Thread: Chapter 7: Techniques of Integration

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Chapter 7: Techniques of Integration

Ok! Chapter 7! Here we go!

Now I'm starting on the infamous "Integration by parts," and when the book says "every differentiation rule has a corresponding integration rule," this really seems to clear a lot of things up for me, and makes it easier for me to really get a grasp on integration techniques. I think, "of course! Why didn't I think of that before?"

So the formula for integration by parts is
∫ f(x) g '(x) dx = f(x) g(x) - ∫ g(x) f '(x) dx
or
∫ u dv = uv - ∫ v du

Its so simple.

Now, I've done a few problems, but I really get the feeling that a couple of these sin and cos problems I did are wrong. Here they are:

I'm sure this one is right:

Evaluate the integral:
6) ∫ t sin 2t dt = A

My work:
u = t
dv = sin 2t
du = dt
v = (-1/2)cos 2t

A = (-t/2) cos 2t + ∫ (1/2) cos 2t
A = (-t/2) cos 2t + (1/4) sin 2t
-----------------------------------------------------------

7) ∫ x²sin πx dx

My work:
∫ x²sin πx dx = A

u = x²
dv = sin πx
du = 2x
v = (-1/π) cos πx

A = (-x²/π ) cos πx - ∫ (-2x/π) cos πx dx
∫ (-2x/π) cos πx dx= B

w = -2x/π
dz = cos πx
dw = -2/π
z = (1/π)sin πx

B = (-2x/π²) sin πx + ∫ (2/π³) sin π x dx
B = (-2x/π²) sin πx - (2/π³) cos πx
A = (-x²/π ) cos πx - B
so
∫ x²sin πx dx = (-x²/π ) cos πx - ( (-2x/π²) sin πx - (2/π³) cos πx )
or
∫ x²sin πx dx = (-x²/π ) cos πx + (2x/π²) sin πx+ (2/π³) cos πx

The book answer is:
(-x²/π) cos πx + (2x/π²) sin πx + (2/π³) cos πx
--------------------------------------------------------------

8 ) ∫ x² cos mx dx

My work:
∫ x² cos mx dx = A

u = x²
dv = cos mx
du = 2x
v = (1/m) sin mx

A = x² sin mx - ∫ (2x/m) sin mx dx
∫ (2x/m) sin mx dx = B

w = (2x/m)
dz = sin mx
du = 2/m
z = (-1/m) cos mx

B = (-2x/m² ) cos mx - ∫ (-2/m² ) cos mx
B = (-2x/m²) cos mx + (2/m³) sin mx
A = x² sin mx - ( (-2x/m²) cos mx + (2/m³) sin mx )
A = x² sin mx + (2x/m²) cos mx - (2/m³) sin mx

It seems to me that a general technique to use with integration by parts is to choose your "u" as a part of the integral that has a positive exponent and your "dv" as a part that has a negative exponent.

I originally missed a part of a substitution in 7, but I fixed it.

This one should be easy shouldn't it?

Evaluate the integral
9) ∫ ln (2x + 1) dx

I've been staring at it, and looking up old information, and staring at it willing the answer to spring forth, and all I have been able to come up with is the definition of ln x which is ∫₁^x (1/t)dt. Maybe what I have to do is change ln (2x + 1) into ∫₁^{2x + 1} (1/t)dt. Can it really be that difficult? This looks like it should be easy. I know I am missing something so small and simple that I should know right away.

Oh, there is a book example that shows this type of problem. So:

∫ ln (2x + 1) dx = A

u = ln (2x + 1)
dv = dx
du = (2 / (2x + 1) ) dx
v = x

A = x ln (2x + 1) - ∫ ( 2x / (2x + 1) ) dx
B = ∫ ( 2x / (2x + 1) ) dx

w = x
dz = ( 2 / (2x + 1) ) dx
dw = dx
z = ln ( 1 / (2x + 1) )

It doesn't look like we are getting anywhere. I think of what I am doing with my choosing u and dv as* moving one part back and one part forward. It doesn't seem like I can move a natural log (meaning ln) any more forward than it already is, only backward, but if it moves backward it doesnt seem solvable.

A = x ln (2x + 1) - ∫ ( 2x / (2x + 1) ) dx
B = ∫ ( 2x / (2x + 1) ) dx

you have to do Integration by parts one more time on B.

First, note that the second integration by parts you did literally undid the work you did with the first integration by parts. Always make sure that, if you're doing a second integration by parts, that you don't use your old v as your new u and your old du as your new dv--this will always undo your work!

Second, can you make a substitution after your first integration by parts? (Or a wacky idea... can you solve ∫ln(x)dx first and then use substitution to do ∫ln(2x+1)dx?)

Third, if you're unsure of your answers, just differentiate and see if you get the integrand back!

It seems that if I keep on going from where I left off I am left with something like this:

B = x ln ( 1 / (2x + 1) ) - ∫ ln ( 1 / (2x + 1) )
C = ∫ ln ( 1 / (2x + 1) )

It seems like I've gone from something more simple like
∫ ln (2x + 1)

to something more complicated
∫ ln (1 / (2x + 1) )

and that if I continue, it will get more so.

Ok, lets start this problem over.

Evaluate the integral
9) ∫ ln (2x + 1) dx

∫ ln (2x+1) dx = A

Substitution set one:
u = ln (2x+1)
dv = dx
v = x
du = (2 / (2x+1) ) dx

A = x ln (2x +1) - ∫ (2x dx ) / (2x +1)

∫ (2x)(2x + 1)^-1 dx = B

Substitution set two:
u = (2x + 1)^-1
dv = 2x
du = -2(2x+1)^-2
v = x²

B = x² / (2x+1) - ∫ (-2x² dx) / (2x + 1)²

This first way is with two successive Integration by parts, and appears to be a continuous loop. This next second way is an attempt to find an appropriate substitution to solve B.

