Integration using U-Substitution
In these next videos I'm going to start talking about some U substitution And U substitution is typically the first integration technique, you learn And the idea with U substitution is you're simply trying to turn an integral into a rather Somewhat complicated integral into one that's a little more recognizable So suppose we have the following problem suppose we have [2x] and then we have x squared Plus 4. Let's make it raised to the 100 power Now if you didn't know anything about U substitution, and you've seen a little bit of integration You could simply multiply this thing out you'd have [a] whole bunch of terms, and then you could integrate those piece-by-piece But probably you wouldn't want to do that at least I wouldn't want to The idea is what we're going to do is we're going to do kind of a relabeling scheme so I'm going to call the stuff on the inside the x squared plus 4 I'm going to re blable that as you so, I'm going to make U equal to the stuff inside the parentheses and The trick with [U] substitution is look for things that are derivatives of each other Notice if you take the derivative of this inside stuff you'll actually get the 2x term It doesn't have to be Exactly the other stuff as as we'll see but that's usually a clue that U substitution works so again I'm calling you the stuff on the inside and basically taking the Derivative of U which means I have to take the derivative of the right side as well The derivative [of] x squared is 2x 4 becomes 0 and then we tack on a DX. What you're actually doing here is calculating a differential and Notice the stuff inside the parentheses now if I go to rewrite this Well, we relabeled x squared plus 4 that's what we called you. So now I have [u] to the 100 Power and Notice my 2x dx that's left over because again.
I've replaced the [U] squared Plus 4 raised to the 100 power the only thing that's left over is the 2x dx term that's exactly what d u is equal to and I'm just going to put my d u out to the right and Now the idea is this is just a basic integration problem. You've got a variable raised to the 100th power It's just like having x raised to the 100 well, you'll get u to the 101 we have to add 1 to the power divide by that new power Plus C. And then the idea is since you started with x's you want to stop with x's right now We have use but simply just use your original substitution and plug that back in So again u is the same thing as x squared plus 4 raised to the 101 power divided by 101 plus C, and That will be your antiderivative so again use substitution is just kind of a trick for relabeling things to make it look like a more basic problem Let's see let's do let's do some more of these Suppose I have cosine to the fourth of theta times sine of theta d-Theta So again the thing I recognize here that kind of makes this stand out to me as well sine and cosine are Derivatives of each other yeah one of them you pick up a negative But that's okay as long as you're only you're getting the correct derivative what they say up to a constant [you] substitution will work remember that cosine of theta to the fourth you can rewrite that as The following and I usually [like] [to] see these a little bit better [I've] gotten used to realizing in my head that they're equivalent, but I used to always rewrite them just to make things a little clearer usually whatever is inside the Parentheses being raised to a power a lot of times that's going to be a good guess [that] that's going to be what you should Equal so I'm going to let U equal cosine of theta well the derivative or the differential is going to be negative sine of theta d-Theta [and] Notice though okay? So I'm calling Cosine of theta U so that'll take care of my u to the fourth and then I have a sine of theta D-Theta left over The differential though says I get negative sine theta D theta well, I can just multiply both sides by [a] negative and get negative d u equals sine Theta D theta, [and] that's now [what] I'll replace the sine theta d theta stuff with so it says that instead of cosine [of] theta that's what I'm calling you so I'll have you to the fourth and Now the sine theta d theta that's left hanging out That's what I'm going to plug in for the I'll plug negative d [U] in for that okay, so you can think about the negative is simply being a negative one and you can pull that all the way out front if You want to? And Again now all you have [to] do is integrate [U] to the fourth becomes U to the fifth over five plus C. And then again, we'll simply plug back in what our original U-substitution was back in for you So it says you was cosine of theta raised to the Fifth [Power] divided by Five plus C and that will be your antiderivative for this original problem, [and] You can always check if you take the derivative of this thing You know you should get what you are integrating back at the very beginning See if we can't find one other One other good problem here.
Let's do a student maybe one Tricky one Okay, so in that problem. I had to manipulate the you let's do one other one here Suppose [I] integrating x over the fourth root of x plus two now [usually] when I see things off by a power like an x squared and an x or maybe x to the ninth and x to The eighth it really stands out to me that it's used substitution But in fact, this is [also] a U-substitution problem Again if you want to rewrite this we can make this x over x plus 2 to the 1/4 power So I'm going to use the same logic in the last one where whatever is being raised to the power That's what will make you equal to so U is equal to x plus 2 the differential d U is simply going to be 1 Dx and if I go to real able things I'm going to leave myself a little space here [because] I'm going to talk about a little bit [more] The x plus 2 is equivalent to you, so I've got you to the 1/4 Dx is equivalent to d. You all right there that is But I still have an x left hanging out that I need to get rid of here well I can I've got a relationship involving U and x I can simply subtract 2 and have that U minus 2 is equal [to] x and That's what I'll plug in for my x on. Top is U minus 2 and Now we'll simply break this up U over u to the 1/4 that's going to become u to the 3/4 minus 2 over U to the 1/4 we'll bring that up and make it to the negative 1/4 and Now we'll simply integrate So you when we add 4 over 4 I'll get to the 7/4 Dividing by 7/4 is the same thing as multiplying by 4/7 minus 2 U to the negative 1/4 if I add 4 over 4 I'm going to get you to the 3/4 I'll have to divide by 3 [for] [werth's] which is the same as multiplying by 4/3 + [C] and again the last thing I need to do here is replace my use with my x plus 2 that, I had found a written from the original substitution [and] if there's any simplification, I'll do that as well, [so] I'll have 4/7 [U] which is x plus 2 to the 7 fourths minus 8/3 you again which is x plus 2 to the 3/4 plus C and? There is your antiderivative, so Again U substitution already becomes a little trickier unfortunately, there's much longer and harder integration techniques than [U] substitution [but] you'll see those soon enough unfortunately, [I] guess so if you want to see some more examples of U substitution I've only done indefinite integrals here.
