## Cylinder/Shell Method – Rotate around a horizontal line

Before reading through this problem, I’d recommend checking out my lesson on finding volumes of rotation using the cylinder shell method. I’m not going to go into quite as much detail here as I did in that lesson. It might help you make more sense of what’s going on if your start there.

Other than that there isn’t much else to add so let’s jump into an example!

## Example 1

Find the area of the solid created by rotating the area bounded between $y= (x-1)^3-3$, $y=-x-2$, and $y=-2$ about the line $y=-1$.

Just as before I’ll use the same 4 step process as in the cylinder method lesson.

### 1. Graph the 2-D functions

As I always say, I suggest starting any problem possible by drawing what is being described to you. Go ahead and start with graphing all of the functions described in the problem. I’ll do this using Desmos. You should end up with something like the graph below. I also went ahead and shaded the bounded region gray to make it a little easier to see (this was not done in Desmos).

### 2. Rotate the 2-D area around the given axis

Again, we want to visualize what the question is asking us to find. We will need to take the shaded region in the above graph and rotate it around the line $y=-1$. Doing this would create a 3-D figure whose volume we’ll need to find. But first let’s draw it.

To do this, imagine the 2-D gray region coming off the paper or screen and rotating around the axis of rotation. Doing this would give us something like the figure below.

### 3. Setting up the integral

I’m not going to go into as much detail to explain where this integral comes from as I did in the cylinder method lesson, but if the following integral confuses you I’d recommend checking that lesson out by clicking on the above link.

Long story short, we want to imagine our 3-D figure is made up of several infinitely thin cylindrical shells. Adding up the volume of all of these shells would result in an integral like this: $$\int 2 \pi r h \ dr.$$

In order to help with coming up with each of these pieces, we need to relate them back to our figure and the functions that created it. In order to visualize this, let’s draw our figure with one of these infinitely thin shells that make up the entire figure. We can consider this one shell and how to represent these dimensions in terms of the given functions.

You can see one of these cylindrical shells represented in the drawing below with a labeled version of the cylinder draw in the upper-right hand corner.

As with all cylinder shell method problems, we need to imagine integrating from the center of the cylinder out to the outer edge. Since our cylinder is laying horizontally, moving from its center to its edge moves up and down. This means we are moving in the y direction. Therefore, we need to integrate in the y direction and represent our integral only in terms of y (we shouldn’t have any x‘s).

So let’s think about each of the three pieces that make up our integral one at a time.

#### Finding r

The radius of this cylinder would simply be the distance between the center of the cylinder and the edge. You can see in the smaller version of the cylinder drawn off to the side that the radius is represented by the red line measuring between the points labeled $(x_2, \ -1)$ and $(x_2, \ y)$.

Since these two points have the same x value, we can find the distance between them by simply finding the distance between their y values. To do this we just need to take the larger value and subtract the smaller one from it. $$r=-1-y$$

#### Finding h

The height of a cylinder will always be measured as the distance between the two flat, parallel faces. Usually they would be the top and bottom, but since our cylinder is sideways, we need the distance between the left side and right side.

Looking at the smaller cylindrical shell off to the side in the drawing above, you can see the height of this cylinder is represented by the red line measuring the distance between the points $(x_1, \ y)$ and $(x_2, \ y)$.

Similar to what we did before, these two points have the same y value. As a result, the distance between them would be the same as the distance between their x values. So we just need to take the larger x value and subtract away the smaller one. $$h=x_2-x_1$$

But remember earlier I said we need everything just in terms of y?

So we need to think about how we can rewrite $x_1$ and $x_2$ in terms of y.

#### Finding $\mathbf{x_1}$$\mathbf{x_1}$

We know that $x_1$ lies on the function $y=-x-2$ so we know that the relationship between $x_1$ and y can be described in the same way $$y=-x_1-2.$$ If we rearrange this to solve for $x_1$ instead of y, we can use this to replace the $\mathbf{x_1}$ in our equation for h. $$y=-x_1-2$$ $$y+x_1=-2$$ $$x_1=-y-2$$

We can use this to rewrite h but replace the $x_1$ with $(-y-2)$ since we know they are equal. $$h= \ x_2- (-y-2)$$ Now we need to do the same thing with $x_2$.

#### Finding $\mathbf{x_2}$$\mathbf{x_2}$

We are going to apply the same idea here as in the previous section. We know that $x_2$ lies on the function $y= (x-1)^3-3$. Therefore, we can describe the relationship between $x_2$ and y as $$y= (x_2-1)^3-3.$$ Now we can solve this equation for $x_2$ and plug this into our equation for h. $$y \ = \ (x_2-1)^3-3$$ $$y+3 \ = \ (x_2-1)^3$$ $$\sqrt[3] {y+3} \ = \ x_2-1$$ $$\sqrt[3] {y+3} +1 \ = \ x_2$$ Now going back to our equation for h, this tells us $$h \ = \ \sqrt[3] {y+3} +1 – (-y-2).$$ And to simplify a bit: $$h \ = \ \sqrt[3] {y+3} +1 + y+2$$ $$h \ = \ \sqrt[3] {y+3} + y+3.$$ Now that we have h and r, we just need to find dr.

#### Finding dr

This is actually the simplest part to find. The dr represents the change in the cylinder’s radius as we go from each shell to the next. Since we move in the same direction of the radius as we integrate to find our volume, the change in r should be the same as the change in y between each step. Therefore, we can say that $$dr=dy.$$

#### Putting it all back into an integral

We already figured out that the volume of our figure can be found by using the integral $$\int 2 \pi rh \ dr.$$ And we just found these three pieces to be $$r=-1-y$$ $$h \ = \ \sqrt[3] {y+3} + y+3$$ $$dr=dy.$$ So we can just plug them into our integral. $$\int 2 \pi \ ( -1-y ) \ \Big( \sqrt[3] {y+3} + y+3 \Big) \ dy$$

Now we need one last piece. We need to add bounds on the integrals.