B = ∫ (2x)(2x + 1)^-1 dx

Substitution set three:
u = 2x + 1
du = 2 dx
x = (u-1)/2

B = (1/2) ∫ (u-1)u^-1 du
B = (1/2) ∫ u/u du - u^-1 du
B = (1/2) ∫ u/u du - (1/2) ∫ u^-1 du
B = (1/2) ∫ du - (1/2) ∫ u^-1 du
B = (1/2) u - (1/2) ln u
B = (1/2)(2x + 1) - (1/2) ln (2x+1)
A = x ln (2x +1) - B
A = x ln (2x+1) - (1/2)(2x + 1) - (1/2) ln (2x+1)

The book answer is (1/2)(2x + 1) ln (2x+1) - x

Your answer is:

x ln (2x+1) - (1/2)(2x + 1) - (1/2) ln (2x+1)

The book's answer is:

(1/2)(2x + 1) ln (2x+1) - x

The difference (literally) between these two is:

(1/2)(2x + 1) ln (2x+1) - x - (x ln (2x+1) - (1/2)(2x + 1) - (1/2) ln (2x+1))

= (1/2)(2x+1-2x-1)ln(2x+1)-x+x-1/2 = -1/2

So your answers differ by a constant. You were solving an indefinite integral, i.e. you were taking an antiderivative. So... what's the deal?

Let me reiterate: if you're unsure of your answer, check by differentiating. If you get the original thing back, you must be right.

Just a little note on my progress. I've completed 40 7.1 Integration by parts problems and 40 7.2 Trignometric Integral problems, and I'm currently working on 7.3 Trignometric Substitution.

Sounds good! It seems that you found 7.2 to be pretty easy?

I got stuck on some problems in the form of sinⁿx tan^mx, and cosⁿx cot^m x, but after working some more on them, I think I figured them out. I used integration by parts and seemed to get the original integral inside the integration by parts side of the equal sign. Going by the examples the book gave in 7.2, they didn't mention that they did this, so I am not sure if that is what I should have done or not.

A lot of these integrals can be done by simple substitution. I'm not seeing how integration by parts will be useful in general--it looks like doing so gives you relationships between different trigonometric integrals, but most of them don't seem to give you back the original integral.

Ok, I wasn't being too specific. Here is an example of how I am doing these types of problems. This is a 7.3 Trigonemetric Substitution problem actually, but it ends up the same way as these cos cot problems.

1) Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle.

∫ 1/(x²(x² - 9)^1/2) dx; x = 3 sec Θ

∫ (9sec² Θ (9sec² Θ - 9)^1/2)^-1 dΘ = A
∫ (9 sec ²Θ(9(sec²Θ - 1) )^1/2 )^-1 dΘ
∫ (9 sec²Θ(9 tan² Θ)^1/2)^-1 dΘ
∫ ( 9 sec² Θ 3 tan Θ )^-1 dΘ
∫ ( 1/9 )( 1 / sec² Θ)( 1/3 )( 1/tan Θ) dΘ
∫ ( 1/27 ) cos² Θ cot Θ dΘ
( 1/27 ) ∫ cos² Θ cot Θ dΘ

This is what I mean by the cos cot type. This is how I work this one out, and I think how I did the others as well.

1/27 ∫ (1 - sin² Θ) cot Θ
1/27 ∫ cot Θ - sin²Θ cot Θ
(1/27) ln (sin Θ) - ( 1/27 ) ∫ sin Θ cos Θ
∫ sin Θ cos Θ = B

Substitution set 1
u = sin Θ
dv = cos Θ dΘ
du = cos Θ dΘ
v = sin Θ

B = sin² Θ - ∫ sin Θ cos Θ
B = sin² Θ - B
2B = sin² Θ
B = (1/2) sin² Θ

Maybe I could have figured that out without using integration by parts.

A = (1/27) ln (sin Θ) - (1/27) B
A = (1/27) ln (sin Θ) - (1/27) (1/2) sin² Θ
A = (1/27) ( ln (sin Θ) - (1/2) sin² Θ )

and now is the part of the problem were I think "what a silly way to solve a problem." Its also the part where I realize that I made a mistake by using x for the whole problem and have to change my previous work to terms of Θ for it to make sense.

x = 3 sec Θ

What I would like to do is get sin Θ alone on one side.

x/3 = sec Θ
x/3 = ( cos Θ )^-1
3/x = cos Θ
3/x = (1 - sin²Θ )^1/2
9/x² = 1 - sin²Θ
1 - 9/x² = sin²Θ
(1 - 9/x²)^1/2 = sin Θ

and now we substitute.

A = (1/27) (ln (1 - 9/x² )^1/2 - (1/2)( 1 - 9/x² )
A = (1/27) ((1/2) ln (1 - 9/x²) - (1/2) (1 - 9/x² )
A = (1/54 ) ( ln ( 1 - 9/x² ) + 9/x² - 1 )

Which of course is not expressed in the same way the book answers:
(x² - 9)^1/2 / (9x)

but I'm sure it is correct.

I also realize that I did not exactly follow the directions.

You have to remember to change the differential when you use trig substitution! In this case, since x = 3 sec Θ, we have dx = 3 sec Θ tan Θ dΘ. So you have to put this into your integral.

*smacks forehead*

What a silly way to solve a problem!

*prepares for another hour of problem solving*

I'm sure I'll get used to it though.

1) ∫ (x²(x² - 9)^1/2)^-1 dx = A; x = 3 sec Θ, dx = 3 sec Θ tan Θ dΘ
∫ x^-2(x² - 9)^-1/2 dx
∫ (3 sec Θ)^-2( (3 sec Θ)² - 9)^-1/2 3 sec Θ tan Θ dΘ
∫ 1/9 sec^-2Θ ( 9 sec² Θ -9)^-1/2 3 sec Θ tan Θ dΘ
∫ 1/3 sec^-1( 9 (sec²Θ - 1) )^-1/2 tan Θ dΘ
∫ 1/9 cos Θ (tan²Θ)^-1/2 tan Θ dΘ
∫ 1/9 cos Θ dΘ
A = 1/9 sin Θ

and we know previously that sin Θ= (1 - 9/x²)^1/2

so
A = 1/9 (1 - 9/x² )^1/2

or
A = 1/9 ( ( x² - 9 ) / x² )^1/2
A = 1/9x (x² - 9 )^1/2
A = (x² - 9)^1/2 / 9x

which is the book answer.

