Question

integrate sin^43x dx

Answer 1

\[\int\limits \sin^{43}(x) dx= - \cos^{43}(x)\]

honestly, this is a weird question lol, so I'm not sure? ._.

Answer 2

+ C ofcourse

Answer 3

sstar, i don't know the answer
but i'm almost sure that's not correct ><

Answer 4

LOL >_< I know, it looks weird

Answer 5

hmm, is it possible to let u = 43? I've never solved such thing lol

Answer 6

meh...

no...this method is funny lol

Answer 7

sstarica the question is\[\int\limits \sin ^{4}3x dx\]

Answer 8

no, you have to substitute
(cos x)^2 = 1 - (sin x)^2

Answer 9

oh that makes everything a lot better....
oneprince USE PARENTHESIS!

Answer 10

ok chia

Answer 11

no wait a min, isn't the sin to the power of 43?

Answer 12

chia.. solve it

Answer 13

was it to the power of 43? or to the power of 2? lol

Answer 14

give me a sec

Answer 15

it's sin(3x)^4

Answer 16

that's the question?

Answer 17

oh lol

Answer 18

ugh...sub 3x for u
and then use the half angle formula to get cosine and sine together, integrate

Answer 19

sstar are you writing it out? caz i don't really want to...

Answer 20

>_< nvm, proceed chia, I misunderstood the question lol

Answer 21

oneprince, did you read what i wrote? is that enough?

Answer 22

AMISTRE!
you can do this! ^_^

Answer 23

Howdy :)

Answer 24

lol

Answer 25

my brain is defunctioning, so I don't think I can write it down even though I took it this semester and it looks SIMPLE! would you amistre? ^_^

Answer 26

[S] sin^4(3x) dx this is missing something lol

this came from something similar to -cos^5(3x) so what do we get when we derive that? and itll tell us what to add...

Answer 27

it's not really missing something, you can let u = 3x and solve normally like chia said

Answer 28

if only I remember lol >_<

Answer 29

Dx(-cos^5(3x)) = 5(3) sin^4(3x) which would be easy to solve right?

So we need a "15" in front of your integral to convert it

Answer 30

lets multiply your original problem by "1"; or rather a convenient for of "1" such as 15/15..

Answer 31

where did you get cos^5?

Answer 32

then we can pull out the bottom "15" and integrate the rest up like normal

Answer 33

sin^4 comes from cos^5; or at least a version of it right?

Answer 34

you can do this :

\[\int\limits (1-\cos^2(3x))^2 dx = \int\limits(1-2\cos^2u+\cos^2u)du\]

and simply solve, right?

Answer 35

where u = 3x ? :)

Answer 36

right? .-.

Answer 37

isn't it simpler that way, lol?

Answer 38

\[\int\limits_{} \frac{5(3)}{5(3)} \sin^4 (3x) dx\]

Answer 39

why?

Answer 40

amistre, i think you're doing it wrong..
and i feel bad caz i'm not actually doing it...

Answer 41

\[\frac{1}{15} \int\limits_{} 5(3) \sin^4 (3x) dx\]

Answer 42

-cos^5 (3x)/15 + C

Answer 43

use the half angle formula

to integrate sin or cos with a power
you MUST have sin * cos

Answer 44

no, amistre, that's definitely wrong T-T

Answer 45

Why don't you make a substitution of \[u=3x \rightarrow dx = \frac{du}{3}\]so that then\[I=\frac{1}{3}\int\limits_{}{}\sin^4 u du\]and now use a reduction formula?

Answer 46

derive it back again, its right.... it should be right lol

Answer 47

using the way I cut it down, you'll get :

\[= \int\limits1 du - 2\int\limits \cos^2u du + \int\limits \cos^2u du\]
then you'll use for cos^2 u = 1/2(1+cos2u)

Answer 48

star is right

Answer 49

^_^ there, I hope it's right

Answer 50

as simple as that :)

Answer 51

sstar is completely right

though i have a question
where's the person who ASKED this question

Answer 52

LOL! watching us, hey oneprince did you understand what we did? ^_^

Answer 53

wait...he left ._.

Answer 54

yeah -_-
but you got it sstar
i have integrating cosine sins T-T
so messy...

Answer 55

\[I=\frac{1}{3}(\frac{1}{32} (12 u-8 \sin(2 u)+\sin(4 u))+c)\]where u = 3x.

Answer 56

o_o....no wait O_O!

Answer 57

that's the final answer?

Answer 58

Sub. u=3x, then du = 3dx --> dx = du/3. Then you have

1/3*int[(sin^4(u) du]

Answer 59

You can use a reduction formula on sin^4(u)...

Answer 60

no need for reduction formula ^_^ + i got you lol :)

Answer 61

what happens when we derive:\[- \frac{\cos^5 (3x)}{15} + C\]
??

Answer 62

why do you want to derive it? that's not the answer lol? ._.

Answer 63

He wants to see if it matches the integrand.

Answer 64

oh

Answer 65

lol.... if its not the answer, then what does it derive to?

Answer 66

there's only one way to find out, T.R.Y ^_^

Answer 67

give me a sec, i'll do it

Answer 68

I won't even try since my page is lagging BADLY

Answer 69

anyhow, I'm off to review for my DM test tomorrow, good luck all ^_^

Answer 70

225 cos(3x)*sin(3x)^4

Answer 71

Omg, I gave you the answer ^^

Answer 72

wait i lied!
i did * 15 instead of / 15
give me another min :P

Answer 73

what answer ._.

Answer 74

ohh i see why you put that 15 there
it's
cos(3x)*sin(3x)^4

Answer 75

cheers :D

Answer 76

Aren't you trying to find the integral of sin^4(3x)?

Answer 77

yeah but amistre is trying to show that he's right
but i showed that he's wrong ^^

Answer 78

lol...not trying to show that im right, just trying to figure out what I did wrong :)

Answer 79

:x sorry i didn't mean it like that
but do you get it now?

Answer 80

I see where my error cropped up ;)

Answer 81

lols kewls