Since we are integrating with respect to y, the bounds of our integrals need to be the range of y values that make up our original 2-D area. Looking back at our original graph, we can see that the original area bounded by the given functions spans over all of the y values between $y=-2$ and $y=-3$. Therefore, we know that the volume of our figure will be $$V \ = \int_{-3}^{-2} 2 \pi \ ( -1-y ) \ \Big( \sqrt[3] {y+3} + y+3 \Big) \ dy.$$

### 4. Solve the integral

Now all we need to do is solve the integral we just found and that will leave us with our volume. This is actually a pretty complicated integral as is it, so let’s start with simplifying it a bit. We’ll do this by pulling out the constant, distributing out through the parenthesis, and combining like terms.

$$V \ = \int_{-3}^{-2} 2 \pi \ ( -1-y ) \ \Big( \sqrt[3] {y+3} + y+3 \Big) \ dy$$ $$V \ = \ 2 \pi \int_{-3}^{-2} \ – \big(y+3 \big)^{\frac{1}{3}} – y \ – 3 -y \big(y+3 \big)^{\frac{1}{3}} – y^2 – 3y \ \ dy$$ $$V \ = \ 2 \pi \int_{-3}^{-2} \ – \big(y+3 \big)^{\frac{1}{3}} -y \big(y+3 \big)^{\frac{1}{3}} – y^2 – 4y -3 \ \ dy$$

Now that we have it in a form that is simplest to integrate we can go ahead and integrate this function one term at a time. I’m not going to show every step of how to do this, but if you’d like to work it out on your own, I’d suggest using u-substitution on the $-(y+3)^{1/3}$ term and using integration by parts on the $-y(y+3)^{1/3}$ term.

$$V \ = \ 2 \pi \Bigg[ – \frac{3}{14} \big( y+3 \big)^{\frac{4}{3}}\big( 2y-1 \big) – \frac{1}{3}y^3 – 2y^2 – 3y \Bigg]_{-3}^{-2}$$

Again, I’m not going to show every step of this. Instead I used Wolfram Alpha from here, but if you evaluate this expression from $y=-3$ to $y=-2$, you’ll see that $$V \ = \ 2 \pi \bigg(\frac{73}{42} \bigg)$$ $$V \ = \ \frac{73 \pi}{21}$$

Hopefully all of this helps you gain a bit of a better understanding of this method, but as always I’d love to hear your questions if you have any. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I’ll send you my bonus FREE calc 1 study guide! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about integrals as well.

## Rotating Volumes with the Cylinder/Shell Method

Similar to using the disk or washer method, we will use the cylinder method to find the volume of a solid. Specifically, it’s used when we rotate a function or region around an axis of rotation. In fact, most problems that require finding the volume of a solid of rotation can use the disk/washer method or the cylinder method. However, one will usually be significantly easier.

I’ll explain what I mean by this with an example.

## Example 1

Find the volume of the solid resulting from rotating the area bound between $y=x^2-2x+2$, $y=0$, $x=1$, and $x=2$ about the y-axis.

Although we are using a different method here we will follow the same 4 step process as I did with the disk method and washer method.

### 1. Graph the 2-D functions

This is generally a good idea with any type of problem: draw out whatever is being described in the problem. This helps to visualize whats going on in the problem and what exactly we are trying to measure. It can also help us decide if the answer we end up with is actually a reasonable one.

I would recommend trying to graph all of the functions listed by hand, but I’ll do this using Desmos. You can see the graph of the functions below with the bounded region shaded gray.

### 2. Rotate the 2-D area around the given axis

Again, we want to visualize the 3-D figure whose volume we are trying to find. To do this we want to imagine rotating the described area around the axis of rotation.

I like to imagine the area actually coming off the page and rotating around our axis of rotation, which is the y-axis in this case. Doing this would create a cylinder-like, round figure. Try sketching this rotation and the resulting figure. It may be helpful in the next step to have this sketch.

If needed, Wolfram Alpha can always be used to create a visualization of this figure.

### 3. Setting up the integral

This is the part where things start to get a bit different using the cylinder method than they were with the disk/washer method.

In order to make sense of the integral we need to set up here, let’s think about what we’re doing differently. With the disk/washer methods we were stacking many very thin disks on top of each other with the same thickness and varying radii to create our figure. This created a stack of cylinders whose volume we could find and add together.

The cylinder shell method is a bit different. Here we need to imagine just the outer shell of a cylinder that is very very very thin. We will stack many of these very thin shells inside of each other to create our figure. Each shell will have the same thickness, but all with different heights depending on where it is in the figure.

What we need to figure out is a formula for the volume of one of these shells, then the integral will be able to go through each shell and add up all of their volumes.

#### What would this look like?

Before we think about creating a function that we will need to integrate I want to take a moment to describe what one of these shells would look like. Imagine forming the outer shell of a cylinder with a piece of paper. All you would need to do is roll up the paper and it would create a cylinder shell. But what’s interesting about this is the cylinder shell came from a rectangle, or more accurately a very very thin rectangular prism. It might look something like this.

Since we are imagining finding the volume of an infinitely thin cylinder, these three figures would have the same volume. The fact that they’re infinitely thin means that the curvature won’t impact the volume of the shell. So to find the volume of the cylindrical shell, we can instead find the volume of a rectangular prism with the same dimensions. This is a much easier exercise to imagine since the volume of a rectangular prism is simply $$V= \ length \cdot width \cdot height.$$

#### But how does this relate to the cylinder?

What we need to think about now is how the dimensions of the rectangular prism translate to dimensions on the cylindrical shell. Looking back up to the drawing above, if you imagine the rectangle curling into a cylinder you can see the long side of the rectangle bends into the top and bottom of the cylinder. This would be the circumference of the cylinder.

In the above drawing we can also see that the shorter side of the rectangle lines up nicely with the height of the cylinder.

And lastly, the very very very thin thickness of the shell and the rectangular prism would clearly correspond with each other.

So as a result, the $$V= \ length \cdot width \cdot height$$ of the rectangular prism would be the same as $$V = circumference \cdot thickness \cdot height$$ for the cylindrical shell.

#### Putting it into an integral

Now that we know how to find the volume of each shell we need to come up with a way to put that into an integral. The reason for this is that the integral adds up the volume of all of the shells that make up the figure to find the total volume.

In order to do this, we will need to think about the formula for the volume in terms of measurments of the cylinders. We know the three pieces we need to find the volume of one of the shells are the circumference, thickness, and height of the cylinders. Typically when we describe a cylinder, we need two measurements to do this: height and radius. So we want to represent the circumference, thickness, and height in terms of height and radius.

First let’s think about the circumference. We know that the circumference of a circle is always going to be $$circumference=2 \pi r$$ where r is the radius.

