Question

Please, please help with substitution rule for indefinite integrals:

Answer 1

\[\int\limits_{}^{} (4z + \sin(3z))e ^{6z ^{2}-\cos(3z)} dz\]

Answer 2

whew, that's a doozy.

Answer 3

So what do you think you want to use for substitution?

Answer 4

\[\int\limits_{?}^{?} \sin(x) d x = -\cos(x) + constant\]

Answer 5

Not sure...possibly something to do with 3z? I have no idea to be honest

Answer 6

I would let u = that yucky exponent on the e

Answer 7

Ignore the two ?'s

Answer 8

Okay but how does that help?

Answer 9

Well.

\[u = 6x^2 - cos(3z)\]
\[\implies du = 12z +3sin(3z)dz = 3(4z + sin(3z))dz\]

Answer 10

Oh so then \[du = 2z ^{3}-\sin(3z)dx\]

Answer 11

Close, but you integrated. You want to take the derivative

Answer 12

oh

Answer 13

Okay. Where do I plug u into?

Answer 14

Well first rewrite dz in terms of du using that last equation I had.

Answer 15

?

Answer 16

\(du = 3(4z + sin(3z))dz \implies dz =\ ?\)

Answer 17

I don't understand - what am I solving for?

Answer 18

Well for this part you're solving for dz.

Answer 19

Err...I divide 3(4z + sin(3z)) on both sides?

Answer 20

Yes

Answer 21

So \[dz = \frac{1}{3(4z+sin(3z))} du\]

Answer 22

So now. Go back to your integral, and replace the exponent of e with u, and the dz with your expression for du, and you'll be pleasantly surprised.

Answer 23

Wait a minute...Why do I replace dz?

Answer 24

Because we are no longer integrating with respect to z, we are integrating with respect to u now. Because it turns out u is nicer.

Answer 25

We are replacing our nasty expression with z for a nice expression with u.

Answer 26

\[(4z + \sin(3z)e ^{u}\]du

Answer 27

But remember that dz does not equal du.

Answer 28

ButI thought we replaced dz

Answer 29

\[dz = \frac{1}{3(4z+sin(3z))} du\]

Answer 30

So you have to put that whole thing in for dz

Answer 31

I'm so lost

Answer 32

Lets try an easy one to start with.

Answer 33

\[\int x*e^{x^2}dx\]

Answer 34

We let \(u = x^2\)

Answer 35

So \[du = (2x)dx \implies dx = \frac{1}{2x}du\]

Answer 36

With me so far?

Answer 37

sure

Answer 38

If not, please ask

Answer 39

yes

Answer 40

Ok. So then

\[\int x*e^{x^2}dx = \int x*e^u(\frac{1}{2x} du)\]

Answer 41

And we can cancel the x in front with the x in the denominator from the du.

Answer 42

to get
\[\frac{1}{2}\int e^udu\]

Answer 43

Which is nice and easy.

Answer 44

\[\frac{1}{2}\int e^udu = \frac{1}{2}e^u + c = \frac{1}{2}e^{x^2} + C\]

Answer 45

So I've got\[1/3 e ^{u}\] for my problem
when do I integrate?

Answer 46

Now.

Answer 47

Before or after I plug in u

Answer 48

Don't switch back your u until after you integrate.

Answer 49

How do I integrate that?

Answer 50

\[\int \frac{1}{3}e^u du = \frac{1}{3}\int e^udu\]

Answer 51

What is the integral of \(e^x\)?

Answer 52

e^x

Answer 53

Right.

Answer 54

\[1/3 *e ^{6z ^{2}-\cos(3z)} \]

Answer 55

+C

Answer 56

oh right

Answer 57

Nice job!

Answer 58

Oh my goodness. Would you please walk me through one more?

Answer 59

Sure, but you'll have to do more of it this time ;)

Answer 60

How's it start?

Answer 61

Okay I'll try:

Answer 62

\[\int\limits_{}^{}(4z ^{3} + 6\csc ^{2}(6z)(z ^{4}-\cot(6z))^{4} dz\]

Answer 63

Ugh

Answer 64

Do I need to FOIL first?

Answer 65

No.

Answer 66

These are all ugly u substitution problems. Foiling will just make you cry.

Answer 67

You want to find a nice thing to pick for a u substitution.

Answer 68

Oh okay

Answer 69

Do you notice anything interesting about these two things?

Answer 70

Not really sure what to look for, actually.

Answer 71

\((4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4\)

You are looking for something in the equation that looks like the derivative of another part of the equation.

Answer 72

Err expression, not equation

Answer 73

Oh! I guess the 4z^3 and z^4

Answer 74

And what about the \(6csc^2(6z)\) and the -cot(6z) ?

Answer 75

AND the csc^2

Answer 76

right.

Answer 77

So what you want to do is pick the one that is not the derivative, and let that be u.

Answer 78

So let\[u = z ^{4} - \cot(6z)\]

Answer 79

Yep, then solve for dz in terms of z and du.

Answer 80

So \[\int\limits_{}^{}(4z ^{3}+6\csc ^{2}(6z))u ^{4}du\]

Answer 81

Nope. but close.

Answer 82

\[u = z^4 - cot(6z)\]
So
\[du =\ ?\]

Answer 83

4z^3 + 6csc^2(6z)

Answer 84

close.

Answer 85

\[du = (4z^3 + 6csc^2(6z))dz\]

Answer 86

So then what does dz = ?

Answer 87

1 over all that mess

Answer 88

yep.

Answer 89

So plug in 1 over all that mess times du for dz in the original equation. and plug in u for what we said it equaled.

Answer 90

\[(4z ^{3} + 6\csc ^{2}(6z))u ^{4} * 1/4z ^{3} + 6\csc ^{2}(6z)\]

Answer 91

simplify!

Answer 92

u^4?

Answer 93

yes

Answer 94

\[(z ^{4} - \cot(6z))^{4}\]

Answer 95

No

Answer 96

\[\int(4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4dz\]
\[ = \int (4z^3 + 6csc^2(6z)) u^4(\frac{1}{4z^3 + 6csc^2(6z)})du \]
\[= \int u^4du =\ ?\]

Answer 97

? I don't know

Answer 98

I can't pull anything out in front of the integral sign