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U-Substitution Integration · Acedra
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U-Substitution Integration
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0:00
in this video i'm going to show you how to integrate using u-substitution so we're going to focus on the definite integrals how can we find the anti-derivative of 4x times x squared plus five raised to the third power so what do we need to do we need to define two things we need to identify the u variable and d u now whatever you select u to be du has to be the derivative if we select u to be 4x the derivative will be 4 and that's not going to get rid of x squared plus 5.
0:42
we need to change all of the x variables into u variables now if we make u equal to x squared plus 5 d u is going to be 2x which can cancel the x and 4x and that's what we want to do so let's set u equal to x squared plus five d u is going to be 2x but times dx
1:08
so what i'm going to do now is solve for dx in this equation so if i divide both sides by 2x dx is equal to du divided by 2x now what you need to do is replace this with u and replace the dx with du over 2x and it will all work out so let's go ahead and do that so we have 4x and then u raised to the third power and then u divided by two x so here we can cancel x four x divided by two x is two so it's two times the antiderivative of u to the third using the power rule is going to be two times
1:55
u to the fourth over four plus c now two over four is one half so it's one half u to the fourth plus c the last thing you need to do is replace u with what it equals and that's x squared plus five so this is the answer
2:24
let's try another problem go ahead and find the integration of 8 cosine 4x dx so what should we make u equal to we need to make u equal to 4x d u will equal 4 dx and if we divide by 4 du over 4 is dx so let's replace 4x with u and let's replace dx with du divided by 4. so this is going to be 8 cosine of the u variable and replace dx with du over four so now we could divide eight by four eight divided by four is two now what is the anti-derivative of cosine the anti-derivative of cosine is sine
3:17
because the derivative of sine is cosine so we have 2 sine u plus the constant of integration c now let's replace u with 4x so the final answer is 2 sine 4x plus c so hopefully you see a pattern emerging when integrating by u substitution the key is to identify what u and d u is going to be once you figure that out you just got to follow the process and it's not going to be that bad so let's work on a few more examples so you can master this technique let's try x cube e
3:56
raised to the x to the fourth so what should we make u equal to x cube or x to the fourth if u is x to the third d u will be three x squared and that will not completely get rid of x to the fourth but if we make u equal to x to the fourth d u will be equal to four x cubed and that can get rid of the x cubed that we see here so let's do that let's make u equal to x to the fourth so d u is going to be 4x cubed dx and then as always solve for dx if you don't do that you can easily make
4:37
a mistake so i recommend in a step isolate dx it'll save you a lot of trouble later on so du is going to be i mean dx is going to be du over 4x cubed now let's replace x to the fourth with u and let's replace dx with d u over four x cubed so this is going to be e
5:05
raised to the u times d u over four x cubed so x cubed will cancel which is good and the 4 we can move it to the front now because it's in the bottom of the fraction it's 1 over 4. now the antiderivative of e to the u is simply e to the u so the final answer well not the final answer but the antiderivative is one fourth e to the u plus c and now let's replace u with x to the fourth so this is the final answer one fourth e raised to the x to the fourth plus c and that's it
5:47
here's another one find the indefinite integral of 8x times the square root of 40 minus 2x squared dx so typically you want to make u equal to the stuff that's more complicated and that is the stuff on the inside of the square root if we make u equal to 40 minus 2x squared d u is going to be the derivative of 40 is zero so we can ignore that and the derivative of negative 2x squared that's going to be negative 4x dx so isolating dx we need to divide both sides by negative 4x so it's
6:28
du over negative 4x so let's replace this with u and this part dx with du over negative 4x so it's going to be 8x times the square root of u times du divided by negative four x eight x divided by negative four x is negative two and i'm going to write that in front the square root of u is the same as u to the one half so now we can use the power rule one half plus one is three over two and then we could divide by three over two or multiply by two over three which is the better option
7:13
so negative two times two thirds that's negative four over three now the last thing we need to do is replace the u variable with 40 minus two x squared so the final answer is negative four over three 40 minus 2x squared raised to the 3 over 2 plus c and that's all we need to do
7:41
let's work on some more problems feel free to pause the video and try this one integrate x cubed divided by two plus x to the fourth raised to the second power now typically it's better to make u equal to the stuff that has the higher exponent four is higher than three so let's make u equal to two plus x to the fourth d u is going to be the derivative of x to the fourth so that's 4x to the third power times dx and as always i recommend that you solve for dx just to avoid mistakes
8:25
so let's replace two plus x to the fourth with u and let's replace dx with this thing that we have here so we have x to the third on top u squared on the bottom and dx is d u over 4 x cubed and if you do it this way as you can see the remaining x variables will cancel out nicely so this 4 is in the bottom let's move it to the front so it becomes 1 4 anti-derivative 1 over u squared d u and so let's move the u squared from the bottom to the top so this is going to be 1 4
9:06
integration of u to the negative two d u and now we can use the power rule so if we add one to negative two that's going to be negative one and then we need to divide by negative one so now let's bring this variable back to the bottom to make the negative exponent positive so it's negative one over four u plus c now let's replace u with what we set it equal to in the beginning so i'm going to need a bigger fraction so u is 2 plus x to the fourth and then plus c so this is the final answer
9:54
let's integrate sine to the fourth of x times cosine of x dx so go ahead and find the anti-derivative of this trigonometric function
10:09
