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### Integration by Parts

Main formula:

Hints and tips:

• Integration by parts never solves the integral completely for you. The point is to reduce a difficult integral to an easier one.

• If you have a polynomial multiplied by an exponential or trigonometric function, you can use the tabular integration shortcut.

• You may have to use parts more than once in the same problem.

• Sometimes, after using parts one or more times, you get the original integral reappearing. Then you can set up an equation for the original integral and find the solution algebraically without doing any more integration.

• Use the mnemonic LIATE (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) to know which part of the expression to make the u.

• Sometimes you have to make one or more substitution before you get a form that works well for integration by parts.

### Integration by Parts

2xcosxdx
• Let u = 2x and dv = cosx
• du = 2dx
• dv = cosxdx
• v = sinx + C
Apply Integration by Parts
2xcosxdx = 2x(sinx) − ∫ sinx(2dx) + C = 2xsinx − 2∫ sinxdx + C = 2xsinx + 2cosx + C
xsin3xdx
• Let u = x and dv = sin3xdx
• u = x
• du = dx
• dv = sin3xdx
• v = [( − cos3x)/3] + C
• Apply Integration by Parts
• xsin3xdx = [( − cos3x)/3]( x ) − ∫ [( − cos3x)/3]dx + C
• = [( − xcos3x)/3] + [1/3]∫ cos3xdx + C
• = [( − xcos3x)/3] + [1/3]( [sin3x/3] ) + C
• = [( − xcos3x)/3] + [sin3x/9] + C
= [( − 3xcos3x + sin3x)/9] + C
lnx2dx
• Use logarithm properties to simplify
• lnx2dx = ∫ 2lnxdx
• = 2∫ lnxdx
• Let u = lnx and dv = dx
• du = [dx/x]
• dv = dx
• v = x + C
• Apply Integration by Parts
• lnx2dx = 2∫ lnxdx
= 2( xlnx − ∫ dx + C ) = 2( xlnx − x + C ) = 2xlnx − 2x + C
ln(x + 1)dx
• Let u = x + 1 and dv = dx
• u = ln(x + 1)
• du = [1/(x + 1)]dx
• dv = dx
• v = x + C
• Apply Integration by Parts
• ln(x + 1)dx = x( ln(x + 1) ) − ∫ x( [dx/(3x + 1)] ) + C
• = x( ln(x + 1) ) − ∫ [x/(x + 1)] dx + C
• Note the alternative form of [x/(x + 1)]
• [x/(x + 1)] = 1 − [1/(x + 1)]
• x( ln(x + 1) ) − ∫ [x/(x + 1)] dx = x( ln(x + 1) ) − ∫ 1 − [1/(x + 1)] dx + C
• = xln(x + 1) − ∫ dx + ∫ [1/(x + 1)] dx + C
= xln(x + 1) − x + ln(x + 1) + C
3x2 exdx
• Let u = x2 and dv = exdx
• du = 2x dx
• dv = exdx
• v = ex + C
• Apply Integration by Parts
• 3x2 exdx = 3∫ x2 exdx
• = 3( exx2 − ∫ ex( 2x dx ) )
• Apply Integration by Parts again on ∫ ex( 2x dx )
• ex( 2x dx ) = ex(2x) + ∫ ex(2dx)
• = 2xex + 2ex + C
Apply substitution
3( exx2 − ∫ ex( 2x dx ) ) = 3( exx2 − 2xex + 2ex + C )
0π (x − 1)sinxdx
• Let u = x − 1 and dv = sinxdx
• du = dx
• dv = sinxdx
• v = − cosx
• Apply Integration by Parts
• 0π (x − 1)sinxdx = − cosx(x − 1) − ∫0π − cosx(dx)
• = (cosx − xcosx + sinx)0π
• = cosπ− πcos π+ sinπ− (cos0 − (0)cos0 + sin0)
• = − 1 + π+ 0 − (1)
= π− 2
010 x exdx
• Let u = x and dv = exdx
• du = dx
• dv = exdx
• v = ex
• Apply Integration by Parts
• 010 x exdx = [ ex(x) ]010 + ∫010 exdx
= e10(10) + e10 − (e0(0) + e0) = e10(10) + e10 − 1 = 11e10 − 1
0e [x/(x + 1)]dx
• Let u = x and dv = [dx/(x + 1)]
• du = dx
• dv = [1/(x + 1)]dx
• v = ln(x + 1)
• Apply Integration by Parts
• 0e [x/(x + 1)]dx = xln(x + 1) − ∫0e ln(x + 1) dx
• From problem 4, we have solved ∫ ln(x + 1) dx
• ln(x + 1) dx = xln(x + 1) − x + ln(x + 1) + C
• xln(x + 1) − ∫0e ln(x + 1) dx = xln(x + 1) − (xln(x + 1) − x + ln(x + 1))
• = [ xln(x + 1) − xln(x + 1) + x − ln(x + 1) ]0e
• = eln(e + 1) − eln(e + 1) + e − ln(e + 1) − ((0)ln(0 + 1) − (0)ln(0 + 1) + 0 − ln(0 + 1)
• = e − ln(e + 1) + ln1
= e − ln(e + 1)
23 ln(3x − 5)3dx
• Use logarithm properties to change form
• 23 ln(3x − 5)3dx = ∫23 3ln(3x − 5)dx
• = ∫02 3n(3x − 5)3dx
• Let u = ln(3x − 5) and dv = 3dx
• du = [3/(3x − 5)]dx
• dv = 3dx
• v = 3x
• Apply Integration by Parts
• 23 ln(3x − 5)dx = xln(3x − 5) − ∫23 [3x/(3x − 5)]dx
• Find the alternative form of [3x/(3x − 5)]
• [3x/(3x − 5)] = [(3x − 5)/(3x − 5)] + [5/(3x − 5)] = 1 + [5/(3x − 5)]
• xln(3x − 5) − ∫02 [3x/(3x − 5)]dx = xln(3x − 5) − ∫23 1 + [5/(3x − 5)] dx
= 3ln4 − 3 − [5/3]ln4 − (2ln1 − 2 − [5/3]ln1) = 3ln4 − [5/3]ln4 − 2ln1 + [5/3]ln1 − 1 = [4/3]ln4 − 1
Prove ∫ sinxcosx = [(sin2x)/2] + C
• Let u = sinx and dv = cosx
• du = cosx
• dv = cosx
• v = sinx
• Apply Integration by Parts
• sinxcosx = sinx(sinx) − ∫ sinxcosx + C
• = sin2x − ∫ sinxcosx + C
• Isolate sin2x
• 2∫ sinxcosx = sin2x + C
sinxcosx = [(sin2x)/2] + C