The thickness of each shell is a bit strange. As we go from one shell to the next throughout the integral we want to think about what is changing. When we are doing this, we will always want to think about integrating throughout the radius. It’s as if we are starting at the center of the figure and integrating in the direction toward the edge of the figure. So when we go from one shell to the next we travel throughout the radius. Therefore, the thickness of each cylindrical shell with be the distance we travel between each shell. This will just be the change in radius between each shell. Therefore we’ll say the the thickness is $$thickness = dr.$$

Lastly, the height will still be described as the height. We don’t really need to do anything here.

Therefore, using these three conversions we know that the volume of the whole figure can be found with the following integral. $$\int 2 \pi r h \ dr$$

#### Relating it back to our figure

So we have a general outline to set up our integral, but now we need to figure out how our figure fits into these pieces. The easiest way to do this is draw it out with one of the cylindrical shells that makes up our 3-D figure.

So remember earlier I said that when we use this method to find the volume, we are integrating in the direction of the radius of the cylinders. We can see in our drawing that if we start in the center of our cylinder and move toward the edge, we would be going in the x direction. Therefore, we need to think about how to represent our integral in terms of x so we can integrate with respect to x.

So we have 3 pieces of our integral that we need to put in terms of x: r, h, and dr.

#### Finding r

You can see in the drawing above I drew an example of one of the cylindrical shells within our figure. There is also a smaller version of this shell in the upper left hand corner which has a few points labeled which will lie on the various functions that created our bounded region.

We can see the radius of our cylinder would be the distance between its center and edge, which is the distance between the two points labeled $(0, \ y)$ and $(x, \ y)$. The x-coordinate of that first point will always be 0 because that point lies on the y-axis. And $(x, y)$ is some point that lies on the function $y=x^2-2x+2$. Since these points have the same y value, the distance between them will just be the distance between their x values. So $$r=x-0$$ $$r=x.$$

#### Finding h

Looking at the labeled cylindrical shell in the drawing above, we can see that the height of the cylinder will be the distance between the points labeled $(x, \ y)$ and $(x, \ 0)$. Again, $(x, \ y)$ is some point on the function $y=x^2-2x+2$. And the y-coordinate of that second point will always be 0 because it sits on the x-axis.

It is also important to notice that these two points will sit on the top and bottom edges of every single shell that makes up this figure. This will be true for the inner most shell, the outer most shell, and every shell between them.

Since these these two points have the same x value, the distance between them will simply be the distance between their y values. So $$h=y-0$$ $$h=y$$

But remember we need everything in terms of x, not y. So we need to think about how to represent this height using x instead. Since our y is just the y-coordinate of some point that lies on the function $y=x^2-2x+2$, we can replace the y with $x^2-2x+2$. So $$h=x^2-2x+2.$$

#### Finding dr

This is actually the simplest part to find. The dr represents the change in the cylinder’s radius as we go from each shell to the next. Since we move in the same direction of the radius as we integrate to find our volume, the change in r should be the same as the change in x between each step. Therefore, we can say that $$dr=dx.$$

#### Putting it all back into an integral

We know that the volume of our figure can be found by using the integral $$\int 2 \pi r h \ dr.$$ And we just found how to represent all of these pieces in terms of x by $$r=x$$ $$h=x^2-2x+2$$ $$dr=dx.$$ So we can substitute all of these pieces into the integral and get something in terms of x that will tell us exactly how to find the volume of our figure. $$V=\int 2 \pi x \big( x^2-2x+2 \big) \ dx$$

But there is actually one more thing we need to consider. Our integral needs bounds. Since we are integrating with respect to x we need to figure out all the x values we want to consider when finding our volume.

To do this we just need to look at the original 2-D region we had bounded by all of our functions. Looking back at our drawings we can see that the entire region goes from $x=1$ to $x=2$. Therefore, these will be our bounds, telling us that $$V=\int_1^2 2 \pi x \big( x^2-2x+2 \big) \ dx.$$

### 4. Solve the integral

Now that we got our integral set up, all we need to go is evaluate the integral and find the volume it represents.

Before doing that I will simplify things a bit by pulling out the constant $2 \pi$ and then simplifying the function by distributing.

$$V=\int_1^2 2 \pi x \big( x^2-2x+2 \big) \ dx$$ $$V=2 \pi \int_1^2 x^3-2x^2+2x \ dx$$ $$V=2 \pi \Bigg[ \frac{1}{4}x^4 – \frac{2}{3}x^3+ x^2 \Bigg]_1^2$$ $$V=2 \pi \Bigg[ \bigg( \frac{1}{4}(2)^4 – \frac{2}{3}(2)^3+ (2)^2 \bigg) \ – \ \bigg( \frac{1}{4}(1)^4 – \frac{2}{3}(1)^3+ (1)^2 \bigg) \Bigg]$$ $$V=2 \pi \Bigg[ \bigg( 4 – \frac{16}{3}+ 4 \bigg) \ – \ \bigg( \frac{1}{4} – \frac{2}{3} + 1 \bigg) \Bigg]$$ $$V=2 \pi \bigg[ \frac{8}{3} – \frac{7}{12} \bigg]$$ $$V=2 \pi \bigg[ \frac{25}{12} \bigg]$$ $$V= \frac{25 \pi}{6}$$

And that’s it! The volume of our 3-D figure is $\frac{25 \pi}{6}$ cubic units.

I do quickly want to circle back to a comment I made a while ago. We could have found this volume using the washer method. However, we would have had to split it into two separate integrals to do so. The reason for this is that the inner radius of the washers on the lower half of the figure is formed by $x=1$. But the inner radius of the washers on the upper half of our figure is formed by $y=x^2-2x+2$. So we’d have to set up one integral for the lower half and another for the upper half, then add the resulting volumes to find the total volume of our figure.

Clearly using the cylindrical shell method is much easier in this case.

If you want more practice on finding volumes of rotation using the shell method, you can find another example here.

Hopefully all of this helps you gain a bit of a better understanding of this method, but as always I’d love to hear your questions if you have any. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I’ll send you my FREE calc 1 study guide! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about integrals as well.

## Integration by parts practice problems

In a previous lesson, I explained the integration by parts formula and how to use it. Sometimes though, finding an integral using integration by parts isn’t as simple as the problem I did in that lesson. So I’d like to show some other more complex cases and how to work through them.

$\mathbf{1. \ \int 4x^2 \ sin(5x) \ dx}$Solution

$\mathbf{2. \ \int xe^{-2x} \ dx}$Solution

$\mathbf{3. \ \int x^{\frac{3}{2}}ln(x) \ dx}$Solution

## Example 1

Evaluate the integral $$\int 4x^2 \ sin(5x) \ dx.$$

I will proceed through this problem following the same steps that I used in the integration by parts lesson.

