0:10

We turn now to one of the best all-purpose methods for integration.

Â This is the integration by parts formula.

Â In this lesson, we'll learn what that is,

Â why it works, and what to do if it doesn't.

Â 0:26

Recall that our method for computing an integral as an anti-derivative

Â is to take the differentiation rule and run it in reverse.

Â In this lesson we'll consider the product rule.

Â Recall the states that d(uv) is udv + vdu.

Â Now by integrating and rearranging this rule, what will we get?

Â Now if we integrate both sides.

Â Then, on the left, taking the integral of u times v yields, of course, u times v.

Â On the right, we have the integral of u dv plus the integral of v du.

Â This leads to the integration by parts formula.

Â The integral of u d v equals u times v minus the integral v d u.

Â Now again, as a reminder, u and v are functions of some third variable, say,

Â acts in the way that this is typically applied is that you are given one of

Â these terms, let's say the integral of udv as your problem.

Â And this is a hard integral, one that you're not sure how to evaluate.

Â What the integration by parts formula does is gives you a hopefully

Â easier integral, in the form of the integral vdu.

Â 2:11

In order to apply the integration by parts formula, we're going to have to identify

Â which term should be u and which should be dv.

Â There are choices to be made.

Â Well, let's proceed by taking a guess and seeing how it works.

Â Let us choose as ux and

Â as dv e to the x, dx.

Â This is going to be a pretty good choice in that when we compute du we get

Â something simple, dx, but when we integrate dv

Â 2:51

to get v, well it's an integral that we can do.

Â The integral of e to the x dx is of course e to the x.

Â Now, in this case, why did we not choose a plus c?

Â An integration constant.

Â Well let's see what happens when we apply the formula without that constant.

Â The integration by parts formula says that the integral of x e to the x dx is u

Â x times v e to the x minus the integral of v,

Â e to the x, times du, dx.

Â And now you see how the formula is supposed to work.

Â This rearrangement gives us a simpler integral to evaluate.

Â Of course, we know that the integral of e to the x

Â Is just e to the x plus a constant.

Â Now, I'll leave it to you to figure out why

Â adding the single integration constant at the end suffices.

Â We don't have to worry about in the middle.

Â 4:22

For another example, compute the integral of log of x, dx.

Â In this case, the choice of u and dv is maybe not so obvious.

Â In general, you want to keep things as simple as possible.

Â Therefore, let u be equal to log of x since we know how to differentiate that.

Â Dv would necessarily be dx, thus computingduU and

Â one over x, dx and v as simply x.

Â The integration by parts formula says

Â that the integral lnxdx is u times v,

Â that is xlnx minus the integral of v du,

Â that is the integral of x times one over x, dx.

Â Notice how that is a simpler integral,

Â yielding x ln x- x + a constant,

Â or if we simplify, x(ln- 1).

Â Again, you'll want to check your work to make sure that the derivative

Â of this Is in fact what we began with is the natural log of x.

Â 5:47

For another example compute the integral of sine of x over x, dx.

Â Well we have a couple of choices here let's try letting u

Â be equal to sine of x and dv being 1 over x dx.

Â In this case, du would be cos x dx, and

Â v would be the antiderivative of 1 over x, that is, in x.

Â 6:20

The integration by parts formula gives what?

Â U times v, that is,

Â ln x sinx minus the integral of v lnx du cosx.

Â Well, this does not look so good.

Â That second integral looks more difficult than the first.

Â I think this counts as a failure.

Â We chose our u and dv incorrectly.

Â There's no guarantee that the integration by parts formula is ever going to work for

Â any choice of u and dv.

Â But let's try again.

Â If we chose you to be 1 over x and dv to be sine of x, dx.

Â 7:13

Then well we can compute du as 1 over x squared

Â dx and we can antidifferentiate sine to get minus cosine.

Â Applying the integration by parts formula with this choice yields what?

Â U times v is negative cosine of x over x

Â then we have to subtract the integral of v du,

Â 7:47

I think this again counts as a fail.

Â That looks like a worse integral than what we started with.

Â What does this mean?

Â Have we chosen wrong?

Â Is there another way to do this?

Â Well, no.

Â In fact, this is not an integral

Â that has a simple, representation for its antiderivative.

Â In fact we call the integral of sine of x over x, the sine integral.

Â It goes under the notation si(x).

Â It does not have a simple antiderivative.

Â Sometimes, the integration by parts method does not work for any choice, du and dv.

Â 8:33

Now let's move on.

Â Let's consider the integral e to the x, sine of x.