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Integration by Parts

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Important Equation 0:07
• Where It Comes From (Product Rule)
• Why Use It?
• Lecture Example 1 1:24
• Lecture Example 2 3:30
• Shortcut: Tabular Integration 7:34
• Example
• Lecture Example 3 10:00
• Mnemonic: LIATE 14:44
• Ln, Inverse, Algebra, Trigonometry, e

### Transcription: Integration by Parts

Ok, we are going to work out some examples on integration by parts.0000

The first example is the integral of arctan(x) dx.0005

Again, the difficult part of integration by parts is deciding what to make u and what to make dv. 0011

In this case, arctan is an inverse trigonometric function.0018

So, we are going to make that the u arctan(x) and our dv is going to be the dx.0024

Then we have to fill in du and v.0038

du is something that hopefully you remember from your Calculus 1 class.0043

The derivative of arctan(x) is 1/x2+1 dx.0049

V is just the integral of dx which is x.0056

Remember your main integration by parts formula, uv - the integral of vdu.0060

This integral converts into u × v is xarctan(x) - integral of vdu.0068

So that is x/x2+1 dx.0079

Now we got this other interval that does not look at all like the first one.0085

Fortunately this one can be solved pretty easily.0090

We do not need to use integration by parts at all.0093

We can do this with a quick substitution.0095

Let us let u = x2+1, then our du will be 2x dx.0098

The point of that is that we practically have du already.0108

We have x dx, so we just need to correct for the fact that x dx is actually 1/2 du.0112

This is x arctan(x) - the integral of 1/2 du all over u.0120

That is x arctan(x) - 1/2, and now the integral of 1/u is just ln(abs(u)).0136

Finally we get x arctan(x) - 1/2, ln we will substitute back, u was x2+1 and as always we attach a constant at the end.0147