so if we make u equal to cosine d u will be negative sine dx that will only cancel one of the sine variables and it's best to make u equal to the trig function that you have more of we have four sines and only one cosine so it's best to make u equal to sine x and d u will be equal to cosine x and since there's only one cosine this will be completely cancelled solving for dx is going to be du divided by cosine x and don't forget that last step always isolate dx
10:49
so let's replace this with you so this is going to be u to the fourth times cosine x and dx is du divided by cosine so we could cancel cosine x and so we're left with the indefinite integral of u to the fourth d u using the power rule four plus one is five and then divide by five so we have one-fifth u to the five plus c and the last thing we need to do is replace u with sine x so it's one-fifth sine raised to the fifth power of x plus c and that concludes this problem
11:39
now what would you do if you have to integrate the square root of 5x plus 4. now in this problem all we could do is set u equal to 5x plus 4. there's nothing else that we can do so let's go ahead and do this the derivative of 5x is going to be 5 and then times dx so isolating dx it's going to be du over 5. so just like before we're going to replace 5x plus 4 with u and d x with d u over five so then this becomes the square root of u and dx is d u divided by five and move the five to the front
12:21
so this is one fifth and then instead of the square root of u we're gonna write it as u to the one-half so now let's use the power rule one-half plus one that's gonna be three over two and if you divide it by three over two what i would recommend is multiply the top and the bottom by two thirds so the threes in the bottom will cancel and the twos will cancel as well so in the end you get one-fifth u to the three-halves times two over three which becomes two over fifteen u to the three-half
13:00
and let's not forget the constant of integration plus c so now to write the final answer all we need to do is replace u with five x plus four so it's two over fifteen five x plus four raised to the three over two plus c and so that's the solution so as you can see u-substitution is not very difficult once you get the hang of it as long as you do a few problems and get used to the method and the techniques employed here it's a piece of cake now this problem is a little bit different from the
13:41
others go ahead and try it i recommend that you pause the video and give this one a go so let's set u equal to the stuff that's more complicated three x plus two now d u that's gonna be the derivative of 3x which is 3 times dx so isolating dx it's going to be du over 3 and this time this x variable will not cancel so notice that the x variables are of the same degree and when you see this situation it indicates that in this expression you need to solve for x so isolating x
14:25
i need to move the two to the other side so i have u minus two is equal to three x and then dividing by three u minus two over three is x which you can write it as you can say x is 1 3 u minus 2 which i think looks a lot better now keep in mind in order to perform u substitution we need to eliminate every x variable in this expression if we replace three x plus two with u and then dx with d u over three this x will still be here and so that's why we need to solve for x in this expression
15:10
so make sure to do that if these two are of the same degree
15:18
so this is going to be one third i'm gonna have to rewrite it because i can't fit it in here so let's replace x first with one third u minus two and then we have the square root of u and dx is d u divided by three so 1 3 times du over 3 that's going to be du over 9. so i'm going to take the 1 9 and move it to the front and then i have u minus 2 and the square root of u is u to the one half and then d u so now we only have the u variable in this form we can integrate it
16:08
now the next thing we need to do is distribute u to the one half to u minus two so u to the first power times u to the one half we need to add one and one half that's going to be u to the three over two and then if we multiply negative two by u to the one half that's negative two u to the one half and then times d u so now we can find the antiderivative of each one so for u to the three halves is going to be three over two plus one which is five over two and instead of
16:41
dividing it by five over two we're gonna multiply by two over five and for u to the one-half one-half plus one is three over two and then we're going to multiply by two thirds and then we need to add plus c
17:00
so now let's distribute one over nine to everything on the inside so 1 over 9 times 2 over 5 that's going to be 2 over 45 times u raised to the 5 over 2. and then we have one knife times this is basically negative four-thirds so that's gonna be negative four over twenty-seven and that's u to the three over two and then plus c so now let's replace u with three x plus two so the final answer is going to be two over forty five three x plus two raised to the five over two
17:43
minus four over twenty seven times three x plus two raised to the three over two plus c and that is it
18:00
now let's work on this example it's very similar to the last one so you can try if you want more practice so let's set u equal to 4x minus 5 which means du is going to be the derivative of 4x that's 4 and then times dx so solving for dx it's du divided by 4. now we need to isolate x in this expression so if we add 5 u plus five is equal to four x and then if we divide by four x is equal to this which we can write it as one fourth u plus five so let's replace x with this expression
18:47
and then 4x minus 5 with u and then dx with du over 4. so what we have is two times one fourth u plus five and then the square root of u or u to the one half and then dx is d u over four so two times one fourth is one half and one half times one over four is one eighth so we can move the one eighth to the front and then we have u to the half times u plus five now let's distribute u to the one half so u to the one half times u to the first power one half plus one that's going to be three over two
19:36
and then plus 5 u to the one half
19:42
so now we need to integrate the expression that we now have
19:52
so this is going to be one over eight u three over two plus one that's five over two and then times two over five and then one half plus one is three over two times two over three and then plus c
20:20
one over eight times two over five that's going to be two over forty and then times u raised to the five over two and then we have five times two over three that's ten over three times one over eight so that's going to be ten over eight times three is twenty four and this is going to be u to the 3 over 2 plus c now 2 over 40 can be reduced to 1 over 20. and let's replace u with 4x minus 5. so this is going to be 4x minus 5 raised to the 5 over 2. and then 10 over 24 you could reduce
21:05
that to 5 over 12 and then it's going to be 4x minus 5 to the 3 halves and then plus c
21:34
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