I've found something I am confused about. With the 7.3 information, when you use trignometric substitution to solve a problem, what do you do when you have to use a double angle identitty to solve an integral, and then are left with just your Θ, and no trignometric operation. This problem 5 is an example:

5) ∫_{sqrt 2}² t^-3(t² -1)^-1/2 dt

t = sec Θ
dt = sec Θ tan Θ

∫ sec^-3Θ tan^-1 Θ sec Θ tan Θ
∫ sec^-2Θ
∫ cos²Θ
∫ 1/2 + 1/2 cos 2Θ
Θ + 1/4 sin 2Θ

I can figure sin out by constructing a triangle, but how do I show the value of Θ in terms of t? Should I use inverse trig to complete this? Should I say that
Θ = arcsec t

Yup, inverse trig functions are the way to go.

Ok, I completed that probelm 5. I seem to have gotten 7 and 9 wrong though. I'll post my work on 9.

9) Evaluate the integral.
∫ (x² + 16)^-1/2 dx

my work:
x = 4 tan Θ
dx = 4 sec² Θ dΘ
tan Θ = x/4

∫ (16(tan² + 1) )^-1/2 4 sec²Θ dΘ
1/4 ∫ sec^-1Θ 4 sec²Θ dΘ
∫ sec Θ dΘ
ln (secΘ + tan Θ)

At this point I am going to kind of abbreviate (x² + 16)^1/2 with a "w".

ln (w/4 + (x/w)(w/4) )
ln (w/4 + x/4)
ln ( (w + x)/4 )

The book answer is ln(w + x). Did I miss a couple two's somewhere?

Your answer is right, as is the book's. Take the derivative of either and you'll see that you get the integrand back. So this begs the questions...

1. When can two functions have the same derivative?

2. Given this, how can the two expressions satisfy whatever condition you find for 1?

Before I answer those two questions I'll say:

I havn't worked out the derivitive for both, but even if I did get the same integrand back, I dont see how these two answers can both be right. It seems obvious to me that ln ( (w+x)/4) could not equal ln (w+x) except for very specific values. If you take my natural log apart you get ln (w+x) - ln (1/4), and ln (1/4) is equal to something around -1.38, not 0, the only value that would make these equal to eachother.

1) Two functions should have the same derivitive only when they have the same rate of change. For example y = 2x + 60 and y = 2x - 90...

Oh... So if I ever integrate any function and get a number that has no variable assigned to it such as that 60 or -90, it really has no signifigance and I can just drop it off. Of course I could just represent the "whatever value" with the old mister C.

Wait, I realize that some of what I just posted doesn't make sense... I guess I am confused for the moment.

Okay, so remember that when you're taking an indefinite integral, you're trying to find all antiderivatives of the integrand. So the answer is only well-defined up to a constant. ln(4) is a constant.

1) Two functions may have the same derivitive when they both have the same rate of change. For example with the equations f(x) = 2x+60, and f(x) = 2x-90, both functions have a derivitive of 2. If you took the integral of 2, you would get 2x + C. This can represent all functions that are represented by 2x plus a number not associated with a variable. If we find that the integral of a particular function is 2x + 5 + C, we could just rewrite that as 2x + C, because C represents all numbers not associated with a variable including our 5.

Ok, I get it. I just had to type it out so I could think slower. hehe

Exactly. You can always absorb constants into the C.

In the middle of problem 15, I believe I am to find the integral of tan²Θ dΘ. I set this integral equal to B. Integral by parts gives me:

Substitution set #1
u = tan² Θ
dv = dΘ
du = 2 tan Θ sec² Θ dΘ
v = Θ

B = Θ tan²Θ - 2∫ Θ tan Θ sec²Θ dΘ
C = ∫ Θ tan Θ sec²Θ dΘ

Substitution set #2
u = Θ sec Θ
dv = tan Θ sec Θ dΘ
du = sec Θ dΘ + Θ sec Θ tan Θ dΘ
v = sec Θ

C = Θ sec²Θ - ∫ (sec²Θ + Θ sec>sup>2</sup>Θ tan Θ) dΘ
C = Θ sec²Θ - tan Θ + ∫ Θsec²Θ tan Θ dΘ
C = Θ sec²Θ - tan Θ + C

oh no, how do I find C?

You flipped a sign in your work:

C = Θ sec²Θ - ∫ (sec²Θ + Θ sec²Θ tan Θ) dΘ
C = Θ sec²Θ - tan Θ - ∫ Θsec²Θ tan Θ dΘ
C = Θ sec²Θ - tan Θ - C

Also, there's an easier way to do tan² Θ: use tan² Θ = sec² Θ - 1.

Wow, can't believe I missed that.

I run into ∫ sec Θ pretty often, and it always takes me a long time to figure it out. I just typed up a proof that I thought would work, but I realized at the end that it failed.

run into ∫ sec Θ pretty often, and it always takes me a long time to figure it out. I just typed up a proof that I thought would work, but I realized at the end that it failed.

Oops. I mean ∫ csc Θ dΘ

I've worked this out a bit and I cant find a way to integrate it. Changing this to sin^-1Θ, and using integration by parts along with the trig identity sin² + cos² = 1 seems to always give an integral that increases the exponent on its theta every attempt and changes between sin^-1 to cos^-1 every attempt. Meaning it seems to never give an integral that you had previously worked out.

Okay, so you know that the derivative of ln|tan x + sec x| is sec x. We can compute this explicitly:

(d/dx)(ln|tan x + sex c|) = (sec² x + sex x tan x)/(sec x + tan x) = sec x

We can construct the antiderivative of csc x using this as motivation and the facts that:

(d/dx) csc x = -csc x cot x

(d/dx) cot x = -csc² x

So we have:

(d/dx)(csc x + cot x) = -csc x cot x -csc² x = -csc x(cot x + csc x)

This looks a lot like what happened above: we take the derivative of a function and get another function times the original function. And this extra function is the function we want to integrate. And so:

(d/dx)(ln|csc x + cot x|) = -csc x(cot x + csc x)/(csc x + cot x) = -csc x

Thus the antiderivative of csc x is -ln|csc x + cot x|.