We do not have any limit values to plug in.0168

The only difficult part there was knowing how to get started.0173

The advice on that is to remember our order of functions.0180

First natural log, then inverse trigonometry.0184

We check these one by one and realize, oh, I have an inverse trigonometric function,0186

So, I am going to make the inverse trigonometric function be...0193

We have one more example for integration by parts.0000

We are going to try to solve the integral of sin(sqrt(x)) dx.0005

This is one that does not lend itself to integration by parts immediately.0008

What we will do is a little substitution.0013

Normally I use u for substitution but since I know we will be using integration by parts later on,0016

I am going to use a different variable here.0023

I will let w = sqrt(x).0031

Whenever you make a substitution you also have to figure out the derivative of the variable.0032

So dw is, if you think of sqrt(x) as x1/2,0035

Then dw is 1/2 x-1/2 dx, which is 1/2 sqrt(x) dx.0041

Which is 1/2w dx.0059

So, that tells you that dx is 2w dw.0065

We are going to make that substitution in here.0072

Now we have the integral of the sin(sqrt(x)) which is converted to w.0075

The dx is converted to 2w dw.0080

A lot of people forget to change the dx when they make a substitution.0086

That is a really important step, when you make the substitution.0093

Now I will pull the 2 outside, and pull the w to the other side and get w sin(w) dw.0099

Now this is kind of a standard integration by parts problem.0112

We have w sin(w).0116

In fact this was done as an example in my first lecture on integration by parts.0119

So, I will not redo it the same way.0125

Instead I will do the tabular integration method.0126

You will see another example of how to do the tabular integration to do something like this quickly.0132

So, I will set up w sin(w).0137

Remember you do derivatives on the left, so the derivative of w is 1, and the derivative of 1 = 0.0143

The integral of sin(w) is -cos(w), and the derivative of -cos(w) is -sin(w).0151

Then we make these diagonal lines with positive and negative signs on the lines, so plus, minus, plus.0160

Then, the answer here is what you get when you multiply along these lines.0172

So, -w cos(w).0178

The second diagonal line has 2 negatives cancelling each other.0183

You get + sin(w), and the third diagonal line has a 0.0188

That just multiplies away to nothing.0195

Now I will substitute back.0198

I get 2 × sqrt(x) cos(sqrt(x) + sin(sqrt(x)).0202

As always, we have to add a constant at the end.0214

So, the trick there was making a little substitution at the beginning.0224

Once we saw the sqrt(x), it looks kind of unpleasant to deal with.0230

So we make this substitution at the beginning that w = sqrt(x).0235

That allows us to convert the integral into something that is very amenable to integration by parts.0240

That is the end of our lecture on integration by parts.0248

Hi there, welcome to educator.com. This is a lecture on integration by parts.0000

The main equation for integration by parts is right here. 0005

The integral of U dV is equal to UV minus the integral of V dU. 0011

Where this comes from is the product rule in reverse. 0016

The product rule is something you learned in Calculus 1, and is a way to take derivatives of products of functions.0025

This is changing around the product rule and using it as an integration formula.0031

The point of integration by parts is that you will be given a hard integral to solve. 0036

What you are going to do, is take the integral that you are given and split it up into two parts, a U part and a dV part. 0044

Then you will invoke this formula to convert it into UV minus the integral of VdU, and if you do that right, then the second integral that you get will be an easier integral. 0054

Then, you can finish the problem by doing that easier integral. 0064

That is the idea of integration of parts, but of course the best way to learn it is to do lots of examples. 0075

Let us go ahead and do some examples.0082

Here is the first example, a very typical integration by parts problem. 0084

We are trying to integrate X sin(X) dX.0086

Remember, the first part is to split this integral up into U and dV and we are going to let U be just X and dV be sin(X) dX. 0091

You always put the dX with the dV part. 0108

Then, we are going to figure out dU and V because those are both parts of the formula before. 0110

dU, if U is X, dU is just dX and V if dV is sin(x), V is the integral of sin(x).0116

The integral of sin(X) is negative cos(X), and remember the integration by parts formula with the integral of U dV is equal to uV minus the integral of VdU. 0126

Now, the integral that we are given, because we have converted it using our substitution, that is now the integral of U dV.0140