### 1. Picking u and dv

Remember, we want to pick the piece of our function that we’d rather differentiate to be u, and the piece we’d rather integrate to be dv. The integral we need to evaluate can clearly be thought of as the product of $4x^2$ and $sin(5x)$. Therefore, we need to take the derivative of one of these and the anti-derivative of the other.

First consider the $4x^2$ piece. If we take the derivative of this function we’ll end up with an x term, but if we take the anti-derivative we’ll end up with an $x^3$ term. This is due to the power rule. Generally it’s best to choose whichever will result in the simplest option. When it comes to polynomials, the simpler one is whichever has the lower power. Therefore, it is preferred to take the derivative of the $4x^2$ piece because the x term resulting from the derivative is simpler than the $\mathbf{x^3}$ term resulting from taking the anti-derivative.

But before we say that we definitely want to assign the $4x^2$ to be u so that we can take its derivative, we want to think about what this will mean for the sin(5x) piece.

What we need to consider is the difference between taking the derivative and the anti-derivative of sin(5x). In either case, we will end up with some constant multiplied by cos(5x) (we can find this using the chain rule for the derivative or u-substitution for the anti-derivative). But you can see that the derivative and the anti-derivative of this piece are equally complex, so it doesn’t make much of a difference whether we say the sin(5x) piece is assigned to u or dv.

Since the derivative of $4x^2$ is much simpler than its anti-derivative, we would rather call it u than dv. And it doesn’t make a difference if sin(5x) is considered to be u or dv. So we will say $$u=4x^2$$ $$dv=sin(5x) \ dx.$$

### 2. Finding v and du

Now to find v we simply need to take the anti-derivative of the dv piece from the previous section. And to find du we need to take the derivative of u.

You can find the anti-derivative of sin(5x) by using u-substitution. I’m not going to show all the steps for this, but we will need to use the fact that the anti-derivative of sin(x) is -cos(x). Knowing this, we can find that the anti-derivative of sin(5x) would give us $$v=- \ \frac{1}{5} cos(5x).$$

Now that we have found v, let’s move onto finding du. This can simply be done by finding the derivative of u from part 1. Finding the derivative of $4x^2$ can simply be found using the power rule. Doing this gives us $$du=8x \ dx.$$

### 3. Plugging it all into the formula

Now that we have found all 4 of the pieces we need, we just have to plug them into the integration by parts formula. To summarize, the 4 pieces we have up to this point are $$u=4x^2$$ $$dv=sin(5x) \ dx$$ $$v=- \ \frac{1}{5} cos(5x)$$ $$du=8x \ dx.$$

So we just need to use the integration by parts formula with these. $$\int u \ dv = uv – \int v \ du$$ $$\int 4x^2 \ sin(5x) \ dx = \Big(4x^2\Big) \bigg(- \ \frac{1}{5} cos(5x) \bigg) – \int – \ \frac{1}{5} cos(5x) \ 8x \ dx$$

Now we can simplify and evaluate the integral on the right side of our equation. $$(1): \ \int 4x^2 \ sin(5x) \ dx = \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{5}\int x \ cos(5x) \ dx$$

But notice, the integral on the right side of our equation is still fairly complex. We still have an integral which is the product of two simpler functions, x and cos(5x). In order to evaluate this integral we’ll actually need to use integration by parts again. So now we need to use integration by parts to evaluate $$\int x \ cos(5x) \ dx.$$ We’ll go ahead and follow the same steps as we did before, but now we have a new integral and will need to reassign our u and dv.

### 4. Picking u and dv

Now that we are going through this process a second time, we don’t really have much of a choice when we pick which piece will be u and dv. The reason for this is that we will need to make this determination based on what we did the first time through. Consider where each of our pieces came from. One of our pieces is x, which came from the $4x^2$ in our original integral. And the other piece is cos(5x) which came from the sin(5x) in the original integral.

Since the x piece in our current integral came from the $4x^2$ piece in the original integral, and we decided that the $4x^2$ piece would be u earlier, we need to follow up by doing the same here. Therefore, we will say $$u=x$$ this time around.

By the same reasoning, we will need to say that $$dv = cos(5x) \ dx$$ this time around since we said assigned the sin(5x) to dv the first time through.

### 5. Finding v and du

Now we just need to take the u and dv from the previous step and use them to find v and du.

To find v we just need to find the anti-derivative of dv. We previously decided that $dv = cos(5x) \ dx$. Just like before, we can find this anti-derivative using u-substitution. Doing this tells us that $$v = \frac{1}{5} sin(5x).$$

And now we just need to find du by taking the derivative of u. Since we know $u = x$, we know that $$du=dx.$$

### 6. Plugging it all into the formula

And finally we just need to plug the 4 pieces we have found into the integration by parts formula. So far we have found $$u=x$$ $$dv = cos(5x) \ dx$$ $$v = \frac{1}{5} sin(5x)$$ $$du=dx.$$

Now putting these into the integration by parts formula we find $$\int u \ dv = uv – \int v \ du$$ $$\int x \ cos(5x) \ dx \ = \ x \bigg( \frac{1}{5} sin(5x) \bigg) \ – \int \frac{1}{5} sin(5x) \ dx$$ $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ – \ \frac{1}{5} \int sin(5x) \ dx$$

And now the integral we need to evaluate is much simpler than what we started with. $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ – \ \frac{1}{5} \bigg( – \frac{1}{5} cos(5x) \bigg)$$ $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ + \ \frac{1}{25} cos(5x)$$

### Wrapping it all together

Now that we have found $$\int x \ cos(5x) \ dx \ = \ \frac{1}{5} x \ sin(5x) \ + \ \frac{1}{25} cos(5x)$$ we can bring this back to our equation (1) back in step 3. And all we need to do is replace the $\int x \ cos(5x) \ dx$ with $\frac{1}{5} x \ sin(5x) \ + \ \frac{1}{25} cos(5x)$. Doing this tells us that $$\int 4x^2 \ sin(5x) \ dx = \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{5}\int x \ cos(5x) \ dx$$ $$= \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{5} \bigg( \frac{1}{5} x \ sin(5x) \ + \ \frac{1}{25} cos(5x) \bigg)$$ $$= \ – \ \frac{4}{5} x^2 \ cos(5x) \ + \ \frac{8}{25} x \ sin(5x) \ + \ \frac{8}{125} cos(5x)$$

And that’s our answer! Clearly a bit more complicated than the first integration by parts example I did, but it isn’t too bad. You essentially just need to apply the same process two time in a row. As long as you stay consistent in your designations of u and dv each time, it should all work out in the end.