This isn't really a constructive way of doing this. But the book must have some calculation of how you can integrate sec x, and I bet you can use this to motivate a method to integrate csc x.

Thanks for that.

Anyway, on problem 24, I believe that I am to, at one point, take the integral of csc Θ sec Θ, but I don't see how. Here is my attempt to do it:

∫ csc Θ sec Θ dΘ = A

Set #1
u = csc Θ
dv = sec Θ
du = -csc Θ cot Θ
v = ln (sec Θ + tan Θ)

A = csc Θ ln (sec Θ + tan Θ) + ∫ csc Θ cot Θ ln (sec Θ + tan Θ) dΘ
∫ csc Θ cot Θ ln (sec Θ + tan Θ) dΘ = B

This looks real messy, but I'll keep on going.

Set #2
u = csc Θ cot Θ
dv = ln (sec Θ + tan Θ)
du = -csc Θ cot² Θ - csc³Θ
v =

Hmm... I dont see how I can do v

This one is a little tricky, but I'd like to let you mull on it before giving you too much of a hint. Here are some things to think about:

1. When you're given a product of trig functions like this, express them in terms of sines and cosines and see if you can do some twiddling about that makes the expression look easier to work with.

2. You can do this by substitution. There are a lot of different potential substitutions, some of which work, and a couple of which work very nicely. If your first attempt doesn't work, try again.

Hmm... I have an idea but I'm not sure if it will work. Let's try it.

∫ sin^-1Θ cos^-1Θ dΘ
2 ∫ 1/2 sin^-1Θ cos^-1Θ dΘ
2 ∫ (2 sin Θ cos Θ)^-1 dΘ
2 ∫ (sin 2Θ)^-1 dΘ
2 ∫ csc 2Θ dΘ
-ln (csc 2Θ + cot 2Θ)

How is that?

That's great! I found another way:

∫1/(sin Θ cos Θ) dΘ =
∫cos Θ/(sin Θ cos² Θ) dΘ =
∫sec² Θ/tan Θ dΘ

Let u = tan Θ, so du = sec² Θ dΘ:

∫(1/u) du = ln|u| + C = ln|tan Θ| + C

Our answers look a bit different, but you can use the trig identity cot Θ = csc 2Θ + cot 2Θ to see they agree.

26) ∫ x²(4x - x²)^-1/2 dx
∫ x²(-(x² - 4x) )^-1/2 dx
∫ x²(-(x² - 4x + 4) - 4)^-1/2 dx
∫ x²(-4 - (x-2)²)^-1/2 dx

z = x - 2
dz = dx
x = z + 2

∫ (z + 2)²(-4 - z²)^-1/2 dz
(1 / i )∫ (z + 2)²(z² + 4)^-1/2

Should I really be creating an imaginary number to solve this problem?

Check your arithmetic when you complete the square.

Oops, I see a mistake.

∫ x²(4x - x²)^-1/2
∫ x²(4x - x² + 4 - 4)^-1/2
∫ x²(4 - (x² - 4x + 4) )^-1/2
∫ x²(4 - (x - 2)² )^-1/2

I've moved onto 7.4: Integration of Rational Functions by Partial Fractions now. 7.3 was pretty difficult for for me, but I feel much more comfortable with it now after completing 30 excersises.

I don't think I've ever seen much of the kind of stuff that's in this next section although it looks like something you would learn in say Algebra 2 or so. I might have some real easy questions.

5a) Write out the form of the partial fraction decompostion of the function (as in Example 7). Do not determine the numerical values of the coefficients.
x⁴/(x⁴ - 1)

Now based on what I've read, it looks like the answer would have to be:
A/(x-1) + B/(x+1) + (Cx + D)/(x²+1)

book answer:
1 + A/(x-1) + B/(x+1) + (Cx + D)/(x²+1)

Why does the book add a one to this answer?

Ok, I see that the first step is really to divide the numerator by the demoninator if the numerator is larger, but the long division example I don't understand too well. I'll do my best to try to represent the problem and why I might have difficulty with it.

f(x) = P(x)/Q(x) = S(x) + R(x)/Q(x)

P(x) = x³ + x
Q(x) = x - 1

Because it is long division with varibles that I am unsure about, I think it is best at this time to write it out like this even though it looks like crap:

.......x²...........
x-1 |x³...... + x
.......x³ - x²
.............x² + x

I would think that at this point you would have -x² + x since I think you are subtracting x³ - x² from x³ + x. Oh I see... nevermind I think I've got it.

Looks like you answered your own question. Just remember that the first step is always to compare the degrees of the numerator and the denominator. If the top is bigger than or equal to the bottom, you've gotta divide!

How about this one?

15) Evaluate the integral.
∫₀¹ (2x + 3)/( (x+1)²)

It doesn't look as if I have to divide since if we expanded out the demoninator, it would be to a larger degree than the numerator, so we can just jump to the next step:
∫ (2x + 3)/( (x+1)²) = (Ax+B)/(x+1) + (Cx+D)/( (x+1)²)

2x+3=(Ax+B)(x² + 2x + 1)+(Cx+D)(x+1)
2x+3=Ax³+Bx²+2Ax²+2Bx+Ax+B+Cx²+Cx+Dx+D
2x+3=x³(A) + x²(B+2A+C) + x(2B+A+C+D) + (B+D)

So then we can make these four equations:
#1) A = 0
#2) B+2A+C=0
#3) 2B+A+C+D=2
#4) B+D=3

Since we know A=0, then we know from equation 2 that:
B+C=0
B= -C

Substituting this into equation 4 we find:
-C + D=3
D = 3+C

Going back to equation 3, we can substitute the values of A, B, and D to find the value of C.
2(-C)+(0)+C+(3+C)=2
-2C + C + 3 + C =2
3=2

So where did I make a mistake?