Using the integration by parts formula, that converts into UV while UV is minus X cos(x), minus the integral of VdU.0151

So, minus the integral of V dU, that is minus cosin(x) dX.0167

OK, I am going to cancel these two negative signs. 0177

Now we have minus X cosin(x).0180

Now, this new integral you can see is just cosin(x).0185

That is a much easier integral to deal with than we started with. 0189

The integral of cosin(x) is just sin(x) and I am going to add on a constant just because you always have a constant for an indefinite integral. 0191

Then, we have our answer is negative X cosin(x) plus sin(x) plus a constant. 0200

Let us try a trickier example. 0208

X2 e3x dX. 0211

Again, we are going to divide it up into a U and a dV.0214

We will let U equal X2 and dV be e3x dX.0218

Again, we have got to figure out dU and V, so dU, if U is X2 is 2x dX. 0228

v is the integral of e3x dX. 0238

That is one third e3x. 0239

Then, I am going to write down the integration by parts formula again.0246

The integral of U dV is UV minus the integral of V dU and we have taken our example integral and we have split it up in to U dV again. 0250

Invoking the formula again, that is UV, so that is 1/3 x2 e3x minus the integral of VdU.0265

VdU is, there is a 1/3 and there is a 2, I will combine those as 2/3, and on the outside is x3x dx. 0280

Now, what we have here is another integral. 0290

It is easier than the first one because it has an x instead of an x squared. 0293

However it is still not an integral that we can do directly. 0297

What we have to do is integration by parts again. 0301

This is a very common issue with integration by parts. 0304

We are going to do integration by parts again on this new integral. 0307

I will let U equal X dV equal e3x dX, and again fill in dU equals dX and V is 1/3 e3x 0310

We still have that first term minus 2/3. 0330

Now we have the integral of U dV, so again using the integration by parts formula, that is UV, the new U and the new V. 0338

So, 1/3 x e3x minus the integral of V dU, minus the integral of 1/3 e3x dX. 0347

I am just going to focus on the stuff on the right. 0364

This is minus 2/9 X e3x and then the 2 minuses give you a plus 2/9. 0370

Now the integral of e3x is 1/3 e3x.0381

If we put all of those parts together we get 1/3 x2 e3x. 0391

Minus 2/9 e3x plus 2/27 e3x.0398

2/27 X e3x plus a constant, and that is the answer.0409

The moral of that example is that sometimes you have to do integration by parts twice in the same problem.0417

First we had to do integration by parts to reduce the original x2 down to an easier integral that just had an X in it. 0425

But that still was not an integral that we could do directly.0434

We had to do integration by parts again to reduce the X, well actually to make the x go away, and give us an integral that we could do directly. 0437

That is a pretty common story with integration by parts, that you have to do it twice. 0446

I want to teach you a secret short cut to doing integration by parts problems. 0449

This is just kind of a book-keeping device, but it can help you do some of these problems really quickly. 0458

It is called tabular integration and I am going to introduce it with an example. 0465

I am going to redo the same problem that I just did. 0469

Remember that problem was x2 e3x dx. 0471

Here is the secret shortcut. 0477

What you do is write x2 e3x, and you make a little table here. 0482

On the left hand side you write down derivatives. 0488

The derivative of x2 is 2x, the derivative of that is 2, and then the derivative of a constant is just 0. 0490

On the right hand side, so those were all derivatives, you take integrals. 0497

The integral of e3x is 1/3 e3x. 0503

The integral of that is 1/9 e3x. 0507

The integral of that is 1/27 e3x. 0511

Then this is just a clever little trick, but say this time we were doing the problem the previous way. 0517

You draw these little diagonal lines and then you put little positive and negative signs on the diagonal lines alternating plus, minus, plus. 0523

Then, what you do is multiply along these diagonal lines. 0533

So, you get x2 times 1/3 e3x. 0543

Now, the next diagonal line is minus 2x/9 e3x. 0553

The next diagonal line is plus 2/27 e3x. 0561

You attach a constant at the end and there is your answer 0567

That is just a clever book-keeping way of suppressing all the grunt work of going through the U and dV stuff.0570

It works really fast for certain kinds of problems.0581

If you have a polynomial like x2, times something like e3x or cos(x) or sin(x) where it is easy to take integrals, then this tabular integration trick works really nicely. 0584

