal to theta of tangent X is secant squared X DX and then isolate the DX divided by secant squared so get DX equals to tu over secant squared X and now we can take this integral into the world this is going to give us in the grow we have the secant squared X that's just put that

down right here right this has nothing to do with you isn't it by then for this part when we multiply by tangent square X tangent X is that you so we can say this multiplied by u square that's the tangent squared in the you world lastly the DX is what we said to be the U over secant squared let's put that down right here as well vu over secant squared X and you see the secant squared X and the secant square X cancel each other up completely and what do we have left this is nothing

but just the integral of U square U and can we integrate this this is just u to a power right definitely sure that's add one so two plus one is three and divide by this new power we just multiplied by one sir so the integral for this is one-third U to the third power what done was the integration part at the end we just have to take this back to the X world so this is going to be one-third U is tangent X so let's put that down right here tangent X to the third power

huh so let's put a third power right here and we are all done so idea plus see this right here is the answer for the second one as you can see they may look similar but sometimes the integral requests different step the different choice for you and now let's look at example number two for example number two I'm going to show you guys how to use use up to deal with definite integrals and that means we have an integral with numbers like this right here and we know the answer to a definite ago is that

we are going to end up with a number right unlike earlier when we did not have these numbers right here the answers were functions but this time we know this is going to be a number likewise this is also going to be a number and I will show you guys how to work out everything in the you world the first one is the integral from 0 to PI over 2 SAG's over 1 plus cosine X DX and then for the second one we have a square right here this is cosine squared X in stealth cosine

to the first power X and everything else stays the same so you see they look so similar isn't it but we all look at this one first and the first question is why should we choose to be the you it seems that we can let u is equal to say X or maybe Cossacks will maybe something else right let me just write it down what happens if I let u is equal to sex maybe works maybe not let me just at least do one type of you guys all right down U is equal to sine

X and then we follow the usual procedure once you read this down be sure you differentiate both sides you can tu any derivative this is going to be cosine X and don't forget the DX and that's I saw lady DX D by positive a cosine X we get DX equal to tu on the top over cosine X and can we take the integral projects world to the you world with all these ingredients let's take a look so now you see this is going to be integral from something to something but then let's focus on a

function part first on the top it's an X but then sine is the U right so we have you on the top over this has nothing to do with you so I will write it down 1 plus cosine X and then the DX is the U over cos x so I like to put that down right here tu over cos x unfortunately of course I X here have another context here they don't cancel each other is it so you see this is a wrong choice for you you should not be psyched because this right here

we still have the X right here X is not invited in the new world and especially when they cannot cancel each other out this is incorrect we can now move up and that means we just have to go back to the question and pick out something else to be that you and now let's see what else can we do for the you this time let's say u equals two Cossacks so I would like to write that down U is equal to cosine X and when I do that I know the derived here cosine X is

negative sine X and when I divided by negative sine X this tax and that's annex will cancel each other but then I'm not going to just let u equal to cos x better yet I can in fact that u equals to the whole denominator I can say U is equal to one plus cos x why because when I differentiate is I will get the U equal to the derivative of 1 the 0 so it doesn't hurt that negative IX it doesn't hurt my earlier plan right so sometimes you may be helpful if you attach the

1 plus just at home doing right here so it doesn't change the diversity that we talked about earlier the U is still equal to negative sine X DX and isolate the DX from here we get DX is equal to u over negative sine X and now we can take this integral into the you world completely let's see we have the integral and then on the top we have the sine X let's just write that down and then let's worry about the function part first okay the whole denominator 1 plus cosine X is 4 we set

it to be the you so the whole denominator is the U and then for the DX is the D u over negative sine X so let's put that down right here bu over negative sine X you see cancel cancel right so that's very nice I don't have X anymore and now as I talked about the earlier I want to work on everything in the you world what can I do well I'm missing these numbers and if you go back to here this is 0 to PI over 2 however doose what the X values this numbers

mean to have X is equal to 0 up to X is equal to PI over 2 I cannot just take 0 and PI over 2 I put it down here because technically they are also not invited in the new world I have to change that correspondingly into the u values how can we do that well let's use this equation here so you'll see when X is equal to 0 or 1 how to do is plug in 0 into this X I can calculate the phone for you so let's do that real quick plug in Z

going to here we will get U is equal to 1 plus cosine X is 0 right this is the work we have to do and you see this is going to be well cosine 0 is 1 1 plus 1 is 2 so we have a 2 right here that means the partner number is U is equal to right and let's look at an up number when X is PI over 2 plug into here we get u is equal to 1 plus cosine of PI over 2 for the x value and that's what this out what's

cosine PI over 2 it's 0 0 plus 1 is just 1 right and you see this is a little bit strange but it may happen you see this right here U is equal to 2 this is the number corresponds to the X is equal to 0 and this number is going to be U is equal to 1 right sometimes it may happen after we do the translations this number is bigger than the top number earlier zero was smaller than PI over 2 right but this time the bottom number here too is bigger than 1 he

