Understanding U-Substitution in Calculus

In this video, I'm going to show you how to integrate using u-substitution. So we're going to focus on definite integrals. How can we find the antiderivative of 4x times x squared plus 5 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 du. 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. We need to change all of the x variables into u variables. Now, if we make u equal to x squared plus 5, du is going to be 2x, which can cancel the x in 4x. And that's what we want to do. So let's set u equal to x squared plus 5. Du is going to be 2x but times dx. 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 du divided by 2x. So here we can cancel x. 4x divided by 2x is 2. So it's 2 times the antiderivative of u to the 3rd. Using the power rule, it's going to be 2 times u to the 4th over 4 plus c. Now, 2 over 4 is 1 half, so it's 1 half u to the 4th plus c. The last thing you need to do is replace u with what it equals, and that's x squared plus 5. So this is the answer. 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. du 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 it with 4. So now we can divide 8 by 4. 8 divided by 4 is 2. Now what is the antiderivative of cosine? The antiderivative of cosine is sine because the derivative of sine is negative. 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 du 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 look. Let's work on a few more examples so you can master this technique. Let's try x cubed e raised to the x to the fourth. So what should we make u equal to? X cubed or x to the fourth? If u is x to the third, du will be equal to x to the third. 3x squared and that will not completely get rid of x to the fourth but if we make u equal to x to the fourth du will be equal to 4x 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 du is going to be 4x cubed dx and then as always solve for dx you If you don't do that, you can easily make a mistake. So I recommend, in this step, isolate dx. It'll save you a lot of trouble later on. So dx is going to be du over 4x cubed. Now let's replace x to the 4th with u. And let's replace dx with du. over 4x cubed. So this is going to be e raised to the u times du over 4x 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 to the U is simply E to the U. So the final answer, well not the final answer, but the antiderivative is 1 fourth E to the U plus C. And now let's replace U with X to the fourth. So this is the final answer. 1 fourth E raised to the X to the fourth plus C. And that's it. 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, du is going to be the derivative of 40 is 0, so we can ignore that, and the derivative of negative 2x squared is going to be negative 4x dx. So, isolating dx, we need to divide both sides by negative 4x, so it's 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 4x. 8x divided by negative 4x is negative 2. And I'm going to write that in front. The square root of u is the same as u to the 1 half. So now we can use the power rule. 1 half plus 1 is 3 over 2. And then we could divide by 3 over 2 or multiply by 2 over 3, which is the better option. So negative 2 times 2 thirds, that's negative 4 over 3. now the last thing we need to do is replace the u variable with 40 minus 2x squared so the final answer is negative 4 over 3 40 minus 2x squared raised to the 3 over 2 plus C and that's all we need to do Let's work on some more problems. Feel free to pause the video and try this one. Integrate x cubed divided by 2 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. 4 is higher than 3. So let's make u equal to 2 plus x to the 4th. du is going to be the derivative of x to the 4th, so that's 4x to the 3rd power times dx. And as always, I recommend that you solve for dx, just to avoid mistakes. So let's replace 2 plus x to the 4th with u, and let's replace dx with this thing that we have here. So we're going to have x to the 3rd on top, u squared on the bottom, and dx is du over 4x cubed. And if you do it this way, as you can see, the remaining x variables will cancel out nicely. This 4 is in the bottom, let's move it to the front, so it becomes 1 fourth, anti-derivative 1 over u squared du. And so let's move the u squared from the bottom to the top. So this is going to be 1 fourth. integration of u to the negative 2 du and now we can use the power rule so if we add 1 to negative 2 that's going to be negative 1 and then we need to divide by negative 1 so now let's bring this variable back to the bottom to make the negative exponent positive so it's negative 1 over for you 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 4th, and then plus c. So this is the final answer. Let's integrate sine to the 4th of x. times cosine of x dx. So go ahead and find the antiderivative of this trigonometric function. So if we make u equal to cosine, du 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 du 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. So let's replace this with u. so this is going to be u to the fourth times cosine x and d-axis du divided by cosine so we could cancel cosine x and so we're left with the indefinite integral of u to the fourth du using the power rule 4 plus 1 is 5 and then divide by 5 So we have 1 5th u to the 5 plus c. And the last thing we need to do is replace u with sine x. So it's 1 5th sine raised to the 5th power of x plus c. And that concludes this problem. 