## Example 2

$$\int xe^{-2x} \ dx$$

We’ll start this by deciding which piece we’ll call u and which piece is dv.

$$u=x$$ $$dv=e^{-2x} \ dx$$

Then we need to use these to figure out du and v.

$$du=1 \cdot dx=dx$$ $$v=-\frac{1}{2}e^{-2x}$$

Now we can plug all 4 of these pieces into the integration by parts formula.

$$\int xe^{-2x} \ dx \ = \ x \bigg( -\frac{1}{2}e^{-2x} \bigg) – \int -\frac{1}{2}e^{-2x} \ dx$$

At this point we are left with a simpler integral to evaluate.

$$= -\frac{1}{2}xe^{-2x} + \frac{1}{2} \int e^{-2x} \ dx$$ $$= -\frac{1}{2}xe^{-2x} + \frac{1}{2} \bigg( -\frac{1}{2} e^{-2x} \bigg)$$ $$= -\frac{1}{2}xe^{-2x} – \frac{1}{4} e^{-2x}$$ $$= -\frac{1}{2}e^{-2x} \bigg( x + \frac{1}{2} \bigg)$$

## Example 3

$$\int x^{\frac{3}{2}}ln(x) \ dx$$

We’ll start this by deciding which piece we’ll call u and which piece is dv.

$$u=ln(x)$$ $$dv=x^{\frac{3}{2}} \ dx$$

Then we need to use these to figure out du and v.

$$du=\frac{1}{x} \ dx$$ $$v=\frac{2}{5}x^{\frac{5}{2}}$$

Now we can plug all 4 of these pieces into the integration by parts formula.

$$\int x^{\frac{3}{2}}ln(x) \ dx \ = \ \frac{2}{5}x^{\frac{5}{2}} ln(x) \ – \int \frac{1}{x} \cdot \frac{2}{5}x^{\frac{5}{2}} \ dx$$

At this point we are left with a simpler integral to evaluate.

$$= \ \frac{2}{5}x^{\frac{5}{2}} ln(x) \ – \ \frac{2}{5} \int x^{\frac{3}{2}} \ dx$$ $$= \ \frac{2}{5}x^{\frac{5}{2}} ln(x) \ – \ \frac{2}{5} \cdot \frac{2}{5} x^{\frac{5}{2}}$$ $$= \ \frac{2}{5}x^{\frac{5}{2}} \bigg( ln(x) \ – \ \frac{2}{5} \bigg)$$

As always, let me know if you have any questions. If anything was confusing here leave a comment or send me an email at jakesmathlessons@gmail.com and I’ll get back to you with an answer. You can also use the contact form below to reach out and I’ll send you my FREE calculus 1 study guide as a bonus! Also check out my other lessons about integrals!

## Washer Method Practice Problem

Once you have the disk method down, the next step would be to find the volume of a solid using the washer method. The washer method for finding the volume of a solid is very similar to the disk method with one small added complexity.

You can think of the main difference between these two methods being that the washer method deals with a solid with a piece of it taken out. Exactly as you would expect from the name, a washer is just a disk with a hole taken out of its center. So let’s jump into an example and I’ll explain the difference as we go.

## Example

Find the volume obtained by rotating the area bounded by $y=x$ and $y= \sqrt{x}$ about the line $y=1$.

## Solution

Like I said, finding the volume of this solid is going to be very similar to finding the volume of a solid using the disk method. Therefore, I’m going to follow the same general process as I did when using the disk method.

### 1. Graph the 2-D functions

Just like before, I’ll do this using Wolfram Alpha. Below you can see the graph of $y=x$ and $y=\sqrt{x}$ and you can see the area that is bounded by these two functions.

### 2. Rotate the 2-D area around the given axis

Now that we have graphed our functions in the 2-D space, we will need to rotate them about our axis of rotation to create a 3-D figure. This is very similar to rotating a single function when using the disk method. The only difference is that the resulting figure will be round figure with something missing out of its center.

Remember, we need to rotate this area between the two functions around the line $y=1$, which is above our functions this time. Again, with Wolfram Alpha, we can see what this figure would look like.

### 3. Setting up the integral

This is where things get a little different as a result of using the washer method. We still want to think about taking infinitely thin cylinders from this figure and measuring their volumes.

But let’s think about what these infinitely thin cylinders look like.

Since our figure is a cone-like shape with a piece missing from the middle, our cylinders are also going to have a piece missing from the middle. This makes them look like washers, which is why we call it the washer method! Since the slices of our figure look like washers we can’t find their volume by just finding the volume of a cylinder. There will be something a little extra.

#### What’s different about a washer?

Let’s take a look at the washer below and consider how we might find its volume.

Clearly this figure takes on the shape of a cylinder. If we knew its radius and its height, we could use the formula for the volume of a cylinder to find its volume. However, there is a piece missing.

If we think about the piece that is missing from the cylinder for a moment you can see that this hole also has a cylindrical shape. In other words, we are starting with a cylinder and taking out a smaller cylinder from the middle of it.

So let’s say the distance from the center of these cylinders to the outer edge of this figure is the large radius, R. And we will say that the distance from the center of this figure to the inner edge is the small radius, r. We can think of r as the radius of the small cylinder that was taken out of the large cylinder.

Clearly both of these cylinders would have the same height, which we will call h, because this washer has the same thickness everywhere. So we can label all of these, giving us something like this.

Then we could use the volume of a cylinder formula to say that the volume of the large cylinder before the middle is taken out is $$\pi R^2 h.$$ And we can also say that the volume of the middle piece that’s taken out is $$\pi r^2 h.$$ So if we started with the volume of the large cylinder and take away the volume of the small cylinder, that means the volume of the washer is $$V= \pi R^2 h \ – \ \pi r^2 h.$$

Then we can simplify this a bit by factoring out like terms to get $$V = \pi h \big( R^2 – r^2 \big)$$

#### But what does this have to do with an integral?

Just like we did when we used the disk method, we will need to add up the volume of all of the washers. This is what the integral accomplishes. In order to write our volume as an integral, we first need to come up with the function we will integrate.