The terms in the numerators should be constant, not linear. You only need a linear term if the denominator is an irreducible quadratic (i.e., quadratic equation with negative discriminant/complex roots) or a power thereof. However, even though you introduced unnecessary terms, you should have been able to proceed and find the right answer. The problem was in this step:

(2x + 3)/(x+1)² = (Ax+B)/(x+1) + (Cx+D)/(x+1)²

2x+3 = (Ax+B)(x²+2x+1)+(Cx+D)(x+1)

The italicized terms are incorrect. Try it again a little more carefully... What happens when you multiply the first equation by (x+1)²?

Ok. I think I am getting it, but then I got this next problem wrong. Why is this?

17) Evaluate the integral:
∫₁² (4y²-7y-12)dy/(y(y+2)(y-3))

My work:
∫₁² (4y²-7y-12)dy/(y(y+2)(y-3)) = A
Since the demoninator is to a larger degree we need not divide first, so we move on to this expansion:
A = ∫ (a/y) + (b/(y+2)) + (c/(y-3))

4y²-7y-12=a(y+2)(y-3)+by(y-3)+cy(y+2)
4y²-7y-12=a(y²-3y+2y-6)+by²-3by+cy²+2cy
4y²-7y-12=ay²-ay-6a+by²-3by+cy²+2cy
4y²-7y-12=y²(a+b+c)+y(-a-3b+2c)+(-6a)

based on this, we can make the following three equations:
#1) a+b+c=4
#2) -a-3b+2c=-7
#3) -6a=-12

from equation 3, we can find that
a=2

we can substitute this into equation 2:
-2-3b+2c=-7
-3b+2c=-5
2c=3b-5
c=(3/2)b-(5/2)

we can then substitute these two values of a and c into equation one:
2+b+(3/2)b-5/2=4
(5/2)b-1/2=4
(5/2)b=9/2
5b=9
b=9/5

going back to our value of c in terms of b we find:
c=(3/2)b-(5/2)
c=(3/2)(9/5)-(5/2)
c=27/10 - 25/10
c=2/10
c=1/5

in summary, our a,b,c values:
a=2
b=9/5
c=1/5

so we can go back to our original integral and enter in these constants in our expansion:
∫ 2/y + (9/5)/(y+2) + (1/5)/(y-3)
∫ 2y^-1 + 9(5y+10)^-1 + (5y-15)^-1

we can then integrate this easily:
2 ln y + 9 ln (5y+10) + ln (5y-15)
ln y² + ln (5y+10)⁹ + ln (5y-15)
= ln (y²(5y+10)⁹(5y-15))
= ln ((5y+10)⁹(5y³-15y²))

we can then enter in the values of 2 and 1 to find the definite integral
ln ((20)⁹(40-60)) - ln ((15)⁹(5-15))
ln ((5*2²)⁹(-5*2²))-ln(5⁹*3⁹*-5*2)
ln (-5¹⁰2²⁰)-ln(-5¹⁰3⁹2)
ln ((-5¹⁰2²⁰)/(-5¹⁰3⁹2))
ln (2¹⁹/3⁹)
ln (2¹⁹3^-9)

The book answer is
9/5 ln (8/3)

I've played around with this a little to get a more comparable answer:
9/5 ln (8/3)
ln (8/3)^(9/5)
ln (2³3^-1)^9/5)
ln ((2^27/153^-9/5)

I know from seeing this that this is a different answer than mine, but I'll continue on anyway:
ln (2¹⁹3^-9)=ln (2^27/153^-9/5)
2¹⁹3^-9=2^27/153^-9/5
2²⁸⁵3^-135= 2²⁷3^-27
2²⁵⁸3^-108=1

And this seems to be more definite proof that my answer is incorrect.

Alright, so your problem is in integrating the terms (5y+10)^-1 and (5y-15)^-1. You're trying to use a shortcut, which is a good thing, but you have to make sure that your rule of thumb is correct. You're right that if we integrate the reciprocal of a linear term, we basically get the log of that term. However, it's generally a constant times that term. You can figure out what it should be by using substitution. Try, for example, integrating the first guy with the substitution u = 5y+10, so that du = 5dy.

Ok. I knew there was something wrong with that, but I thought the process I used was right, so I wasn't sure why. I wasn't giving enough thought to dy.

So the trick you were using is that integrating something of the form f(ax+b) is as easy as integrating something of the form f(x). If F(x) is an antiderivative of f(x), we always have F(ax+b)/a is an antiderivative of f(ax+b)--you can check either by integrating with substitution or by taking derivatives and using the chain rule. (In fact, substitution and the chain rule are really "opposite processes" in that you can show substitution from the chain rule and vice versa.) In any case, you just have to remember to divide by the coefficient on x.

I don't believe I've made the same mistake on 19, but it seems I have made a mistake. I just don't know what it is.

19) Evaluate the integral.
∫ dx/((x+5)²(x-1))

My work:
∫ dx/((x+5)²(x-1)) = A
A = ∫ a/(x+5) + b/(x+5)² + c/(x-1)
1=a(x+5)(x-1)+b(x-1)+c(x+5)²
1=aX²+4ax-5a+bx-b+cx²+10cx+25c
1=x²(a+c)+x(4a+b+10c)+(-5a-b+25c)

#1)a+c=0
#2)4a+b+10c=0
#3)-5a-b+25c=1

a+c=0
a= -c

4a+b+10c=0
-4c+b+10c=0
b+6c=0
b=-6c

-5a-b+25c=1
5c+6c+25c=1
31c=1
c=1/31

a= -c
a= -1/31

b= -6c
b= -6/31

A = ∫ (-1/31)(x+5)^-1 + (-6/31)(x-5)^-1+(1/31)(x+1)^-1
A = (1/31)(-ln(x+5)+6(x+5)^-1+ln(x-1))
A = (1/31)( ln ((x-1)/(x+5)) + 6(x+5)^-1)

book answer is:
(1/36)(ln((x-1)/(x+5)) + 6(x+5)^-1)

It would seem to me that the book answer suggests that my constant values are incorrect.

5+6+25=31?

no

31) Evaluate the integral.
∫ dx/(x³ -1)

I don't know how to start this one. I could split the denominator like this:
∫ dx/( (x^3/2+1)(x^3/2-1) )

but then I'm not really sure how to do the expansion with fractional exponents.