may happens then just how to work out this carefully all right now let's integrate this that before that pay attention to anything from here we do have a negative 1 in the denominator 1 over negative 1 is negative 1 by some constant multiple we can take the outside so we can negative one right here and then we have the integral going from U is equal to 2 to U is equal to 1 okay all this is in terms of U now and we'll finish up everything you this is 1 over u the U and can

we integrate this well we have to have the negative 1 and the antiderivative of this that we can use is Ln absolute value of U right 1 over u is going to give us Ln absolute all of you and then be sure we plug in the values going from 2 to 1 and be sure we plug in the number on the top right here first into this right here and then subtract plugging this in here so do everything carefully plug in 1 2 here we will end up with negative 1 times the Ln so I'll

just write down negative Ln absolute value of 1 this is the first part isn't it at the - I had to plug in 2 into here I will end up with L this is negative 1 so this is negative Ln absolute value to me here so this is what we have okay so that's work this out but one is up the Ln Ln 1 is 0 so this is just 0 negative times negative is positive so we pretty much have positive Ln of absolute value - but - is passive anyways so the answer is just

Ln 2 right this is 0 this is parsnip so altogether we have Ln you don't need the absolute value because 2 is parsley already so the answer to this is Ln 2 tip all right let's do cat this right here when we have the cosine squared X instead of the first power Machoke - can we still use the same strategy the good thing is that we still have the sides on the top right so hopefully you can somehow connect with or with it just now with this hmm but then can I let you equals to

the whole denominator and once again let me write this down for you guys maybe works maybe not if you want to let u equals to the whole denominator I will get 1 + cosine squared X and out to differentiate both sides right away right so I get D u equals 2 derivative 1 0 but what's the derivative cosine square X don't forget to bring your 2 to the front and then in test dates the same which is still cosine X and then 2 minus 1 is 1 but the inside function is cosine X so we

have to multiply by the derivative cos x because the chain rule right or well multiply by negative sine X and of course you can touch the DX after that but you see if you let u equals 2 the whole denominator D u it's all this 2 times cos x times negative sine X DX this is not going to be as nice as this because on the top we only have a sex right so we cannot cancel everything else that we want therefore this is not the correct choice for you so what can we do let's

talk about the inside function again especially when you have cosines where X this you can look at this as well cosine X and then raised to the second power right so as you can see an inside function right here is just the cosine X so I'm going to just let u equals 2 Cossacks and let me just draw a line between right here for you guys now this is the time to work this out correctly I will let u equal to just the inner function cosine X I don't have the right to put on 1

plus because the inside function is just the Cossacks right okay if not this you see tu is going to be negative sine X DX and you see that's pretty much the same thing over there very nice and then I want to isolate the X so DX is going to be vu over this the U over negative sine X and now I will take this integral into the you world so let's see what we end up with this is going to be the integral but then I want to take out these numbers these are the X

values right so when X is zero that plug in you get u it's equal to cosine 0 and this is going to give us 1 cosine 0 is 1 so this number right here is U is e put 1 and when X is equal to PI over 2 plug into here we get U is equal to cosine of PI over 2 cosine PI over 2 is 0 so the bottom here is 1 and then the top is U is equal to 0 and once again the number is big on the bottom and then small on

the top but even happen right ok so that's C on the top here we have sine x over 1 plus and then this part we have cosine X and a square but then cosine X is that you so we have u square right u square and then for the DX is this the U over negative sine X and then science xx cancels each other out that's very nice so we can continue the majority down right here we have a negative 1 right here let's take it out and this time just going to black negative because

negative 1 times the integral it's the same as negative you go negative integral 1 to 0 and we have 1 over 1 plus u square u and now looking at this we have to ask ourself the rooty of what will give us 1 over 1 plus u squared the answer to die is inverse tangent so this part is going to give us inverse tangent you okay and then don't forgive still have the negative the front so we have negative inverse tangent u and then we are done with the integration part and then by the way

i forgot to mention is earlier from what dealing with they have an integrals we don't need to worry about a plus C we can just write this down the plus C is not needed because when you do the subtraction that is the C will always cancel anyways so this is all we need and then don't forget the bottom number is 1 and then the part number is 0 and then a plug in 0 first the number on the top first always okay so let's do that the first part is going to be negative inverse tangent

of 0 and then minus negative inverse tangent of 1 all right what's the inverse tangent of 0 this is just 0 because tangent 0 is 0 so the inverse tangent of 0 is also 0 this part is gone 0 that's nice and then negative times a negative is positive however what's the inverse tangent of 1 this is something you guys definitely have to know ok because the original tangent of PI over 4 is 1 therefore the inverse tangent of 1 is PI over 4 negative negative is positive and this is positive over Co altogether anyway

and this right here is the answer to this right here okay and you see this is why I'd really like to show you guys two integrals at the same time and ever have one more to show you guys okay so that's see example number three we have the integral x times the square root of x squared minus 1 DX and then for the second one we have the integral x times the square root of x minus 1 DX what's the difference you'll see this right here it has x squared but then for this one here

we only have X to the first power which of this do you think is easier the answer to die is the first one right here and I will to tell you guys sometimes when were two integrals the more the better and that's true in this situation why because I can just let u equals to the inside and the derivative of x squared minus 1 is 2x right you see that X and this X well cancel each hat out that it's very very nice and now let's put up all the work right here let u equal