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 dx with du over 5. So then this becomes the square root of u, and dx is du divided by 5. And move the 5... to the front so this is 1 5th and then instead of the square root of you we're going to write as u to the 1 half so now let's use the power rule 1 half plus 1 that's going to be 3 over 2 and if you divide it by 3 over 2 what I would recommend is multiply the top and the bottom by 2 thirds So the 3's in the bottom will cancel, and the 2's will cancel as well. So in the end, you get 1 5th u to the 3 halves times 2 over 3, which becomes 2 over 15 u to the 3 halves. 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 5x plus 4. So it's 2 over 15, 5x plus 4, raised to the 3 over 2, 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 others go ahead and try it I recommend that you pause the video and give this one a go so let's set you equal to the stuff that's more complicated 3x plus 2 Now, du, that's going to 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. When you see this situation, it indicates that in this expression, you need to solve for X. So isolating X, I need to move the 2 to the other side. So I have u minus 2 is equal to 3x. And then divided by 3, u minus 2 over 3 is x. Which you can write it as, you can say x is 1 third 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 3x plus 2 with u, and then dx with du over 3, this x will still be here. And so that's why we need to solve for x in this expression. So make sure to do that if these two are of the same degree. So this is going to be 1 third. I'm going to have to rewrite it because I can't fit it in here. So let's replace x first with 1 third, u minus 2, and then we have the square root of u, and dx is du divided by 3. So 1 third times du. over 3 that's going to be DU over 9 so I'm going to take the one ninth and move it to the front and then I have you minus 2 and the square root of u is u to the 1 half and then do you so now we only have the u variable in this form we can integrate it now the next thing we need to do is distribute you to the 1 half to you minus 2 so you to the first power times you to the one half we need to add one and one half that's going to be you to this 3 over 2. And then if we multiply negative 2 by u to the 1 half, that's negative 2 u to the 1 half. And then times du. So now we can find the antiderivative of each one. So for u to the 3 halves, it's going to be 3 over 2. plus 1 which is 5 over 2 and instead of dividing it by 5 over 2 we're going to multiply by 2 over 5 and for u to the 1 half 1 half plus 1 is 3 over 2 and then we're going to multiply it by 2 thirds and then we need to add plus C So now let's distribute 1 over 9 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 1 knife times... this is basically negative 4 thirds so that's going to be negative 4 over 27 and that's u to the 3 over 2 and then plus c so now let's replace u with 3x plus 2 so the final answer It's going to be 2 over 45, 3x plus 2, raised to the 5 over 2, minus 4 over 27, times 3x plus 2, raised to the 3 over 2, plus c. And that is it. Now let's work on this example. It's very similar to the last one, so you can try it 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 5 is equal to 4x and then if we divide by 4 X is equal to this which we can write it as 1 4th u plus 5 So let's replace x with this expression, and then 4x minus 5 with u, and then dx with du over 4. So what we have is... 2 times 1 fourth, u plus 5, and then the square root of u, or u to the 1 half, and then dx is du over 4. So 2 times 1 fourth. times 1 4th is 1 half. And 1 half times 1 over 4 is 1 8th. So we can move the 1 8th to the front. And then we have u to the half times u plus 5. Now let's distribute u to the 1 half. So u to the 1 half times u to the first power. 1 half plus 1, that's going to be 3 over 2. And then plus 5, u to the 1 half. So now we need to integrate the expression that we now have. So this is going to be 1 over 8 u. 3 over 2 plus 1, that's 5 over 2, and then times 2 over 5. And then 1 half plus 1 is 3 over 2 times 2 over 3, and then plus C. 1 over 8 times 2 over 5. That's going to be 2 over 40. And then times u raised to the 5 over 2. And then we have 5 times 2 over 3. That's 10 over 3 times 1 over 8. So that's going to be 10 over... 8 times 3 is 24. 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 can reduce that to 5 over 12. And then it's going to be 4x minus 5 to the 3 halves. And then plus c.