Consider the following drawing of the area described in this problem rotated around $y=1$ with one of the washer slices depicted.

The green washer in the figure represents one of the infinitely thin washers that we are slicing the figure into. Since we are trying to make this integral represent the sum of all of these washers, we need to think about the volume of each washer in particular.

You can imagine if we were to look at a different washer to the left or right of this one, the inner and outer radius would be different. As we change the x value where the washer sits, both radii would need to change. The inner radius would always reach to the part of this figure resulting from rotating $y=\sqrt{x}$ around $y=1$. And the outer radius would always reach out to the part of this figure resulting from rotating $y=x$ around $y=1$.

We already figured out that the volume of a washer would be $$V = \pi h \big( R^2 – r^2 \big).$$ But now we need to apply this to this specific washer in the drawing above.

Let’s look up at the green washer in the above drawing. In order to find its volume, we will need its inner radius, outer radius, and height (or thickness).

#### Finding the inner radius

The inner radius of this washer would be the distance from the center of the washer to the inner edge. In the drawing above, we can see that this would be the distance between the axis of rotation (the green line labeled $y=1$) and the the function which is closer to this axis (the red function labeled $y=\sqrt{x}$).

Off to the side of these functions you can see a closer look at this washer with three important points labeled. In this larger version the inner radius is the distance between the point $(x, \ 1)$ and $(x, \ y_1)$. We know that the center of our disk will always have a y-coordinate of 1 because we rotated our function around the line $\mathbf{y=1}$. And we will leave the x-coordinate as the variable x because we are trying to find the volume of any washer along this figure with all different x values, not just the one disk drawn above.

Let’s take a second to think about the other point I mentioned. This point is labeled $(x, \ y_1)$. This is just meant to be any (x, y) combination that sits on our function $\mathbf{y=\sqrt{x}}$. This point will always have the same x-value as our other labeled point, so the distance between these two points will simply be the distance between their y-coordinates, which are $y_1$ and 1. To find the distance between these two values, we just need to do the larger value minus the smaller one. Looking at the drawing, you can see that $y=1$ is above $y=\sqrt{x}$ over the entire domain of the area we care about. Therefore, the larger y value minus the smaller one, is $$1-y_1.$$

But we need this to be in terms of x. Remember we realized earlier that the inner radius of our washer will depend on x, so we want everything in terms of x. We know that our point $(x, \ y_1)$ is some point that lies on the function $y=\sqrt{x}$. So we know $\mathbf{y_1=\sqrt{x}}$. Therefore, if the inner radius of our washer is $1-y_1$, we can also say that the radius is $1-\sqrt{x}$. So the inner radius of this washer will be $$r=1-\sqrt{x}.$$

We could plug in any x value into this to find the inner radius of the washer that corresponds with that specific x value. So when we integrate across a range of x values, we will be taking into account the inner radius of the washers with all of the x values in that range. But let’s not get ahead of ourselves. We have a bit more work to do before we can do that.

#### Finding the outer radius

We will find the outer radius very similarly to how we found the inner radius. The outer radius of this washer would be the distance from the center of the washer to the outer edge. In the drawing above, we can see that this would be the distance between the axis of rotation (the green line labeled $y=1$) and the the function which is farther from this axis (the blue function labeled $y=x$).

Again, let’s take a look at the washer drawn off to the side of our figure. This time we are looking for the outer radius. The outer radius would be shown here as the distance between the points $(x, \ 1)$ and $(x, \ y_2)$.

Since these two points have the same x value, the distance between them will be the same as the distance between their y values. To find this we just need to take the larger one and subtract the smaller of the two. The point $(x, \ 1)$ will always lie on $y=1$ and the point $(x, \ y_2)$ will always lie on $y=x$. Since $y=1$ is above $y=x$ between $x=0$ and $x=1$, we know that 1 will be larger than $y_2$. Therefore, the distance between these two points is $$1-y_2.$$

But again, we need this radius to be written in terms of x. We know that our point $(x, \ y_2)$ is some point that lies on the function $y=x$. So we know $\mathbf{y_2=x}$. Therefore, if the outer radius of our washer is $1-y_2$, we can also say that the radius is $1-x$. So the outer radius of this washer will be $$R=1-x.$$

#### Finding the height (or thickness)

The height of our infinitely thin cylinders or washers is actually quite simple. Just like when we integrate a 2-D function to find the area under the curve, our slices here are all the same width. We don’t have to worry about each washer, or cylinder, having a different height.

The height of each cylinder will just be how far we always move over before taking another slice. Since we are moving over in the x direction as we imagine the next slice, this can simply be our change in x between the slices. Change in x is always represented as dx. So we can simply say the height of each cylinder is $$h=dx.$$

#### Back to the integral

Like I said before, all the integral will do is go through all the x values in our domain and add up the volumes of all of the infinitely thin washers. In order for it to achieve this, we need to put a function for the volume of each washer that depends on x. We already know that the volume of a washer in general would be $$V = \pi h \big( R^2 – r^2 \big).$$

This means that our integral might look something like this $$\int \pi h \big( R^2 – r^2 \big).$$

But this doesn’t really have any meaning on its own. In order to give this meaning we need to represent this volume in terms of x and give the integral a domain of x values to integrate over.

Remember we also found the inner radius, outer radius, and height of the washers that make up our figure to be $$r=1-\sqrt{x},$$ $$R=1-x,$$ $$h=dx.$$

Putting all these into our integral, along with the fact that our figure takes on all x values between $x=0$ and $x=1$ tells us that the volume of this 3-D figure is $$V = \int_0^1 \pi (dx) \Big( \big( 1-x \big)^2 – \big( 1-\sqrt{x} \big)^2 \Big).$$

But we should rewrite this in a form that is more in line with how integrals are usually formatted. And we can also pull our constant $\pi$ outside of the integral. $$V = \pi \int_0^1 \big( 1-x \big)^2 – \big( 1-\sqrt{x} \big)^2 \ dx$$

### 4. Solve the integral

We made it through the hard part! Now that we created our integral to represent the volume, we just need to evaluate the integral. Before we integrate this, let’s start with simplifying it.