to the inside x squared minus 1 and let's go ahead the friendship outside T you will be just 2x DX and then let's isolate the DX D pappas about 2 X we can e^x equals to D u over 2 X and now I can take this integral into the you world so this is going to be integral this X and then square root but then the inside X square minus 1 is the U right so put on you inside DX is d u over 2 X so I put down T u over 2 X and

you see this X and X cancel each other out that's very nice we have a 1/2 right here so that's put down the outside of integration and then welcome to integrate square root of you how come do that well let me put on Interpol side we are going to integrate instead of the square root of U we look at this s you to the one-half power you this is what we do whenever we have a square root or any kind of radical or roots right write that as a power and this is in the New

World so that's going integrator we can use the power rule that do it backwards we add one first one half plus one is three half and this is the new exponent and we have two divided by this right here which is the same as multiplied by the reciprocal so I'm going to put down two over three right reciprocal and you see this to an a two will cancel each other out then one third this is you to the three half power and then we are done with the integration step and now we can just write

this back in terms of X so we have one third I can open a parenthesis for the U which is x squared minus one and then oh this raised to the 3 power and this is the gentlemen you can just put plus C and present this for the answer but sometimes you may also like to write this in terms of square root because you have the half power here the over two right here is the square root so I can also reduce this one sir on the outside right this is one third this is on

the outside and then the half power is the square root and then you can put down x squared minus one and don't forget we still have that third power here this is also okay and then plus C either way which one that you prefer so that's it for that one and now you are wondering okay for this one we have the integral x times the square root of x minus 1 DX this is only to the first power earlier when you have x squared I differentiate that I get 2x the minus 1 becomes you're sold

doesn't matter but then this is just X if I differentiate I get 1 so one cannot consult this X do you think this still works so now here is another thing what to study this is right here is that sometimes we want to in use up he may not work as nicely as these kind of things we may have to do more algebra in order to make it work nicely so in fact we can still use u-substitution and I'm still going to that u equals to the inside X minus 1 and we still follow the

same procedure even though I cannot cancel this X right away but this is the algebra that we can do so let me show you still differentiate both sides on the left hand side we get bu but another of this derivative X is 1 and then this is 0 so it's just pretty much 1 DX I can just put down DX and that's all then use the same as DX DX is just u so we are ready and let's take a look on how does this look like in the new world we have the integral right

here and then X is just X right because this is what we have at the moment my focus was I'll change the X minus 1 to u so at the square root right here and then this was the you and then the DX was the you unfortunately this axis still here this part I don't have any additional X in the denominator to cancel out this X and that's why I said sometimes the more the better right so what can we do can we integrate this already or not no because this right here is an X

and keeping my X is not invited in the you world however we can still continue with this because right here we have an X and we do have a nice connection between U and X and let me write this down again for you guys U is equal to X minus 1 and why don't we just add 1 on both sides because this way we can say X is equal to u plus 1 and then instead of this X we can just change this to u plus 1 then every single B in terms of U everybody

happy right so now this is going to be integral X is U plus 1 and be sure to use the parentheses so we have the parentheses U plus 1 and we still have the square root of U U and finally this integral is completely in the new world and the pond right here is that sometimes we may have to do some algebra to make sure things happen don't just give up right here look back to the u equation to see what we can do okay how can we integrate this can we just use the reverse

power rule with that you know because this U and the school of you are affecting each other right so to make sure we multiply things out so we're still just doing algebra right now so that's right there you as you to 1/2 power for the square root right here so this is integral u plus 1 and this is U to the one-half power you and take this multiply multiply right just algebra for now in the world u times du to the one-half power this is like you to the first power and when you multiply we

add the powers right so when you will multiply we get to the 1 plus 1/2 which is three 1/2 and then this times that Plus u to the one-half power u so far just algebra and now we can finally do the reverse power rule integral u to this tree have power what we are going to do is add 1 to the power 3 half plus 1 is X a 3 half plus 2 over 2 3 plus 2 is 5 so this is 5 half and divided by Phi half it's the same as saying x 2

over 5 so the first part is going to be 2 over 5 u to the 5 over 2 and then the next part we have u to the one-half that's add 1 1 half plus 1 is 3 half and then T 5 as we have is the same as multiplying by 2/3 and this is positive so we add to serve you to the 3 half power and this was the integration step and we are done for that finally we just have to write this back in terms of X well well this is going to be

2/5 and U is X minus 1 so let's go down X minus 1 raised to the 5 over 2 power and then we add it with 2/3 this U is X minus 1 also and then this is 3 half power you can present this answer or you can write the half power as square root just that how we did it right here so this is going to be 2 over 5 I will write this as a square root X minus 1 in the parenthesis and then raised to the 5th power and then plus 2/3 the

half power is a square root and then the X minus 1 raised to the third power and Plus see so either way which one that you want to use to present your answer did all right I hope this video helps if he does please hit that like button and please recommend my channel to your friends classmates students and teachers thank you so much I'll see you guys next time