Welcome to Calculus. I'm Professor Grist. We're about to begin Lecture
22 on Integration By Parts. We turn now to one of the best
all-purpose methods for integration. This is the integration by parts formula. In this lesson, we'll learn what that is, why it works, and
what to do if it doesn't. Recall that our method for computing
an integral as an anti-derivative is to take the differentiation rule and
run it in reverse. In this lesson we'll
consider the product rule. Recall the states that d(uv) is udv + vdu. Now by integrating and
rearranging this rule, what will we get? Now if we integrate both sides. Then, on the left, taking the integral of
u times v yields, of course, u times v. On the right, we have the integral
of u dv plus the integral of v du. This leads to the integration
by parts formula. The integral of u d v equals u
times v minus the integral v d u. Now again, as a reminder, u and v
are functions of some third variable, say, acts in the way that this is typically
applied is that you are given one of these terms, let's say the integral
of udv as your problem. And this is a hard integral,
one that you're not sure how to evaluate. What the integration by parts formula
does is gives you a hopefully easier integral,
in the form of the integral vdu. Let's look at this in
the context of an example. Compute the integral of xe to the x dx. In order to apply the integration by parts
formula, we're going to have to identify which term should be u and
which should be dv. There are choices to be made. Well, let's proceed by taking a guess and
seeing how it works. Let us choose as ux and as dv e to the x, dx. This is going to be a pretty good choice
in that when we compute du we get something simple, dx,
but when we integrate dv to get v,
well it's an integral that we can do. The integral of e to the x
dx is of course e to the x. Now, in this case,
why did we not choose a plus c? An integration constant. Well let's see what happens when we
apply the formula without that constant. The integration by parts formula says
that the integral of x e to the x dx is u x times v e to the x
minus the integral of v, e to the x, times du, dx. And now you see how the formula
is supposed to work. This rearrangement gives us
a simpler integral to evaluate. Of course,
we know that the integral of e to the x Is just e to the x plus a constant. Now, I'll leave it to
you to figure out why adding the single integration
constant at the end suffices. We don't have to worry
about in the middle. Now, we should probably rearrange terms,
simplify a little bit to get a final answer of quantity x minus one
times e to the x times a constant. In this case you should always
check your answer to make sure that when you differentiate it you
get back the original integrant. For another example,
compute the integral of log of x, dx. In this case, the choice of u and
dv is maybe not so obvious. In general, you want to keep
things as simple as possible. Therefore, let u be equal to log of x
since we know how to differentiate that. Dv would necessarily be dx,
thus computingduU and one over x, dx and v as simply x. The integration by parts formula says that the integral lnxdx is u times v, that is xlnx minus the integral of v du, that is the integral of
x times one over x, dx. Notice how that is a simpler integral, yielding x ln x- x + a constant, or if we simplify, x(ln- 1). Again, you'll want to check your work
to make sure that the derivative of this Is in fact what we began
with is the natural log of x. For another example compute
the integral of sine of x over x, dx. Well we have a couple of choices
here let's try letting u be equal to sine of x and
dv being 1 over x dx. In this case, du would be cos x dx, and v would be the antiderivative of 1 over x,
that is, in x. The integration by parts
formula gives what? U times v, that is, ln x sinx minus the integral
of v lnx du cosx. Well, this does not look so good. That second integral looks
more difficult than the first. I think this counts as a failure. We chose our u and dv incorrectly. There's no guarantee that the integration
by parts formula is ever going to work for any choice of u and dv. But let's try again. If we chose you to be 1 over x and
dv to be sine of x, dx. Then well we can compute
du as 1 over x squared dx and we can antidifferentiate
sine to get minus cosine. Applying the integration by parts
formula with this choice yields what? U times v is negative cosine of x over x then we have to subtract
the integral of v du, we get a cosine x over x squared dx. I think this again counts as a fail. That looks like a worse integral
than what we started with. What does this mean? Have we chosen wrong? Is there another way to do this? Well, no. In fact, this is not an integral that has a simple, representation for
its antiderivative. In fact we call the integral of sine
of x over x, the sine integral. It goes under the notation si(x). It does not have a simple antiderivative. Sometimes, the integration by parts method
