Section 11: Methods of Integration2001 by Karl Hahn 
As you drive down a highway, you can sometimes get a good idea of what the scenery looks like to drivers heading in the opposite direction by looking in your rear view mirror. That is especially true if the scenery is simple. The same is true of finding indefinite integrals. If the function is simple, you can sometimes figure out an antiderivative using only what you know about derivatives. Indeed we had a table of derivatives of simple functions. Clearly if the function you want to find the indefinite integral of appears in the "derivative" (righthand) column of this table, then an antiderivative of that function appears in the lefthand column.
Let's start with something very easy. Suppose you want to find the indefinite integral,
e^{x} dx 
You already know that if f(x) = e^{x}, then f'(x) = e^{x}. That is the special property of the e^{x} function  that it is its own derivative. So clearly it must be its own antiderivative as well. The only other thing we have to do here is stick in the undetermined constant, and we will have the indefinite integral of e^{x}.
e^{x} dx = e^{x} + C eq. 11.21 
Now suppose that you had the same problem except with the additional complication that the e^{x} is multiplied by the constant, A. From your knowledge of derivatives you recall that if f(x) = Ae^{x}, then f'(x) = Ae^{x}. Clearly the function, Ae^{x}, has that same property when A is constant  that it is its own derivative. So again it must be its own antiderivative as well. From that you have
Ae^{x} dx = Ae^{x} + C eq. 11.22 
Indeed if you took any function, f(x), for which you already knew an antiderivative, F(x), and you multiplied f(x) by a constant, A, wouldn't it be true that AF(x) was always an antiderivative of Af(x)? Think about it. Remember that F'(x) = f(x). Recall the rule about taking the derivative of something that is multiplied by a constant (check the page of rules if you have to). The rules force you to conclude taking the derivative of AF(x) will give you AF'(x) = Af(x). Since AF(x) is always an antiderivative of Af(x), it follows that if you know the indefinite integral of a function, then you can easily find the indefinite integral of a constant times that same function. In symbols it looks like this (where F(x) is an antiderivative of f(x))

Look at equation 11.23 carefully. Ignore the middle expression and just regard it as setting the lefthand expression equal to the righthand expression. That means that whenever you see a constant multiplied times the entire integrand, you can move that constant multiplier outside the integral sign. (If you remember that an integral is a limit of a Riemann sum, the previous statement is simply the equivalent of the distributive law over that sum)
Let's see how we can use equation 11.23 to work backwards on our old friend, x raised to a power. Always remember that finding an indefinite integral means finding an antiderivative. We know from our study of derivatives that the derivative of f(x) = x^{p} is f'(x) = px^{p1}. This works even when p is negative and when it is not an integer. So from this we know that an antiderivative of px^{p1} is x^{p}.
Now suppose that you would like to find the indefinite integral
x^{s} dx eq. 11.24a 
where s is a constant. How do we run the process of taking the derivative of x^{s} backward? We have to munge the integrand into px^{p1}. We know the antiderivative of that expression. For our integrand in equation 11.24a to look like px^{p1} we would have make p1 = s and equivalently p = s+1 in order to make the exponent come out right. If you make this substitution, the indefinite integral that we're after becomes
x^{p1} dx eq. 11.24b 
It still isn't quite right because it isn't multiplied by p. But if p is not zero, then it's always true that p/p = 1. So the expression in equation 11.24b is the same as:
p 
Now recall the rule we just learned in equation 11.23. Since 1/p is a constant multiplier over the entire integrand, we can factor it to outside the integral sign.
1 
px^{p1} dx eq. 11.24d 
Now the integrand is indeed the derivative of something that we know. It is the derivative of x^{p}. So the indefinite integral is
x^{p1} dx = 
1 
^{ } 1 px^{p1} dx = 
Did you catch me playing fast and loose with the algebra here?
In equation 11.24e, the right hand side ought to be
1C is an undetermined constant. If that is the case, then isn't C/p also an undetermined constant? Does it matter what we call it? Remember that any constant will do. So the way we have it in equation 11.24e is correct even though I did not strictly adhere to the rules of algebra. And typically that is the way you will do these problems. You won't add the undetermined constant in until you have done all the other steps first. Then stick it in as the last step and you'll be good to go. 

