Professor Dave here, let’s talk strategy. In our journey through calculus, we started
with differentiation because it’s quite simple and straightforward. We learned the
power rule to take the derivative of a polynomial, and then we learned the sum and difference
rules, product and quotient rules, and then the chain rule. From these, we found that
we could take the derivative of just about anything, and even if it got a little messy,
there was no question as to how to go about doing it, we just have to do the work. Integration
doesn’t work like this, unfortunately. There is no rigid algorithm to follow. We sometimes
have to look at an integrand and come up with a strategy for integration, and this strategy
could involve one of the techniques we have discussed, or something different still. This
is what makes integration so difficult, but also so exciting. It’s like a sport. There
is no guaranteed way to win, you just have to learn as many tips, tricks, and methods
as you can, practice a lot, and over time you will become more and more adept. If you
haven’t watched my tutorials that go over the substitution rule, integration by parts,
and trigonometric substitution, go back a bit in the series and check those out now,
as in this clip, we will assume that these methods are already understood. If you’re
all caught up, let’s go ahead and use this opportunity to synthesize everything we’ve
learned about integration so far. We will look at some examples where it’s not immediately
obvious what approach will be the one to use, but we will talk through the logic that should
be employed in order to get to the answer. First things first, it will either be the
case that you have access to a table of integration formulas, or we will have to memorize as many
of them as possible. Either way, we will need a table like this to refer to while determining
a strategy for integration. In this table we can see the formula that allows us to integrate
each term in a polynomial, followed by all of the special formulas we should know. For
example, we must recall that the integral of one over x is the natural log of x. Then
there is one for an exponential function. Then all of the trigonometric functions. These
are useful, because if we can get an integrand to look like something we see here, we will
probably have an easy time integrating, but we may need to do some substitution to get
there. We will also want to memorize, or at least be able to recognize, the formulas that
allow us to perform integration by parts, as well as trigonometric substitution. Armed
with all of this, let’s see what we can do.
Alright, let’s give this first one a shot. How about the integral of x over root (x squared
plus four), dx. We immediately recognize that this is too complicated to be integrated directly,
so we need to use one of our techniques. With this example, we will learn right away that
sometimes, multiple techniques could be applied. We could be tempted to use trigonometric substitution,
noticing that this radical expression allows it. We absolutely can use this technique,
however, we might also notice that direct substitution will be much easier, since we
see that there is an expression for which a form of its derivative is also present.
If we take u to equal x squared plus four, this part becomes root u, and then du will
equal two x dx. We have x dx, so that’s no problem, we can just rearrange slightly
to make x dx equal to du over two, plug that in, and we are left with du over two root
u. If we modify slightly, we can pull one half out of the integral, change the root
u in the denominator into u to the negative one half, and now things look nice and easy.
U to the negative one half becomes u to the one half over one half, or two u to the one
half, plus C. The twos cancel, u to the one half is simply root u, and we change u back
to the other form. This leaves us with a nice and tidy answer, root x squared plus four,
plus C. So as we just saw, there may be more than one path to an answer, and it’s not
a big deal if you end up taking a route with a couple extra steps, provided you still arrive
at the correct answer. And in fact, you can always check that your answer is correct by
taking its derivative and verifying that the result is equivalent to the integrand, which
means you can walk out at the end of your integration exam already knowing that you
got one hundred percent. But to summarize the lesson learned from this example we just
completed, we should carefully note that apart from basic situations where simplifying the
integrand will allow it to be integrated, like distributing something across a sum or
difference, expressing roots as fractional exponents, or applying trig identities, it
is the case that direct substitution will usually be the easiest technique if a special
technique must be applied. So it may be a good strategy to check if this is possible
first, and only if it isn’t should you then see which of the other two techniques might
apply. How about this one, the integral of cotangent
x times the natural log of sine x, dx. This is a product of functions, so integration
by parts may come to mind, but remember, let’s first check to see if one of these functions
is the derivative of the other. Cotangent is cosine over sine, that would require the
quotient rule, maybe we could try that if we get to that point, but how about the other
function? This derivative will require the chain rule. The derivative of natural log
of sine of x will involve the derivative of natural log first, which is one over whatever
it’s operating on, so we get one over sine x, then we multiply by the derivative of the
inner function, and the derivative of sine is cosine, so we get cosine x over sine x.
Well that is cotangent, so substitution should work out perfectly here. We make LN sine x
equal to u, and then we find that cotangent x dx is equal to du. That will give us u du,
we integrate to get one half u squared, switch u back to what it was before, and then finish
things off with plus C. Easier than you expected, I bet.
Let’s try another. How about the integral of cosine root x dx. This seems like it could
be straightforward, but it’s possible that no strategy jumps out at you immediately.