So what do we need to do? We need to define two things. We need to identify the u variable and du.

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. We need to change all of the x variables into u variables. Now, if we make u equal to x squared plus 5, du is going to be 2x, which can cancel the x in 4x. And that's what we want to do. So let's set u equal to x squared plus 5. Du is going to be 2x but times dx.

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 du divided by 2x. So here we can cancel x. 4x divided by 2x is 2. So it's 2 times the antiderivative of u to the 3rd.

Using the power rule, it's going to be 2 times u to the 4th over 4 plus c. Now, 2 over 4 is 1 half, so it's 1 half u to the 4th plus c. The last thing you need to do is replace u with what it equals, and that's x squared plus 5. So this is the answer.

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. du 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 it with 4. So now we can divide 8 by 4. 8 divided by 4 is 2. Now what is the antiderivative of cosine? The antiderivative of cosine is sine because the derivative of sine is negative. 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 du 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 look. Let's work on a few more examples so you can master this technique. Let's try x cubed e raised to the x to the fourth. So what should we make u equal to? X cubed or x to the fourth?

If u is x to the third, du will be equal to x to the third. 3x squared and that will not completely get rid of x to the fourth but if we make u equal to x to the fourth du will be equal to 4x 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 du is going to be 4x cubed dx and then as always solve for dx you If you don't do that, you can easily make a mistake. So I recommend, in this step, isolate dx.

It'll save you a lot of trouble later on. So dx is going to be du over 4x cubed. Now let's replace x to the 4th with u. And let's replace dx with du. over 4x cubed.

So this is going to be e raised to the u times du over 4x 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 to the U is simply E to the U. So the final answer, well not the final answer, but the antiderivative is 1 fourth E to the U plus C. And now let's replace U with X to the fourth.

So this is the final answer. 1 fourth E raised to the X to the fourth plus C. And that's it. 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, du is going to be the derivative of 40 is 0, so we can ignore that, and the derivative of negative 2x squared is going to be negative 4x dx. So, isolating dx, we need to divide both sides by negative 4x, so it's 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 4x. 8x divided by negative 4x is negative 2. And I'm going to write that in front.

The square root of u is the same as u to the 1 half. So now we can use the power rule. 1 half plus 1 is 3 over 2. And then we could divide by 3 over 2 or multiply by 2 over 3, which is the better option.

So negative 2 times 2 thirds, that's negative 4 over 3. now the last thing we need to do is replace the u variable with 40 minus 2x squared so the final answer is negative 4 over 3 40 minus 2x squared raised to the 3 over 2 plus C and that's all we need to do Let's work on some more problems. Feel free to pause the video and try this one. Integrate x cubed divided by 2 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. 4 is higher than 3. So let's make u equal to 2 plus x to the 4th.

du is going to be the derivative of x to the 4th, so that's 4x to the 3rd power times dx. And as always, I recommend that you solve for dx, just to avoid mistakes. So let's replace 2 plus x to the 4th with u, and let's replace dx with this thing that we have here. So we're going to have x to the 3rd on top, u squared on the bottom, and dx is du over 4x cubed.

And if you do it this way, as you can see, the remaining x variables will cancel out nicely. This 4 is in the bottom, let's move it to the front, so it becomes 1 fourth, anti-derivative 1 over u squared du. And so let's move the u squared from the bottom to the top. So this is going to be 1 fourth. integration of u to the negative 2 du and now we can use the power rule so if we add 1 to negative 2 that's going to be negative 1 and then we need to divide by negative 1 so now let's bring this variable back to the bottom to make the negative exponent positive so it's negative 1 over for you 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 4th, and then plus c. So this is the final answer. Let's integrate sine to the 4th of x.

times cosine of x dx. So go ahead and find the antiderivative of this trigonometric function. So if we make u equal to cosine, du 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 du 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. So let's replace this with u. so this is going to be u to the fourth times cosine x and d-axis du divided by cosine so we could cancel cosine x and so we're left with the indefinite integral of u to the fourth du using the power rule 4 plus 1 is 5 and then divide by 5 So we have 1 5th u to the 5 plus c.