Equation 11.24f works even when s = 0. Note that when you put in zero for s you have x^{0} = 1 as your integrand. Hence

Indeed equation 11.24f works with all values of s except s = 1. Can you see why it fails then? You'd have a division by zero. But when s = 1, the integrand is really 1/x. So in that case the indefinite integral you are trying to find is
1 
dx 
Note the second way of writing this indefinite integral. Most textbooks will use this nomenclature when the integrand is a quotient. That is they will put the dx in the numerator of the quotient. But the two ways of writing it shown in equation 11.26a mean exactly the same thing.
So where have we seen 1/x as the derivative of something? Check the table. Doesn't the derivative of F(x) = ln(x) fill the bill? Since ln(x) is an antiderivative of 1/x, you have
1 
dx 
This indeed is the property of the natural log that causes it to come up so frequently in physics, chemistry, and engineering. Any model of any phenomenon that causes you to take the integral of a 1/x style function will result in a natural log being part of the model.
But what if x is negative? You can't take the log of a negative number. Does that mean that there is no indefinite integral of 1/x for that domain? You only have to make a minor mod to equation 11.26b to make it work for negative x as well.

That is the natural log of the absolute value of x. Look at what happens when you take the derivative (using the chain rule) of F(x) = lnx. Remember that the derivative of x is sgn(x) (where sgn(x) = 1 for x > 0 and sgn(x) = 1 for x < 0). Hence
dlnx 1 1 F'(x) =So lnx is also an antiderivative of 1/x, but this antiderivative will admit negative values of x.=sgn(x) =eq. 11.27 dx x x
We already saw what the indefinite integrals of e^{x} and Ae^{x} were. What about
e^{bx} dx eq. 11.28a 
where b is a constant? Recall that the derivative of f(x) = e^{bx} is f'(x) = be^{bx}. Now lets play the same game we played before. Whenever b is not zero, you have
e^{bx} dx = 
b 
e^{bx} dx = 
1 
be^{bx} dx eq. 11.28b 
And when b is zero, you don't care because e^{0x} = 1, which you already know how to integrate. In equation 11.28b, the righthand integrand is the derivative of something we already know  that is it is the derivative of e^{bx}. Hence
e^{bx} dx = 
1 
e^{bx} + C eq. 11.28c 
This pattern works with any function. If you know that F(x) is an antiderivative of f(x), then it is always the case (when b is a nonzero constant) that

You can see why by taking the derivative of the indefinite integral expression shown above using the chain rule.
Suppose you wanted to find the indefinite integral of cos(x). Another glimpse of the table ought to convince you that sin(x) is an antiderivative of cos(x). So clearly
cos(x) dx = sin(x) + C eq. 11.210a 
But what if you wanted to find the indefinite integral of cos(x+φ), where φ is a constant. Notice that when you have any function, f(x+φ), and you take the derivative (using the chain rule), you end up with f'(x+φ). It doesn't take a whole lot of vision to see that this same arrangement will work in reverse. So we see that
cos(x+φ) dx = sin(x+φ) + C eq. 11.210b 
or in general, whenever F(x) is an antiderivative of f(x).

for any constant, φ. Or even more generally (combining the three rules for constants we learned on this page), whenever A, b and φ are constants and F(x) is an antiderivative of f(x), then

provided that b is not zero.
Suppose you needed to know the indefinite integral of 3 + e^{x}. From equations 11.23 and 11.25 you should be able to see easily that
3 dx = 3x + C eq. 11.212a 
And from equation 11.29 you should be able to see that
e^{x} dx = e^{x} + C eq. 11.212b 
So what happens if you take the derivative of the sum of these two results? Try it. Doesn't that get you back to the original function of 3 + e^{x}? And doesn't that mean that
(3 + e^{x}) dx = 3x  e^{x} + C eq. 11.212c 
(Notice again the treatment of the undetermined constant, C. By rights, you should have 2C in the above, but if C is an undetermined constant, then so is 2C. So why bother with the 2)
Remember the sum rule for derivatives? It is that the derivative of a sum is the sum of the derivatives. Can you see how this would work for indefinite integrals as well? If F(x) is an antiderivative of f(x), and G(x) is an antiderivative of g(x), then it is always true that

Or in other words, the indefinite integral of the sum is the sum of the indefinite integrals. Another way to put this is