I've also got a lot of videos with Definite integrals involving new substitution along with pretty much every other Integration technique as well on my website so feel free to hop over there and shoot me an email you got any questions.
I've replaced the [U] squared Plus 4 raised to the 100 power the only thing that's left over is the 2x dx term that's exactly what d u is equal to and I'm just going to put my d u out to the right and Now the idea is this is just a basic integration problem. You've got a variable raised to the 100th power It's just like having x raised to the 100 well, you'll get u to the 101 we have to add 1 to the power divide by that new power Plus C. And then the idea is since you started with x's you want to stop with x's right now We have use but simply just use your original substitution and plug that back in So again u is the same thing as x squared plus 4 raised to the 101 power divided by 101 plus C, and That will be your antiderivative so again use substitution is just kind of a trick for relabeling things to make it look like a more basic problem Let's see let's do let's do some more of these Suppose I have cosine to the fourth of theta times sine of theta d-Theta So again the thing I recognize here that kind of makes this stand out to me as well sine and cosine are Derivatives of each other yeah one of them you pick up a negative But that's okay as long as you're only you're getting the correct derivative what they say up to a constant [you] substitution will work remember that cosine of theta to the fourth you can rewrite that as The following and I usually [like] [to] see these a little bit better [I've] gotten used to realizing in my head that they're equivalent, but I used to always rewrite them just to make things a little clearer usually whatever is inside the Parentheses being raised to a power a lot of times that's going to be a good guess [that] that's going to be what you should Equal so I'm going to let U equal cosine of theta well the derivative or the differential is going to be negative sine of theta d-Theta [and] Notice though okay? So I'm calling Cosine of theta U so that'll take care of my u to the fourth and then I have a sine of theta D-Theta left over The differential though says I get negative sine theta D theta well, I can just multiply both sides by [a] negative and get negative d u equals sine Theta D theta, [and] that's now [what] I'll replace the sine theta d theta stuff with so it says that instead of cosine [of] theta that's what I'm calling you so I'll have you to the fourth and Now the sine theta d theta that's left hanging out That's what I'm going to plug in for the I'll plug negative d [U] in for that okay, so you can think about the negative is simply being a negative one and you can pull that all the way out front if You want to? And Again now all you have [to] do is integrate [U] to the fourth becomes U to the fifth over five plus C. And then again, we'll simply plug back in what our original U-substitution was back in for you So it says you was cosine of theta raised to the Fifth [Power] divided by Five plus C and that will be your antiderivative for this original problem, [and] You can always check if you take the derivative of this thing You know you should get what you are integrating back at the very beginning See if we can't find one other One other good problem here.
Let's do a student maybe one Tricky one Okay, so in that problem. I had to manipulate the you let's do one other one here Suppose [I] integrating x over the fourth root of x plus two now [usually] when I see things off by a power like an x squared and an x or maybe x to the ninth and x to The eighth it really stands out to me that it's used substitution But in fact, this is [also] a U-substitution problem Again if you want to rewrite this we can make this x over x plus 2 to the 1/4 power So I'm going to use the same logic in the last one where whatever is being raised to the power That's what will make you equal to so U is equal to x plus 2 the differential d U is simply going to be 1 Dx and if I go to real able things I'm going to leave myself a little space here [because] I'm going to talk about a little bit [more] The x plus 2 is equivalent to you, so I've got you to the 1/4 Dx is equivalent to d. You all right there that is But I still have an x left hanging out that I need to get rid of here well I can I've got a relationship involving U and x I can simply subtract 2 and have that U minus 2 is equal [to] x and That's what I'll plug in for my x on. Top is U minus 2 and Now we'll simply break this up U over u to the 1/4 that's going to become u to the 3/4 minus 2 over U to the 1/4 we'll bring that up and make it to the negative 1/4 and Now we'll simply integrate So you when we add 4 over 4 I'll get to the 7/4 Dividing by 7/4 is the same thing as multiplying by 4/7 minus 2 U to the negative 1/4 if I add 4 over 4 I'm going to get you to the 3/4 I'll have to divide by 3 [for] [werth's] which is the same as multiplying by 4/3 + [C] and again the last thing I need to do here is replace my use with my x plus 2 that, I had found a written from the original substitution [and] if there's any simplification, I'll do that as well, [so] I'll have 4/7 [U] which is x plus 2 to the 7 fourths minus 8/3 you again which is x plus 2 to the 3/4 plus C and? There is your antiderivative, so Again U substitution already becomes a little trickier unfortunately, there's much longer and harder integration techniques than [U] substitution [but] you'll see those soon enough unfortunately, [I] guess so if you want to see some more examples of U substitution I've only done indefinite integrals here.
I've also got a lot of videos with Definite integrals involving new substitution along with pretty much every other Integration technique as well on my website so feel free to hop over there and shoot me an email you got any questions.
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