And the last thing we need to do is replace u with sine x. So it's 1 5th sine raised to the 5th power of x plus c. And that concludes this problem.

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 dx with du over 5. So then this becomes the square root of u, and dx is du divided by 5. And move the 5...

to the front so this is 1 5th and then instead of the square root of you we're going to write as u to the 1 half so now let's use the power rule 1 half plus 1 that's going to be 3 over 2 and if you divide it by 3 over 2 what I would recommend is multiply the top and the bottom by 2 thirds So the 3's in the bottom will cancel, and the 2's will cancel as well. So in the end, you get 1 5th u to the 3 halves times 2 over 3, which becomes 2 over 15 u to the 3 halves. 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 5x plus 4. So it's 2 over 15, 5x plus 4, raised to the 3 over 2, 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 others go ahead and try it I recommend that you pause the video and give this one a go so let's set you equal to the stuff that's more complicated 3x plus 2 Now, du, that's going to 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. When you see this situation, it indicates that in this expression, you need to solve for X. So isolating X, I need to move the 2 to the other side. So I have u minus 2 is equal to 3x.

And then divided by 3, u minus 2 over 3 is x. Which you can write it as, you can say x is 1 third 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 3x plus 2 with u, and then dx with du over 3, this x will still be here. And so that's why we need to solve for x in this expression. So make sure to do that if these two are of the same degree.

So this is going to be 1 third. I'm going to have to rewrite it because I can't fit it in here. So let's replace x first with 1 third, u minus 2, and then we have the square root of u, and dx is du divided by 3. So 1 third times du. over 3 that's going to be DU over 9 so I'm going to take the one ninth and move it to the front and then I have you minus 2 and the square root of u is u to the 1 half and then do you so now we only have the u variable in this form we can integrate it now the next thing we need to do is distribute you to the 1 half to you minus 2 so you to the first power times you to the one half we need to add one and one half that's going to be you to this 3 over 2. And then if we multiply negative 2 by u to the 1 half, that's negative 2 u to the 1 half. And then times du.

So now we can find the antiderivative of each one. So for u to the 3 halves, it's going to be 3 over 2. plus 1 which is 5 over 2 and instead of dividing it by 5 over 2 we're going to multiply by 2 over 5 and for u to the 1 half 1 half plus 1 is 3 over 2 and then we're going to multiply it by 2 thirds and then we need to add plus C So now let's distribute 1 over 9 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 1 knife times... this is basically negative 4 thirds so that's going to be negative 4 over 27 and that's u to the 3 over 2 and then plus c so now let's replace u with 3x plus 2 so the final answer It's going to be 2 over 45, 3x plus 2, raised to the 5 over 2, minus 4 over 27, times 3x plus 2, raised to the 3 over 2, plus c. And that is it.

Now let's work on this example. It's very similar to the last one, so you can try it 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 5 is equal to 4x and then if we divide by 4 X is equal to this which we can write it as 1 4th u plus 5 So let's replace x with this expression, and then 4x minus 5 with u, and then dx with du over 4. So what we have is...

2 times 1 fourth, u plus 5, and then the square root of u, or u to the 1 half, and then dx is du over 4. So 2 times 1 fourth. times 1 4th is 1 half. And 1 half times 1 over 4 is 1 8th. So we can move the 1 8th to the front.

And then we have u to the half times u plus 5. Now let's distribute u to the 1 half. So u to the 1 half times u to the first power. 1 half plus 1, that's going to be 3 over 2. And then plus 5, u to the 1 half.

So now we need to integrate the expression that we now have. So this is going to be 1 over 8 u. 3 over 2 plus 1, that's 5 over 2, and then times 2 over 5. And then 1 half plus 1 is 3 over 2 times 2 over 3, and then plus C. 1 over 8 times 2 over 5. That's going to be 2 over 40. And then times u raised to the 5 over 2. And then we have 5 times 2 over 3. That's 10 over 3 times 1 over 8. So that's going to be 10 over...

8 times 3 is 24. 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 can reduce that to 5 over 12. And then it's going to be 4x minus 5 to the 3 halves. And then plus c.

Transcript for:Understanding U-Substitution in Calculus

Transcript for:
Understanding U-Substitution in Calculus