Â Let's let u be sine of x, dv be e to the x, dx.

Â Computing du as cosine to xdx and v, e to the x,

Â 9:05

Well again, this looks like one of those situations where this formula did

Â not simplify things.

Â However, we should not give up.

Â Let's consider applying the method again to the integral of e to the x,

Â co sin of x, dx.

Â Let's choose our u to be cosine of x, dv to be e to the x dx.

Â Then computing du and v, much as before, we can apply the integration by parts

Â formula to this integral to yield u times v,

Â that's e to the x, cosine of x, minus the integral of vdu, and

Â that yields o plus integral of e to the x, sin of x, dx.

Â And what do you we have?

Â Oh, dear.

Â We have our original integral again and it seems that though we are caught in a loop.

Â This certainly seems like a failure however,

Â let's be careful let us call this integral of E to the X, sign of X, I.

Â What have we computed?

Â We've computed that this I is really e

Â to the x sin of x minus some other integral.

Â Let's call the integral of e to the x cos of x, J.

Â Then what we've computed in our last

Â attempt was that I = e to the x sine x- J.

Â And J = e to the x cosine x + I.

Â This is a system of two equations with two unknowns.

Â Let's substitute the latter into the former.

Â And what do we obtain?

Â We get that I = e to the x sine of x- e to the x cosine of x- I.

Â 11:09

Be careful with your signs.

Â In this case, solving for I yields an answer it

Â yields one half quantity E to the X sine of X minus E to the X cosine of X.

Â So with two applications of integration by part we can solve this problem.

Â The moral is don't quit too early.

Â 11:39

But this yields the question.

Â How many times do we have to apply this formula in order to get an answer?

Â Well, I have a bit of bad news for you.

Â You don't know.

Â There are examples where you have to apply this

Â many times in order to get the correct answer.

Â However, this might not be so bad.

Â Sometimes one can obtain what is called a reduction formulae.

Â Let's look at an in depth example.

Â Consider the integral of x to the n cosxdx,

Â where n is a positive integer.

Â In this case we're going to choose our u

Â to be something that will simplify under differentiation.

Â Let's let it be x to the n.

Â So that dv is co sin of x dx.

Â In this case, du becomes n times x to the n minus one.

Â Dx and v is of course sin of x.

Â The integration by parts formula give what?

Â U times v, that's x to the n sin of x minus the integral of v d u and

Â that gives n times x to the n minus one times sin of x dx.

Â Pulling out the n gives us something that is maybe a little simpler,

Â but it's still of the form x to some power times sine of x.

Â So, let's try the general case of the integral of x to the n sine of x dx.

Â We'll again apply the integration by parts formula and

Â make the same choices as we did before.

Â So that du is nx to the n-1 dx, and v is negative cosine of x.

Â Applying integration by parts gives what.

Â The integral of x to the n sin of x is x to the n times

Â minus cos of x + the integral of n x to the n- 1 cos x dx.

Â Now we see something we see something, the same form as what we began.

Â But notice that the powers of x are reduced by one.

Â This allows us to get a reduction formula.

Â 14:15

Continuing, here's the crucial step.

Â X to the n, cosine of x, dx, integrates to X to the n times sine of x minus n times.

Â Well, let's see, if we take the integral that we had on the previous slide and

Â use the second integration by parts application

Â we get quantity negative x to the n minus one.

Â Cosine of x plus the integral.

Â 14:45

N minus 1 times x to the n minus 2, cosine of x, dx.

Â This simplifies, a little bit, into a formula that looks somewhat complicated.

Â But note, one ends with an integral of the form

Â x to some power times cosine of x d x where that power of

Â x is two less than the original n that you began with.

Â Likewise, if we compute the integral of x to the n sine of x.

Â And follow the algebra of substituting in

Â the integration by parts formula that we derived on our last slide.

Â What do we get?

Â We get in the end an integral of x to the n minus 2 times sine of x dx.

Â These final formulae are called reduction formulae.

Â They take a difficult integral, something like x to the n cosine of x,

Â and reduce it to a simpler integral, x to the n-2, cosine of x.

Â By applying this inductively, eventually you get down to x to the zero

Â times cosine of x, or maybe x to the one times cosine of x, and these are solvable.

Â 16:15

Integration by parts is one of the key tools for computing integrals.

Â You're going to want to have that well practiced and memorized but

Â it's not the end of our techniques.

Â In our next lesson, we'll present a technique of integration,

Â based on substitution, but tuned for trigonometric variables.

Â