In order to simplify this function that we need to integrate, the first step would be to F.O.I.L. out each portion. Remember $(1-x)^2$ is NOT $(1^2-x^2)$. Instead we need to treat $(1-x)^2$ as if it were $(1-x)(1-x)$.

$$V = \pi \int_0^1 \big( 1-x \big)^2 – \big( 1-\sqrt{x} \big)^2 \ dx$$ $$V = \pi \int_0^1 (1-x)(1-x) – \big(1-\sqrt{x})(1-\sqrt{x} \big) \ dx$$ $$V = \pi \int_0^1 \big(1-2x+x^2 \big) – \big(1-2\sqrt{x}+x \big) \ dx$$ $$V = \pi \int_0^1 1-2x+x^2 – 1+2\sqrt{x}-x \ dx$$ $$V = \pi \int_0^1 x^2 -3x +2\sqrt{x} \ dx$$ $$V = \pi \int_0^1 x^2 -3x +2x^{\frac{1}{2}} \ dx$$

Now we have a function that is fairly simple to integrate. All we need to do to integrate this is use the power rule.

$$V = \pi \Bigg[ \frac{1}{3}x^3 – \frac{3}{2}x^2 + \frac{4}{3}x^{\frac{3}{2}} \Bigg]_0^1$$ $$V = \pi \Bigg[ \bigg( \frac{1}{3}1^3 – \frac{3}{2}1^2 + \frac{4}{3}1^{\frac{3}{2}}\bigg) – \bigg( \frac{1}{3}0^3 – \frac{3}{2}0^2 + \frac{4}{3}0^{\frac{3}{2}} \bigg) \Bigg]$$ $$V = \pi \bigg( \frac{1}{3} – \frac{3}{2} + \frac{4}{3} \bigg)$$ $$V = \pi \bigg( \frac{2}{6} – \frac{9}{6} + \frac{8}{6} \bigg)$$ $$V = \frac{1}{6} \pi = \frac{\pi}{6}$$

So we know the volume of this solid is $\frac{\pi}{6}$! Hopefully this has helped you with the washer method, but if there’s still a topic you’d like to learn about take a look at some of my other lessons and problem solutions about integrals. You can also get some more practice with the washer method here. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

I also encourage you to join my email list! Just enter your name and email below and I’ll send you my calc 1 study guide as a FREE welcome gift!

## Rotating Volumes with the Disk Method

Rotating functions around an axis to create a 3-D shape then finding its volume is one of the more common applications of integrals. This is commonly referred to as finding a volume using the disk method. It seems like a complicated type of problem, but if you think about what you are actually measuring it isn’t so bad.

Let’s think about a specific example. Imagine taking the function $y=x^2$ between $x=0$ and $x=2$ and rotating it around the x-axis then finding the volume of this solid using the disk method.

## 1. Graph the 2-D function

The first thing I would recommend doing with a problem like this is to graph the function that’s given to you. Here we are graphing $y=x^2$ within our given domain of $0 \leq x \leq 2$. Using Wolfram Alpha we can see this graph below.

## 2. Rotate the 2-D function around the given axis

Once you graph the function on the 2-D x-y-plane we need to imagine rotating it around the axis given in the problem. This will result in creating a 3-D figure whose volume we need to find.

In this case we need to rotate this portion of our function around the x-axis. Another way to say this is that we are rotating around the line $y=0$. Again, using Wolfram Alpha we can see what this figure would look like.

I would always recommend drawing out the 2-D graph and 3-D rotation anytime you need to find the volume of a solid like this. It helps to visualize the solid whose volume you are trying to measure and it makes it much easier to make sure you are setting up your problem correctly.

## 3. Setting up the integral

As with any problem where we need to find a volume using the disk method, what we want to imagine here is having infinitely thin cylinders stacking up to create our 3-D figure. When we add up the volumes of all of these infinitely thin cylinders, get the volume of the entire figure. This is all the integral is doing.

You can see in this drawing our function has been rotated around the x-axis to create a round cone-like 3-D figure. The green cylinder in the figure represents one of the infinitely thin disks that we are slicing the figure into.

Since we are trying to make this integral represent the sum of all of these disks, we need to think about the volume of each disk in particular. Clearly each disk is a very thin cylinder. So in order to find their volumes, we should start with the volume of a cylinder, which is

$$V=\pi r^2 h.$$

Thinking back to our example of rotating $y=x^2$ around the x-axis, let’s determine the radius and height of our cylinders.

### How do we find the radius?

Considering that each cylinder will be a different size, it seems clear that the radius of each cylinder will depend on which cylinder we’re considering. In fact, if you look at our drawing, you can see that the radius of each cylinder will simply depend on the x value where it’s sitting.

You can see in the drawing above that I drew a copy of the disk we are considering down below the function. Imagine we are trying to find the distance between the two points we labeled. The center point of our disk is labeled (x, 0). We know that the center of our disk will always have a y-coordinate of 0 because we rotated our function around the line $\mathbf{y=0}$. And we will leave the x-coordinate as the variable x because we are trying to find the volume of any disk along this figure with all different x values, not just the one disk drawn above.

Now consider the upper point. This point is labeled (x, y). This is just meant to be any (x, y) combination that sits on our function $y=x^2$. This point will always have the same x-value as our other labeled point, so the distance between these two points will simply be the distance between their y-coordinates, which are y and 0. To find the distance between these two values, we just need to do the larger value minus the smaller one, or $y-0$, which is just y.

But we need this to be in terms of x. Remember we realized earlier that the radius of our disk will depend on x, so we want everything in terms of x. We know that our point (x, y) we were looking at is some point that lies on our function. So we know $\mathbf{y=x^2}$. Therefore, if the radius of our disk is y, we can also say that the radius is $x^2$. So,

$$r=x^2.$$

### How do we find the height?

The height of our infinitely thin cylinders is actually quite simple. Just like when we integrate a 2-D function to find the area under the curve, our slices here are all the same width. We don’t have to worry about each disk, or cylinder, having a different height.

The height of each cylinder will just be how far we always move over before taking another slice. Since we are moving over in the x direction as we imagine the next slice, this can simply be our change in x between the slices. Change in x is always represented as dx. So we can simply say the height of each cylinder is

$$h=dx.$$

If we were instead rotating our function around a vertical line (like $x=0$) the height of our disks would be dy. This is only if we are using the disk method and would not necessarily be the case when using the shell method.