Yeah, partial fractions only work with integral (i.e., integer) exponents.* And we only know how to do partial fractions if the denominator is a product of linear and irreducible quadratic terms. Well, we have the tremendous fact:

Any polynomial with real coefficients factors into linear and irreducible quadratic polynomials with real coefficients.

So now all you have to do is factor the denominator into said things. Since our denominator is cubic, it must either split as:

(linear)(linear)(linear)

or:

(linear)(irreducible quadratic)

In any case, it must have a linear factor, i.e. it must have a real root. So... can you think of a root of x³-1? After you factor out this linear term, you're left with a quadratic, which you can solve via the quadratic formula (and, if the roots are complex, you just leave it alone).

-------------------------------------------------------------------------------

*Actually, this might not be true, but for our intents and purposes, it is.

Well... I knew that one root had to be (x-1), but I wasn't sure what the rest of it would be. Could I have used that algebraic division to find the rest of it?

I did figure this problem out though just now. I had an idea to use a couple substitutions. The first one, since I was working with
x³-1

I tried to make x equal to something that had a lone integer of 1 so they would cancel eachother out. I figured that I could do this:
x=z+1

so that
x³-1

turns into
z³+3z²+3z
z(z²+3z+3)

So I expanded my integrand to
A = ∫ a/z + (bz+c)/(z²3z+3)

and found
a = 1/3
b = -1/3
c = -1

Entering in these values:
A = ∫ (1/3)/z + ((-1/3)z -1)/(z²+3z+3)

then I thought I would have to rearrange the numerator of the second fraction:
A = (1/3)ln z - (1/6)∫(2z+6)/(z²3z+3)
A = (1/3) ln z - (1/6)∫(2z+3+3)/(z²+3z+3)
= (1/3) ln z - (1/6)∫(2z+3)/(z²+3z+3)+3/(z²+3z+3)
= (1/3) ln z - (1/6)ln(z²+3z+3)-(1/6)∫3/(z²+3z+3)
= (1/3) ln (x-1) - (1/6) ln (x²+x+1) - (1/2)∫1/(x²+x+1)
= (1/3) ln (x-1) - (1/6) ln (x²+x+1) - (1/2)∫1/( (x+(1/2) )² + (3/4) )

u = x+(1/2)
du=dx

= (1/3) ln (x-1) - (1/6) ln (x²+x+1) - (1/2)∫1/(u² + (3/4) )
= (1/3) ln (x-1) - (1/6) ln (x²+x+1) - (1/(3^1/2) tan^-1(2u/(3^1/2) )
= (1/3) ln (x-1) - (1/6) ln (x²+x+1) - (1/(3^1/2) tan^-1((2x+1)/(3^1/2) )


The book answer is:
(1/3) ln (x-1) - (1/6) ln (x²+x+1) - (1/(3^1/2) ) tan^-1( (2x+1)/(3^1/2) )

It looks like they agree. I am surprised at myself for being able to do that one.

and, if the roots are complex, you just leave it alone

33) Evaluate the integral.
∫₂⁵ (x²+2x)dx/(x³+3x²+4)

x³+3x²+4 does have a real root, but it looks like this:
-1 - 1/( (3 - 2(2^1/2) )^1/3 - (3 - 2(2^1/2) )^1/3

That's kind of nuts, should I really use this?

Making the substitution x = z+1 was a great idea. Unfortunately, this won't work if you're dealing with something which has two linear factors. You're right in suggesting that division is the best way to go. Personally, I find polynomial division to be kind of fun, so I always jump at the chance to do it. Seriously.

(Oh, and a little algebraic tip: a³ - b³ always factors as (a - b)(a² + ab + b²).)

Don't be surprised that you made it through the problem unscathed--take it as encouragement! This shows that your knowledge and skills are improving! You were able to come up with a neat trick to overcome the hurdle of factoring; you successfully navigated the partial faction decomposition; and you managed to integrate each term. Integrating the irreducible quadratic term usually throws people for a loop, but you saw to complete the square to make it look like the derivative of arctan.

Thanks. And for problem 33 I see that I made a mistake that seems common to myself after I finish a problem that seemed difficult. My brain automatically tries to find a difficulty tricky solution to the next problem, however for this next one, the solution is much easier. I just multiply by three and it can be integrated.

Yeah, I guess that might be "step 0" for partial fractions: see if you can just do a substitution first.

I do have a question that I should know about natural logs but don't.

If I have
ln x - ln y

I know that I could write that as
ln (x/y)

but if I had
a ln x - b ln y

would that mean:
(a/b) ln (x/y)

or would I first have to distribute the exponents?

Do the exponents first. You should get ln(x^a/y^b).

So, I just finished 50 of the 7.4 problems and I'm moving on to the next section, 7.5 Strategy for Integration, but I need a long break first. Some of these partial fraction problems take quite a lot of effort it seems.

The partial fractions problems really require a lot of techniques--polynomial division, factoring polynomials, solving simultaneous equations, substitution, completing the square, trig substitution, trig integrals... It's really a good example of how math builds upon itself, and it shows why you need to get each idea down before moving to the next.

Take a breather! The next section is basically review of the previous ones, except you'll be expected to figure out which technique you need to use without prompting, and you'll also be expected to use multiple techniques in one integral.

14) ∫ (1+ln x)^1/2dx/ (x ln x)

I've used a variety of methods and narrowed it down to this:

lnx + 2x (ln x)^1/2 + ∫ 1/(ln x)^1/2

but is it possible to integrate ∫ 1/(ln x)^1/2? The book says that there are some functions that can't be integrated with what I know now. One of the functions it mentions is ∫ dx/ln x.

Assuming everything you did to get to this step is correct, then yes, you must be able to integrate that term. Off the top of my head, I can't think of a way to integrate it, though. If your work is right, then it's a real nasty integral, as I was able to compute the original integral and, well, it involves functions which look a lot different from what you have.

So how did I manage to do it? When you have something really ugly like this integrand, the first thing to look for is usually a substitution. It's tempting to try some sort of integration by parts (which is what I'm assuming you did at some point, perhaps after a substitution), but doing IBP first doesn't make sense, as you'll definitely have to use a substitution (or the chain rule, which is the opposite of substitution) to calculate your du and v. Since you're going to have to substitute at some point, you may as well do it first.