does not work for any choice, du and dv. Now let's move on. Let's consider the integral e to the x,
sine of x. Let's let u be sine of x,
dv be e to the x, dx. Computing du as cosine to xdx and
v, e to the x, yields, by the integration by parts
formula, e to the x sin of x, that's u times v, minus the integral
of e to the x cosine of xdx. Well again, this looks like one of
those situations where this formula did not simplify things. However, we should not give up. Let's consider applying the method
again to the integral of e to the x, co sin of x, dx. Let's choose our u to be cosine of x,
dv to be e to the x dx. Then computing du and v, much as before,
we can apply the integration by parts formula to this integral
to yield u times v, that's e to the x, cosine of x,
minus the integral of vdu, and that yields o plus integral of e to the x,
sin of x, dx. And what do you we have? Oh, dear. We have our original integral again and it
seems that though we are caught in a loop. This certainly seems
like a failure however, let's be careful let us call this
integral of E to the X, sign of X, I. What have we computed? We've computed that this I is really e to the x sin of x minus
some other integral. Let's call the integral of
e to the x cos of x, J. Then what we've computed in our last attempt was that I = e to the x sine x- J. And J = e to the x cosine x + I. This is a system of two
equations with two unknowns. Let's substitute the latter
into the former. And what do we obtain? We get that I = e to the x sine
of x- e to the x cosine of x- I. Be careful with your signs. In this case, solving for
I yields an answer it yields one half quantity E to the X
sine of X minus E to the X cosine of X. So with two applications of integration
by part we can solve this problem. The moral is don't quit too early. But this yields the question. How many times do we have to apply this
formula in order to get an answer? Well, I have a bit of bad news for you. You don't know. There are examples where
you have to apply this many times in order to
get the correct answer. However, this might not be so bad. Sometimes one can obtain what
is called a reduction formulae. Let's look at an in depth example. Consider the integral
of x to the n cosxdx, where n is a positive integer. In this case we're going to choose our u to be something that will
simplify under differentiation. Let's let it be x to the n. So that dv is co sin of x dx. In this case,
du becomes n times x to the n minus one. Dx and v is of course sin of x. The integration by parts
formula give what? U times v, that's x to the n sin of
x minus the integral of v d u and that gives n times x to the n
minus one times sin of x dx. Pulling out the n gives us something
that is maybe a little simpler, but it's still of the form x
to some power times sine of x. So, let's try the general case of
the integral of x to the n sine of x dx. We'll again apply the integration
by parts formula and make the same choices as we did before. So that du is nx to the n-1 dx,
and v is negative cosine of x. Applying integration by parts gives what. The integral of x to the n
sin of x is x to the n times minus cos of x + the integral
of n x to the n- 1 cos x dx. Now we see something we see something,
the same form as what we began. But notice that the powers
of x are reduced by one. This allows us to get a reduction formula. Continuing, here's the crucial step. X to the n, cosine of x, dx, integrates to
X to the n times sine of x minus n times. Well, let's see, if we take the integral
that we had on the previous slide and use the second integration
by parts application we get quantity negative
x to the n minus one. Cosine of x plus the integral. N minus 1 times x to the n minus 2,
cosine of x, dx. This simplifies, a little bit, into
a formula that looks somewhat complicated. But note,
one ends with an integral of the form x to some power times cosine
of x d x where that power of x is two less than the original
n that you began with. Likewise, if we compute the integral
of x to the n sine of x. And follow the algebra of substituting in the integration by parts formula
that we derived on our last slide. What do we get? We get in the end an integral of x
to the n minus 2 times sine of x dx. These final formulae
are called reduction formulae. They take a difficult integral,
something like x to the n cosine of x, and reduce it to a simpler integral,
x to the n-2, cosine of x. By applying this inductively,
eventually you get down to x to the zero times cosine of x, or maybe x to the one
times cosine of x, and these are solvable. Integration by parts is one of
the key tools for computing integrals. You're going to want to have that
well practiced and memorized but it's not the end of our techniques. In our next lesson,
we'll present a technique of integration, based on substitution, but
tuned for trigonometric variables.