### Back to the integral

Now we got our height and radius of each disk we get from slicing this figure. Putting this together, we can say that the volume of a single disk can be represented as

$$V=\pi \big( x^2 \big)^2dx.$$

But this of course is just one disk. We need to add up the volume of all of the disks to get the volume of the full figure. This is exactly what the integral accomplishes. Before doing this, remember we are only rotating this function between $x=0$ and $x=2$. So these will be the bounds of our integral. Therefore, the volume of our entire figure can be found with the following integral.

$$\int_0^2 \pi \big( x^2 \big)^2dx$$

## 4. Solve the integral

We made it through the hard part! Now all we need to do is solve this integral and we will have the volume of our figure. First let’s simplify this integral a little, then we can integrate using the power rule and evaluate at the given bounds.

$$\int_0^2 \pi \big( x^2 \big)^2dx$$

$$\pi \int_0^2 x^4 dx$$

$$\pi \Bigg[ \frac{1}{5} x^5 \Bigg]^2_0$$

$$\frac{\pi}{5} \Big[ (2)^5 – (0)^5 \Big]$$

$$\frac{\pi}{5} \Big[ 32 \Big]$$

$$\frac{32 \pi}{5}$$

So we found that the volume of the solid is $\frac{32 \pi}{5}$! Hopefully this has helped you with the disk method, but if there’s still a topic you’d like to learn about take a look at some of my other lessons and problem solutions about integrals. Once you know and understand the disk method, another good application of integrals to check out would be the washer method. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

I also encourage you to join my email list! Just enter your name and email below and I’ll send you my free calculus 1 study guide as a bonus for joining me!

## Integration by parts

Integration by parts is another common technique used to find complex antiderivatives. This method tends to be a little more straight forward in its application than u-substitution. The main reason for this is that it requires the use of a formula, and if you can follow the formula you should be able to work through the rest.

First let’s introduce the formula, then I’ll explain how to use it. If you already know how to do these and you’re looking for extra practice problems, click here.

$$\int u \ dv = uv- \int v \ du$$

All this formula is really saying is that if we need to integrate some function which can be thought of as the product of two pieces, u and dv, then we can rewrite our integral in this other form. Notice we still would have an integral to solve after using this formula. But the hope is that $\int v \ du$ is easier to find than $\int u \ dv$.

## But how do you use the formula?

Using the integration by parts formula can be broken down into 3 simple steps and is going to start out somewhat similarly to integrating with u-substitution.

### 1. Picking u and dv

The first thing we need to do to use this formula is decide which piece of our function will be called u and which piece will be called dv. As we work through this problem, we will eventually need to work with the derivative of u and the antiderivative of dv. Therefore, to decide which piece we want to be u and dv, we should also consider the derivative and antiderivative of the pieces.

Let’s consider the following integral which we will find using integration by parts.

$$\int xsin(10x) \ dx$$

Clearly we can see that we are being asked to integrate some function which is the product of two smaller functions. It is the product of x and sin(10x). Therefore, between x and sin(10x), we will need to call one of these u and the other will be dv.

#### Does it matter which is which?

Yes, it does matter. You will need to take the derivative of u and the antiderivative of dv. So you want to pick one to be u and the other to be dv so that the derivative of u and the antiderivative of dv are easiest to work with.

Consider this: the sin(10x) term can be either u or dv. The reason for this is that whether you take the derivative or the antiderivative of sin(10x), the result will be some constant multiplied by cos(10x). As a result, it doesn’t make much of a difference whether we call sin(10x) the u or the dv.

Let’s think about the x term. If we call it u and have to work with its derivative, we’ll make things pretty easy on ourselves. I say this because the derivative of x is just 1. Alternatively, if we make x be dv and take it’s antiderivative, we will need to work with an $\mathbf{x^2}$ term (due to the power rule). Therefore, it will be a lot easier to work with the derivative of x than it will be to work with it’s antiderivative. This tells us that it’ll be easiest to call x the u piece.

Since it doesn’t matter what we call the sin(10x) term, but it’ll be a lot easier to make x be the u piece, we will say

$$u=x$$

$$dv=sin(10x) \ dx.$$

### 2. Finding v and du

Now that we have determined our u and dv, we need to use these to calculate v and du. To find du we just need to take the derivative of u.

$$\frac{du}{dx} = \frac{d}{dx} \big[ x \big]$$

$$\frac{du}{dx} = 1$$

Now we can just imagine multiplying both sides by dx to find

$$du=dx.$$

And to find v we just need to take the antiderivative, or the integral, of dvYou can do this using u-substitution with $u = 10x$, but I will use WolframAlpha.

$$v = \int sin(10x) \ dx$$

$$v = – \frac{1}{10} cos(10x)$$

### 3. Plugging it all into the formula

Once you have laid out all four of the pieces we need, we can plug them all into the integration by parts formula. Just so we have everything in one place, let’s list out everything we have up to this point.

$$u=x$$

$$du=dx$$

$$v= – \frac{1}{10} cos(10x)$$

$$dv= sin(10x) \ dx$$

Now going back to the integration by parts formula I mentioned earlier, we can plug all of these in to the formula.

$$\int u \ dv = uv- \int v \ du$$

$$\int xsin(10x) \ dx = (x)\bigg( – \frac{1}{10} cos(10x) \bigg) – \int \bigg(- \frac{1}{10} cos(10x) \bigg) \ dx$$

Before integrating, let’s simplify this as much as we can by pulling the constant out of the integral.

$$\int xsin(10x) \ dx = \ – \frac{1}{10} xcos(10x) + \frac{1}{10} \int cos(10x) \ dx$$

Notice, the integral we need to compute now is much simpler than the integral we started with. This will be similar to the integral we computed to find v earlier. We can use u-substitution to find this by using $u=10x$. I’m not going to show these steps, but I encourage you to work this out on your own!

$$\int xsin(10x) \ dx = \ – \frac{1}{10} xcos(10x) + \frac{1}{10} \bigg( \frac{1}{10} sin(10x) \bigg) + c$$

$$\int xsin(10x) \ dx = \ – \frac{1}{10} xcos(10x) + \frac{1}{100} sin(10x) + c$$

### Some additional comments

And that’s it! With some, more complex, integration by parts problems you may have to apply this formula more than one time. Once you get to step 3, you might find that the simpler integral is still somewhat complicated and requires the use of integration by parts again.

In these cases, you can simply treat this integral like a sub-problem to find our main integral and go through these 3 steps with that integral. You can see an example of this here.

Overall, integration by parts isn’t terribly complicated once you know the formula and understand how to apply it. Take a look at some of my other lessons about integrals for some extra practice. And don’t forget, email me at jakesmathlessons@gmail.com if you can’t find the lesson or problem you’re looking for!

## Extra Examples

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