Now there's no obvious substitution that makes things real pretty. I mean, there is one obvious substitution, which is u = ln x, which gives you du = dx/x, and this will change your integrand to sqrt(1+u) du/u, which already looks a lot nicer, but this still isn't something we immediately know how to integrate. You can start playing around with this prettier integral--here it may actually make sense to try IBP or something, or you may have to do another substitution.

Or you can use the following strategy: do substitution, and pick u to be the ugliest part of your integral. This often works well if your integral is pretty well behaved except for one pesky term. The 1/x ln x part is pretty well behaved--this function alone is pretty easy to handle. The thorn in our side is the sqrt(1+ln x) term. So I'd suggest trying u = sqrt(1+ ln x). WARNING: the substitution is not a cakewalk. You're going to have to work to compute du, or rather, you're going to have to work to make du look like a piece of your integral. And you may have to work to get rid of all of the x's in your integral even after you figure out how to deal with du. And, after all's said and done, the integral will still require some more massaging.

Your question now is probably, "How do I know when to stop going down one route and trying another?" Well, both you and I manage to make our integrals look nicer via some sort of IBP or substitution. In my work, I manage to get it down to something simple I know how to integrate. In your case, you get it down to something simple you don't know how to integrate. If you had to do a lot of work to get to a simple-looking integrand, and if you don't see how to deal with that term, it's likely that working out that integral itself is a massive task, and it's likely that doing so will involve undoing all of the work you did up to that point. Personally, I tried a couple of quick IBP methods on 1/sqrt(ln x), noticed that the new integral wasn't getting any easier, and decided to start over.

This is hard. This is frustrating. You're likely to reach a couple of dead ends on any given problem in this section before hitting upon a method that works. You just have to get down the techniques in previous sections, study the strategies suggested by Stewart, do a lot of problems, and persevere while doing each one. You'll build intuition for what works and what doesn't. You'll begin to see what sequence of steps you're likely to have to take in order to calculate an integral. I don't mean to scare you with all of this... I imagine it's easy to get bogged down by this working on your own, though, so I figure preemptively telling you that it won't necessary be easy will at least give you the motivation to keep working if you don't seem to be getting anywhere.

Well... what I saw first was that I might be able to make things easier by trig substitution on the
(1+ln x)^1/2

and turn it into a
sec Θ

This seems to turn the problem into
2∫ sec³Θ csc Θ = A

which seems much easier. Then I used a trig identity to make
A = 2∫ tanΘ sec²Θ + secΘ cscΘ
A = tan²Θ + 2∫ secΘ cscΘ

then I saw how part of this would work out nicely if I converted back to x
A = ln x + 2∫ (ln x + 1)/((ln x)^1/2)
∫ (ln x + 1)/((ln x)^1/2) = B

then I did another substitution, this time
z = (ln x)^1/2

and got
B = 2∫ z²e^{z^2}dz + 2∫ e^{z^2}dz
2∫ z²e^{z^2}dz =C

I saw that C could be greatly simplified using IBP and got
C = ze^{z^2} - ∫ e^{z^2}dz

Previously I knew that
B = C + 2∫ e^{z^2}dz

so then
B = ze^{z^2} + ∫ e^{z^2}dz

I was really hoping that those would cancel eachother out, but no such luck. I thought maybe if I converted back again to x, that I might see something that works, and that's where I arrive to
A = ln x + 2B

or
A = ln x + 2x( ln x)^1/2 x + ∫ dx/((ln x)^1/2)

What I could do is concentrate on integrating
∫ sec Θ csc Θ

I am pretty sure I've done that before, I just don't remember how. I can convert it into
∫ sin^-1 2Θ dΘ

but I would need a cos from somewhere to finish it.

I found the post where I figured it out.

∫ sin^-1 2Θ dΘ
∫ csc 2Θ dΘ
(1/2) ln (csc 2Θ - cot 2Θ)

so if i go back to where I had
A = tan²Θ + 2∫ sec Θ csc Θ dΘ

then I get
A = tan²Θ + ln (csc 2Θ - cot 2Θ)

and now its just a matter of changing the 2Θ's back to Θ's, and converting back to x.

Originally Posted by Demen Tolden

Well... what I saw first was that I might be able to make things easier by trig substitution on the
(1+ln x)^1/2

and turn it into a
sec Θ

This seems to turn the problem into
2∫ sec³Θ csc Θ = A

I got an exponent of 2 instead of 3 on the secant (or, alternatively, I got a cotangent instead of a cosecant). In any case, though, this is a pretty fantastic idea--you're really combining a lot of methods into one step, which shows a lot of ingenuity and understanding of the integration techniques.

Well, I've done 64 of these 7.5 problems. I seem to be going through them much faster and it is getting easier I think. I've been stuck on this next one though for about 2 hours, and I can't seem to figure it out.

65) dx / ( (x+1)^1/2 + x^1/2 )

I've been trying hard to find a way to combine the denominator into something more workable, but I have not been successful. The most promising I have done with this is:
x = tan² Θ

which makes:
∫ ( 2 tan Θ sec Θ dΘ) / (1 + sin Θ)

but I don't see how I can go any further this way.

This one's kind of a trick. I don't know if any of the techniques you learned in the previous sections will help, and the way I found involves something you learn prior to calculus. Do you remember "rationalizing the denominator"?

Wow, thanks a lot. I looked up "rationalizing the denomiator," and understood what you meant and finished the problem in about 30 seconds.

Just so you know, I tried a number of substitutions before noticing the trick. And then I slapped my head.

Yup, I do that a lot it seems.

I finished all 81 of the 7.5 integrals. 7.6 is "Integration Using Tables and Computer Algebra Systems." The book claims to have a table of integrals contained within its reference pages, but either the previous book owner removed this, or this is well hidden somewhere within the book. There is no observable evidence that there have been any removed pages. Perhaps I should skip this section if I can't find these tables?

That would bring me to 7.7, Approximate Integration, which seems to be further work with the Reiman Sum idea. Is this useful? Should I skip this as well?