When in doubt, try substitution. Here the most obvious thing to do is to make u equal
to root x, that way we no longer have a composite function. And if u will be root x, we need
to get du. Well du over dx will be the derivative of x to the one half, and that’s one half
times x to the one half minus one, or negative one half. The negative means we put it on
the bottom of a fraction, and one half means square root, so du over dx is one over two
root x. Now normally, we would bring dx up here, and then we would look for one over
two root x, dx in the integrand, so that we could replace it with du. That’s how the
substitution method typically works. Unfortunately we don’t see that in the integrand, but
here’s another trick. If you’re trying substitution and you don’t see g prime x
dx in the integrand, try solving for dx instead. Here, that will be two root x du, so let’s
replace dx with this expression. It may be the case that the terms with x in them cancel
out any remaining x terms in the integrand, which will solve our problem, or failing that,
they may be terms that can also be expressed in terms of u. In this case, u is root x,
and we have a root x here, so why not just change that to u instead? Now we can pull
the two out, and we are left with the integral of cosine u times u du. This is something
that can very easily be integrated by parts. So let’s put a box around this to mark our
progress, as we don’t want to get lost, and once we solve this integral we can come
back to where we were. Now when we integrate by parts, we assign a u and a dv, and then
find du and v. Here we have u’s already, so in order to avoid confusion, let’s change
this to some other variable, like z. So now we have cosine z times z dz. Now let’s integrate
by parts. Remember, we pick a function for u that when in the form of du in the new integral,
will make it much easier to integrate, so that should definitely be z, because that
will become one. So u equals z, and dv equals cosine z dz. That means du is simply dz, and
v is sine z, since sine is the integral of cosine. Now we use the formula. The first
integral is equal to uv, which is z sine z, minus the integral of v du, which is sine
z dz. The integral of sine is negative cosine, so the negatives cancel, leaving us with z
sine z plus cosine z, plus C. Now that we are all done with the integration, we are
safe to return the z’s back to u’s, because we only changed them so that we could use
the integration by parts method without confusion, so that’s now u sine u plus cosine u, plus
C. Let’s also recall that when we did the original substitution, we arrived at two times
this integral, which we then integrated by parts, so we have to bring back the two now,
since we are done integrating. Now we just remember that we initially used u to represent
root x, so let’s bring back root x, and we get two times this quantity. Then we just
distribute to get two root x sine root x, plus two cosine root x, plus C. So that may
seem like it was terribly complicated in retrospect, and our answer does look far more complex
than the integrand, but that’s the answer, and all we really did to get there was a simple
substitution, followed by integration by parts. So let’s take a breathe here. We’ve seen
that sometimes it’s best to just try a substitution and see what happens. It may go as planned,
or we may get to something else that can be integrated by parts. This is because integration
by parts requires a product of functions in the integrand, and this approach works best
when one function is a power of x, or a polynomial, and the other is some transcendental function,
like a trig function, or an exponential or logarithmic function, and we can assign u
and dv according to the strategy we described when we learned this technique. Sometimes,
as we just saw, we need to do some kind of substitution first in order to get things
to a place where integrating by parts makes sense, and if that’s the case, then that’s
just what we have to do. Sometimes the substitution isn’t obvious, so just try something. If
it doesn’t work, try something else. Unfortunately there will indeed be cases when even these
techniques don’t apply, and if so, we have to be clever and try something else. Sometimes
we have to manipulate the integrand in particular way. Take something like the integral of dx
over one minus cosine x. None of our techniques will work here. But with a difference in the
denominator, we might be able to think of something else to try. We know that we can
simplify a sum or difference in the denominator of a fraction by multiplying by its conjugate.
So what if we multiply by one plus cosine x over one plus cosine x? Remember that this
is useful because then on the bottom when we foil, we get one plus cosine minus cosine
minus cosine squared, so the cosine terms cancel. With squared trig functions, an identity
can usually be used, and in fact one minus cosine squared is indeed sine squared. This
was useful because with a single term in the denominator, we can now split this up into
simpler fractions. One over sine squared is cosecant squared, and the other fraction can
remain as cosine over sine squared. This can be split up into two integrals, the first
of which is easy to integrate, because cosecant squared simply becomes negative cotangent.
Then with the other one, it is fairly easy to spot that substitution will be our best
approach. If we put u in place of sine x, we get u squared on the bottom, and du will
be cosine x dx, which we have here, so du over u squared, or u to the negative two du.
This becomes negative one over u, or negative one over sine x, which is cosecant x, so we
end up with negative cotangent x minus cosecant x plus C.
We have learned a lot about integration over the past few tutorials. We learned what it
means, how to do it, and a few tricks for when it becomes more complicated. Sometimes
we have to substitute twice, sometimes we have to substitute and then integrate by parts,
or vice versa. With practice, you will only become better and better at the art of integration,
and you will be able to apply ingenuity to figure out an appropriate strategy, allowing
you to integrate expressions you’ve never even seen before. We will now be moving on
to other concepts involving integration, but first, let’s check comprehension.