Not only that, but just as equation 11.213b breaks an integrand that is the sum of two things into the sum of two integrals, if the integrand is the sum of three things, you can break it into the sum of three integrals. Indeed if it is the sum of n things, you can break it into the sum of n integrals.
In the table that follows you can see that we have essentially taken the table from the section on derivatives and turned it around. It is useful to know these indefinite integrals at a glance, so consider memorizing this (which should be easy if you have already familiarized yourself with the table of derivatives). These together with the rules we covered above will form the core of knowledge you need in more complex integration problems after you have applied other tools to the integrand.
function indefinite integral 

dx x + C 

1 x dx 

1 x^{s} dx [s ≠ 1] 

dx 
1 

dx^{ } 
^{ } 1 1^{ } 1 x^{s} dx [s ≠ 1]  



e^{x} dx e^{x} + C 

cos(x) dx sin(x) + C 

sin(x) dx cos(x) + C 

dx 
Be sure you understand how we arrived at all the indefinite integrals in the table above from the table of derivatives. These are the indefinite integrals you can infer simply by knowing the derivatives of some elementary functions. The goal of all the other methods we will learn from now on will be to reduce the problem of taking the indefinite integral to one or more of the integrals in this table. It is also very useful for you to memorize equation 11.211b and equation 11.213b. They form your first shelf of tools for reducing an indefinite integral problem into an entry on the table.
Find the indefinite integrals of the following (and don't forget to add the undetermined constant on each one. Most instructors mark you down a point or two if you leave that out. And don't forget to check your answers by taking the derivative of the function you ended up with and matching that back to the original integrand):
1) 
______ √3x + 2 dx 
2) 
(sin(πx) + cos(3πx)) dx 
3) 
2 
where n is a constant. Notice that on this one the integrand is a sum, but there are no parentheses enclosing it before you get to the dx. Many books will write it this way because the integral is assumed to be over everything between the integral sign and the dx, even if you leave the parentheses out.
4) 
x^{4} + 8x^{3} + 12x^{2}  4x  3 dx 
5)