The table of integrals should be at the very back of the book, printed on slightly thicker paper. It's likely that the previous owner ripped them out, as they're pretty useful. You can look for "list of integrals" or something on Wikipedia, though, and you should be able to find all or more of the integrals listed in Stewart. Or you can search the net for scans of the Stewart tables. We never cover 7.6, but it could be informative to plug in some of the integrals into Maple or whatever and see what they give.

We do cover 7.7. It's not the most interesting, and it's not really the most useful--if you need to approximate a hard integral, you're probably going to let your computer do it. But your computer may do it... using methods described in this section. And it's good to understand what's going on behind the scenes. So I'd at least skim it.

7.8, however, is definitely important!

Yea, 7.8 does seem extemely necisary, but there is something that bothers me about it. I understand that if you work out the integral of x^-1 that it comes out to be f(x) = ln x, and if you assign an infinitely large number to x, the function f(x) gets infinitely large, meaning the area between the x axis and y = x^-1 is infinite, but it doesn't seem to make much sense to me that this can be possible when you are not dealing with a function that has no rate of change. Certainly if the y values of the x axis and y = x^-1 get closer and closer the larger the value of x is, the area between the two much approach something.

Oh well, I guess I just have to accept it anyway.

It's kind of unintuitive, but this basically says that, as x goes to infinity, 1/x goes to 0 kind of slowly. Think of this: this region extends infinitely to the right, so why would you expect it to be finite in area? There's a tug and pull between the infinite length of integration and the fact that 1/x goes to 0 as x goes to infinity. In the end, the length of the interval wins out in this case.

15) ∫_2π^∞ sin Θ dΘ

This works out to be
-1[cos Θ]_2π^∞

which is
1 - cos ∞

I understand that, intuitivly, this area cannot be determined because cos ∞ will continually range between 1 and -1, and therefore is divergent, but it seems there should be a more accurate description for this.

Not a big deal though.

Right, it's examples like this which show why it's important to use limit notation instead of just "plugging in infinity". So the way I would write this out is:

_2π∫^∞ sin Θ dΘ =
lim_t->∞ _2π∫^t sin Θ dΘ =
lim_t->∞ -cos Θ dΘ _2π|^t =
lim_t->∞ 1 - cos t

Now this limit does not exist, as cos t oscillates between -1 and 1, and so the limit and integrals diverge.

This is just an interesting little integral I've come across and I'm wondering if there is a way to use the following knowledge to integrate it. I know I could probably solve it eventually just by finding the right substitutions, but I guess I want to see if seeing this pattern might be useful.

26) ∫ ( x arctan x ) (1 + x²)^-2 dx

Now, this is what I see:
arctan x = f(x)

∫ f ' '(x) f(x) (f '(x) )² dx

It seems like it is almost a three part chain.

Well, x is not quite f ' '(x) but is is very close.

Ok, I have a stupid question, but here goes.

(0 - 1)^4/5

Does this equal 1 or a version of i? It must be a version of i since if I had (-1)^1/2, which I know should be i, I could turn that into (-1)^2/4, square the -1 to make 1, and then take the fourth root of 1 to get 1. There must be a rule that says when considering exponents, you must divide first.

-------------------------------------------------------
I actually moved this post to be in another thread since it was a more general question.

Originally Posted by Demen Tolden

This is just an interesting little integral I've come across and I'm wondering if there is a way to use the following knowledge to integrate it. I know I could probably solve it eventually just by finding the right substitutions, but I guess I want to see if seeing this pattern might be useful.

26) ∫ ( x arctan x ) (1 + x²)^-2 dx

Now, this is what I see:
arctan x = f(x)

∫ f ' '(x) f(x) (f '(x) )² dx

It seems like it is almost a three part chain.

Well, x is not quite f ' '(x) but is is very close.

I'm kind of sick and not at full mental acuity, but I don't see how you got that integral expression in terms of derivatives of arctan.

Did you manage to solve this? It looks to me like an integration by parts, which itself requires a substitution, and then... after I did some work, it looks like a partial fractions decomposition finishes the rest.

I used about 3 pages of paper and, when I found a process that solved it, four substitutions, some trignometric, some not.

If
f(x) = arctan x

Then
f '(x) = 1/(1+x²)

and if
g(x) = 1+x²

which, in terms of f(x) would be
( f '(x) )^-1

then
g '(x) = 2x

and... err.. hmm... There must be a connection...

( ( f '(x) )^-1) ' = -1 ( f '(x) )^-2f ''(x)

wich would be...
-1( (1 + x²)^-1)^-22x(1+x²)

or
4x(1+x²)

I don't know, it just seems very closely related to me, like its nearly a three part chain.

Here's one in the Chapter Review that looks like it should be easier, but I'm having trouble with it.

12) ∫ sin x/(1+x² )

I recognize that if you squared the hypotenuse on a triangle with the sine of x you would have the denominator, but anyway, I have only successfully been able to represent the integrand in the way that the problem starts out as and this:

∫ sin ( tan Θ) dΘ

If I substutute for the tan Θ, I just end up with the original problem, and if I substitute tan Θ for x, I just get this second integrand. I don't really see any other substitutions that would do much.

Here is another one:

29) ∫ x⁵ sec x dx

If I use any kind of substitution for sec x, it seems I have to deal with arcs, which would make the problem more difficult. If I use Integration By Parts, it looks like I have to choose to either continually increase the exponent on x, or turn the sec x into ln (sec x + tan x) and not be able to go further.

Yeah, I don't know how to attack those. The best I can say at the moment is keep trying!

Well, I think I'll be moving on. I didn't finish those last couple up, but I think I am pretty well off still. I've done hundreds of integrals and spend nearly an entire month on this chapter, and I think I'm ready for more material. Doing this chapter took me two to four times more time than I have yet spent on a chapter in this book.

I think it's worth spending a lot of time on this chapter. It's very technical, and a lot of the rest of the book depends on you being able to do it. In general, you're expected to be able to handle whatever derivatives and integrals are thrown at you. Derivatives are pretty easy to handle in general, but integrals aren't, as you have seen. But now they should be more second nature to you.