I want to do a more complicated example, where we have ex cosin(x) dX. 0595

Again, this is one where we can not do the previous shortcut, the tabular integration idea, because we do not have a polynomial.0608

If we took derivatives of either one of these, ex or cosin(x), we would never get down to 0.0612

The shortcut is not available here. 0620

Let U equal e3x, or sorry, ex, and dV equal cosin(x) dX, and fill in dU is ex dX. 0625

A lot of students leave off the dX when they are doing these problems. 0633

It is really important to include the dX. 0643

It makes all the notation work out and helps you to track where you are going late on, so do include the dX.0646

If dv is cosin(x) dX, then V is the integral of cosin(x), which is just sin(x). 0651

This integral, again using our formula UV minus the integral of V dU, is UV. 0657

So, ex sin(x) minus the integral of V dU. 0667

That is ex sin(x) dX. 0677

That gives us a new integral ex sin(x) and we are going to do integration by parts again. 0681

We are agian going to let U equal ex. 0688

This time dv is going to be sin(x) dx, and dU is ex dX. 0694

V is the integral of sine which is negative cosin(x).0700

Invoking our integration by parts formula again, this is UV. 0709

Negative ex cosin(x) minus the integral of V dU, that is negative cosin(x). 0718

I will combine the negatives and that becomes positive there. 0730

ex cosin(x) dx.0733

I am just going to get rid of the brackets and bring the negative sign through so we get ex times sin(x) plus ex cosin(x). 0741

Minus the integral of e x cosine(x) dx.0752

Here, we have got a strange thing happening because what we did was integration by parts twice. 0757

If you look at this integral that we ended up with, which was supposed to be getting easier, it is exactly the same as the integral that we started with. 0764

That seems like we are spending a lot of time to just go around in circles, because we have done a lot of work and we have come up with the integral that we started with. 0771

In fact, we can use this to finish the problem. 0782

The way we do that is we let i be this integral that we started with. 0789

That means that i is equal to all these calculations that we did. 0794

Then this integral that we get at the end is i again.0798

So what we can do is we can move i over to the other side of the equation. 0805

What we get is 2I is equal to ex sin(x) plus ex cosin(x). 0810

That means we can solve for i. 0820

I is just one half the quantity ex sin(x) plus ex cosin(x).0824

And then of course we have to add on a constant, as always.0836

All of a sudden, even though it looked like we were going around in circles, we have solved our integral. 0841

That is the answer right there. 0844

So that is another common pattern that can happen with integration by parts. 0852

It happens a lot when you have combinations of ex and cosin(x).0857

You have eax, e to some number x, and then either sine, or cosine, of some number times x. 0861

You can do this trick of doing integration by parts twice and then you get an expression that you can solve for your original integral.0871

I want to show you a little memory trick that can help a lot with integration by parts.0880

The difficult part with integration by parts is that you will be given an integral of a whole lot of stuff. 0890

What you have to do is decide how to break up that integral into a U part, and a dV part. 0898

That is what make integration by parts tricky, deciding what to make U, and what to make dV. 0914

The idea is that after you go through the integration by parts formula, you want to get an integral that is easier than the one you started with, not harder. 0925

That can often be difficult to predict ahead of time. 0932

There is this little mnemonic that can help you remember how to split up integrals that way.0938

Just remember these five letters. L.I.A.T.E., lee-ah-tay. 0942

What those stand for is the functions that you should use for U, if you have them. 0949

Let U be the following functions if you have them. 0960

First of all L stands for ln(x). 0966

If you see a ln(x), that is your U. 0970

I stands for inverse functions. 0973

If you see an inverse trigonometric function, for example arcsine, arctan, and those are also written as sine inverse and tan inverse. 0976

If you see those, you know that is going to be your U. 0991

A stands for algebra. 0994

If you see something like x or x2, that is going to be your U. 0998

T stands for trigonometry.1003

Something like sin(x) or cosin(x). 1010

E stands for exponential, so ex.1015

Work through these functions in order, and whichever one of these you see first, that is going to be your U.1020

That is a very effective way of solving integration by parts problems.1026

As you work through your homework and try this out on different problems, keep this in mind and try it out. 1032

I hope it works out for you. 1039

So, that is the end of the first lecture from educator.com on integration by parts.1040