n
∑ B_{k}(kx + 1)^{k} dx
k=1

where each B_{k} is a constant. This problem is shown in a funny color because it is intended to challenge you. So consider it extra credit. But keep in mind that everything you need to know in order to do this problem is on this page of the tutorial. Don't let the summation sign scare you. The rule given in equation 11.213b applies to sums even when they are shown using the sigma notation. Your answer will have to use the sigma notation as well. Use the rule given in 11.211b to integrate the expression inside the summation. You can treat k as a constant for the purposes of the rule, because k does not vary with x. Your answer will be the summation of whatever you end up with after applying the rule and taking the integral.
Long ago in a galaxy far away...© 2001 by Karl Hahn 
the technological terror known as the Death Star aimed its weapons at the peaceful planet, Alderaan, and, just for jollies, shattered it into countless fragments that still drift endlessly through the universe to this very day. You probably recall the exact scene of Alderaan's destruction from the movie, Star Wars. Surely the Death Star had enormous energy resources at its disposal. But the question that was undoubtedly gnawing at the minds of many a viewer after having seen the movie was, "Exactly how much energy does it take to fragment an entire planet?" Well thanks to what you've learned in last few sections, along with a little basic physics that I will tell you now, you have become equipped to answer this mystery.
The basic physics is this: The energy required to lift an object of mass, m, from the surface of a planet of mass, M, to a point where it is free of the gravity of that planet is
GMm E =where r is the radius of the planet, and G is the universal constant of gravity. If you measure mass in kilograms, distance in meters, and energy in wattseconds, then G = 6.7 × 10^{11} wattsecondmeters per kg^{2} (from Astronomy Data Book by J. Hedley Robinson).eq. 11.214 r
Here's how we figure it. Imagine that you have a perfectly spherical planet. For simplicity we shall assume that its mass density is uniform all the way through, with a value of ρ kilograms per meter^{3}. We will use the physics formula to figure out how much energy it takes to remove a very thin shell of the planet from its surface and free it from the planet's gravity. Let the thickness of the shell be dr. Then the volume of the shell (not the volume enclosed by the shell, but the volume sandwiched between the outer and inner surfaces of the shell) will be very nearly the surface area of the planet times the thickness, dr (the thinner the shell, the more accurate this estimation of its volume is). The mass of that shell, dm, will be ρ times the shell's volume. Since we know the surface area to be A = 4πr^{2},
dm = 4πr^{2}ρ dr eq. 11.215aWe also know that the planet's mass is its volume times its density. The volume we know from V = (4/3)πr^{3}. So we have
4 M =If you put the expressions for M and for dm into the physics formula given in equation 11.214 and gather the terms, you find that the energy, dE, needed to remove the thin shell of thickness dr from the planet is:πr^{3}ρ eq. 11.215b 3
16G dE =Of course once you have removed the outer shell, the planet is just a wee bit smaller. In other words, r is just a little less. To know how much energy we need to remove all of the planet's volume, we need remove thin shell after thin shell, until we get down to nothing, and we need to add up all the dE's that result. That is where integration comes in.π^{2}r^{4}ρ^{2} dr eq. 11.215c 3
If E(r) is the amount of energy that it takes to dismember a planet of radius, r, then
dE = E(r)  E(r  dr) eq. 11.216This equation helps us get our signs right when we do the definite integral. It shows that we need to take the energy function of the larger radius minus the energy function of the smaller radius. In other words, we will take the definite integral from the smaller radius to the larger radius.
But what is the energy function? Doesn't adding up all those dE's look a whole lot like taking a Riemann sum? So we do it by taking the integral. Remember that the volume of the shell in equation 11.215a was only an approximation? But the integral takes the limit as the shell's thickness, dr, goes to zero. And that in that limit, the approximation is perfectly accurate.
Since we are taking the sum over a bunch of different r's, the independent variable that we integrate over must be r. The indefinite integral of equation 11.215c is
E(r) = 
16G 
But 16/3, G, π, and ρ are all constants. By the rule established in equation 11.23, the above integral is the same as.
16G E(r) = 
r^{4} dr eq. 11.217b 
We've seen this form of integral before in equation 11.24f. Applying that formula to this we get
16 E(r) =This, remember, is only the indefinite integral. We determined earlier that we have to take the definite integral from the smaller radius to the larger radius. Since we want to dismember the entire planet and have nothing left when we are done, the smaller radius is zero. And since we start with the entire planet, the larger radius is the radius of the undisturbed planet, which I will call r_{A} for the radius of Alderaan.G π^{2}ρ^{2}r^{5} + C eq. 11.217c 15
r_{A} E = 0 
16G 
_{ } 
16 
r_{A} 0 
16 = 
Instead of carrying around all those symbols, we can use equation 11.215b to simplify the expression we have on the right. That equation gives us the mass of the planet as a function of r. If we square both sides of 11.215b we find that
^{ } 16 M^{2} =If M_{A} is the mass of Alderaan and r_{A} is its radius, the above becomesπ^{2}r^{6}ρ^{2} eq. 11.219a _{ } 9 9 M^{2}= 1 eq. 11.219b 16 π^{2}r^{6}ρ^{2}
9 M_{A}^{2}Since you can multiply anything by one and not change it, I will multiply the right side of equation 11.218 by the form of one that we have here in equation 11.219c. We get lots of good cancellations that way (getting rid of the π^{2} and the density and leaving us with a simple expression in terms of radius and mass):= 1 eq. 11.219c 16 π^{2}r_{A}^{6}ρ^{2}
3 GM_{A}^{2} E =This gives you the energy that the Death Star had to deliver in order to blow up Alderaan. I don't happen to know the mass or radius of Alderaan, but since I have always been deeply concerned that the Death Star might show up some day and smash our beautiful world, I have worked the numbers for the earth. From Astronomical Data Book by J. Hedley Robinson:eq. 11.220 5 r_{A}
M_{e} = 6.0 × 10^{24} kilograms (mass of the earth) r_{e} = 6.4 × 10^{6} meters (radius of the earth)My calculator tells me that for the earth, E = 2.3 × 10^{32} wattseconds. To give you some idea of how much energy that is, the total radiated output of the sun (not just what falls on the earth, which is only one part in 2.2 × 10^{9} of the sun's total output, but what goes off in all directions into space as well) is 3.9 × 10^{26} watts. So it would take about 590,000 seconds for the sun to radiate that much energy. That's about 4 hours short of a week. Looking at it another way, according to Einstein's famous equation, E = mc^{2}, (where c is the speed of light, about 3 × 10^{8} in meters per second), the Death Star would have to convert 2.5 trillion^{[1]} tons of matter completely to energy in order to destroy the earth. I doubt if the Death Star even weighs that much.
What it boils down to is that the numbers having to do with things astronomical (such as gravitational binding energy of a planet) are  well  astronomical.
I hope that puts your mind at ease. The earth is way too tough a planet for the Death Star. So stuff that into your shiny black helmet, Darth Vader.
^{[1]}That's the American version of a trillion, which is 10^{12} or equivalently a million millions.
Move on to Section 11.2: Simple Substitution of Variables
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