Question

Need help please(last question I have):
Find the indefinite integral of:
sin^(3)(18x)*cos^(9)(18x) dx

Answer 1

Do you use this rule:
Sin A Cos B = 1/2[sin(A-B) + sin (A+B)]?

Answer 2

no, strip out a sin(18)x

Answer 3

sin(18x) I mean...

Answer 4

sin^(3)(18x)*cos^(9)(18x) dx=sin(18x)*sin^(2)(18x)*cos^(9)(18x) dx

Answer 5

now note that sin^2x=1-cos^2x

Answer 6

okay I see what you did

Answer 7

yeah

Answer 8

ya it's not so hard ..i will tell u how to solve this types of problem

Answer 9

now it's all gonna be cos after simplifying, each multiplied by sin
easily done from here with a couple simple u-subs

Answer 10

so you'll have:
sin(18x) *(1-cos^(2)x) * cos^(9)(18x)? dx

Answer 11

yes

Answer 12

simplify

Answer 13

what do you do next? Im not so good at this :(

Answer 14

distribute and what do you get?

Answer 15

trying to find my notes for problems like these

Answer 16

sin(18x) - cos^(2)xsin(18x)

Answer 17

and the cos^(9)(18x)

Answer 18

after that, do u do a U substitution of sin 18x?

Answer 19

write out what you get after distribution completely so I can see where you are

Answer 20

sin(18x) - cos^(2)(x)sin(18x)*cos^(9)(18x) dx

Answer 21

\[\sin(18x)(1-\cos^2(18x))(\cos^9(18x))\]\[=\sin(18x)(\cos^2(18x)-\cos^{11}(18x))\]

Answer 22

okay I see how you got that

Answer 23

notice that sin(n) is almost the derivative of cos(n)

Answer 24

yea

Answer 25

you have to be a little careful here, there is a chain rule element to this u-sub

Answer 26

yeah, you get cos(8x)(8) right?

Answer 27

i mean 18x

Answer 28

why did I write down 8 on my paper -.-.
It becomes: cos(18x)(18)

Answer 29

then du/18 = cos(18x)dx right?

Answer 30

why do you want to make u=sin(18x) ?
how are you going to represent the cos^2 and such in the interval

Answer 31

?

Answer 32

integral*

Answer 33

hmm since du/18 = cos(18x)dx, cant u say (du/18)^(11) is the same as cos^(11)(18x)?

Answer 34

yeah, and what the heck is the integral\[\int\frac{du^{11}}{18}\]gonna be ?

Answer 35

what does \(du^n\) mean?
nothing, in this context, it's undeined

Answer 36

what do you suggest U to be?

Answer 37

you should always consider what's going to happen when you integrate (fx))^n
in these cases you should a,ost always try to choose the function that is being raised to a power when it is multiplied by a constant multiple of its derivative

Answer 38

almost*

Answer 39

so in this case, it would be what Chlorophyll suggested

Answer 40

I mean u = cos(18x)
-> du = ?

Answer 41

no, it would be the function inside the power

Answer 42

I thought you knew that @Chlorophyll you beat me to correcting you ;)

Answer 43

oh ok, so cos(18x) = -sin(18x)(18)

Answer 44

you should always write the dx, that illustrates exactly what you are subbing

Answer 45

yeah sorry

Answer 46

\[u=\cos(18x)\implies du=-18\sin(18x)dx\]

Answer 47

u = cos(18x)
du = -sin(18x)(18) dx
du / 18 = -sin(18x)

Answer 48

and dx at the end*

Answer 49

you ought to put that - sign on the other side as well
we are subbing for sin(18x)dx, not -sin(18x)dx

Answer 50

true, -du/18 = sin(18x) dx

Answer 51

now sub that into what we had way back up there

Answer 52

(-du/18) (u^(2) - u^(11))

Answer 53

yep :)

Answer 54

wait we missed a 9

Answer 55

( 1 - u² ) u^9

Answer 56

I sensed Chlorophyll correcting me :)

Answer 57

xD what do you mean a 9? where?>

Answer 58

@TuringTest Sorry !

Answer 59

oh I just distributed wrong being the lazy slob that I am

Answer 60

ah I see

Answer 61

oh by all means please, it's great to have more than one pair of eyes on a problem @Chlorophyll

Answer 62

so how did we get (1-cos(18x))cos^(9)(18x)?

Answer 63

@TuringTest I'm too too lazy, that's why I'm sort of shadow better person :)

Answer 64

(1-cos^2(18x))cos^(9)(18x)

Answer 65

yeah I forgot the ^2

Answer 66

sin^3x*cos^9x=sinx*sin^2x*cos^9x=sinx*(1-cos^2x)*cos^9x

Answer 67

strip out a sin to get an even power that we can convert to cosine is the overarching theme of these problems I'd say

Answer 68

hmm I see how you got what you have now, but what do you do now? -du/18 is sin(18x)dx

Answer 69

Do we do a u substitution for Cos(x)?

Answer 70

sinx*(1-cos^2x)*cos^9x
Pretend X = 18x
Let u = cos 18x

Answer 71

-> - du/18 = sin 18x
=> -1/18 ∫ ( 1 - u² ) u^9 du

Answer 72

so then u get:
-1/18[(u-(u^(3) / 3)) (u^(10/10)], correct?

Answer 73

Note: ( 1 - u² ) u^9 = u^9 - u^11

Answer 74

yeah so u^(-2)

Answer 75

No, you need to do integral one by one term!

Answer 76

ahhh nvm, i just confused myself. Yeah.

Answer 77

so now we have:
1/18 integral of (u^(9) - u^(11))

Answer 78

taking the integral, you get:
1/18 [u^(10)/10 - u^(12) / 12 ] right?

Answer 79

Yes!

Answer 80

= u^(10) / 180 - u^(12) / 216

Answer 81

Now, put cos18x back!

Answer 82

= cos^(10)(18x) / 180 - cos^(12)(18x) / 216

Answer 83

we also forgot a +C

Answer 84

Watch out the negative sign ( - 1/18)

Answer 85

Oops.

Answer 86

okay so:
-cos^(10)(18x) / 180 + cos^(12)(18x) / 216

Answer 87

+ C xD, forgot it again

Answer 88

Fully correct now :)

Answer 89

Got it!
-((cos^(10)(18x))/180)+ cos^(12)(18x) / 216 is the correct answer.
Thank you so much!!

Answer 90

I'm glad the team work works out :)

Answer 91

ahaha :)