Hi this video is a continuation of the previous video which discusses integrals Now we will talk about one of the integration techniques, namely the X substitution method Hi for those who don't understand what integrals are or want to know the basics of integrals first, how can you watch this video first Now we will discuss the substitution method in the previous video we have discussed integrals of a power number for example x to the power of n can be expressed as one percent plus one times x to the power of N + 1 + C this is a general formula that we should have already known well the substitution method is useful to manipulate the form so that the problem we have can yield to this general formula for example let's say we have an integral of 2x times x squared + 5 ^ 5 DX in the previous video we have agreed that we cannot integrate this directly because there are no properties that can cover this This is the multiplication of two functions, if we want to integrate it directly, we have to explain it first, where x squared plus 2 to the power of 5 will be very difficult if explained, so here we can use the substitution method What is done in the method substitution is assumed to be u-disk this problem we choose as is x squared + 5 by assuming x squared + 5 y as uh then we can Express this integral as integral x squared plus five yes replaced with UU so it becomes uh power-5 well but there is 2xdx that we don't know we should not be anything in Uh so simply the survey method is that we can see it as changing the variable that we initially Expressed in X and Now we want to Express in us Muara Karang how this expresses the remaining expression 2xdx as deu we can see ui7 if it is differentiated with X it becomes 2x then dxc can be as if multiplied to the right so d u = 2x DX so that 2xdx this problem we can replace with deu until here understand if the form is this makes it similar to the general formula We only differ in the variables the difference is expressed in UU this just integrate normally which becomes m is 5 so it becomes one sixth Uh power-6 plus C well but remember the problem we are expressed in X so we return it back to X so it becomes one-sixth of x squared + 5 to the power of -6 plus C the next example is the integral of 6 x squared + 6 divided by the root of x to the power of 3 + 3x + 1dx well which becomes the UD of this problem becomes Uye is X ^ 3 + 3 x + 1 so we will get the DPRD x is 3 x squared plus three then deu = 3 x squared plus 3dx so we can write this integral as the integral of the root of x to the power of 3 + 3 x + 1 which below we replace it with the root poe-in ah now what about 6 x squared plus 6dx what we have until now is Hi squared plus 3dx if we want to form 6 x squared plus 6dx that is right We just multiply it by two Okay we will get 6 x squared plus 6dx that is 2D so yes we can replace this with 2D u-kiss tidy up the shape of the root is concentrated half if below it becomes minus then it can be written as two the callus as 2D potassium to the power of half Now we just integrate the n which becomes Min half Min half plus one is half then we get two times one and a half times uh the power of half to C we tidy up the 2 first if like above becomes 4 then turn it back to X ^ 3 + 3 x + 1 to the power of half + c to the power of half that means the root so we get the answer as follows have you started to get an idea of how the substitution method is done So the first thing we have to get the idea is which one we Turn on as How to choose Wooyoung right at number 1 and number 2 So what we choose as that is what if we lower it can be the rest the rest in quotation marks okay example number one x squared plus five if lowered 2xdx yes then the remaining 2xdx that we haven't assumed we can replace with deu Later in number two if we lower it it becomes 3X ^ 2 + 3 it's different from 6 x squared plus six but it's only a multiple we can multiply by two to make it the same that's so once again to determine which one should be uh that is the one that if it is derived can be the remainder or maybe a multiple of the remainder until here hopefully understand yes let's try another example question Now use the limit suppose there is an integral x root x squared plus 3dx the limit is from one to the root of 6 by approximately which one so have four coins here is x squared + 3 Why because the derivative of the UU is 2x so we can get it first = 2x DX well until here we can Express the root of x squared + 3 as the root uh well xdx as what we have is 2xdx so it becomes xcxn we just multiply by half then we will get half do = x DX so we can replace the xdx in the question with half deui now we have got an integral expression in us mua well what about the limit the limit if we just enter one to the root of 6 that's wrong we can't just use it because one to the root of 6 this is the limit for X Not For Me so we have to change the limit So if x y = 1 How many do you have If x = root 6 how many do you have here uh = x squared + 3 so if x is with one we just put XC into Have or not Have will be 4 while if x = root 6 have will be 9 then the limit which was initially one will be four which was initially because then it becomes 9 up to here okay if it's like this we just integrate we can tidy up half of it first just put it outside the root Wow is uh ^ half-half if added one becomes three two then we will get half times 1/3 two in Kaliwungu ^ 3/2 we try to tidy up and enter the limit we will get one third multiplied by 9 ^ 3/2 minus 4 ^ 3/2 we Simplify again 9 is the same as three to the power of 2 and 4 is two to the power of two so we can get the final result is 19/3 Well this is roughly the basic description What is the substitution method and How to use it in the question when to use the substitution method yes when we can't integrate the integral directly we don't can use this formula directly so we have to manipulate the shape first so that the shape becomes like this one of the most common methods that we can use to change the shape so that it can be like this basic formula is called the substitution method hopefully this video is useful to provide an understanding of the substitution method basically the integral does need intuition because in essence the integral is guessing the basic function So it really takes a lot of practice in questions like this so that intuition and analysis are honed to determine Wooyoung right more interesting questions related to the substitution method we will discuss in the next videos Thank you for those who have liked subscribe Turn on the notification button please share this video if it is useful don't forget to leave a comment and if you want to learn mathematics Cut can contact the contact in the description column