integrate sin^43x dx

- anonymous

integrate sin^43x dx

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- anonymous

\[\int\limits \sin^{43}(x) dx= - \cos^{43}(x)\]
honestly, this is a weird question lol, so I'm not sure? ._.

- anonymous

+ C ofcourse

- anonymous

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

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## More answers

- anonymous

LOL >_< I know, it looks weird

- anonymous

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

- anonymous

meh...
no...this method is funny lol

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok chia

- anonymous

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

- anonymous

chia.. solve it

- anonymous

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

- anonymous

give me a sec

- anonymous

it's sin(3x)^4

- anonymous

that's the question?

- anonymous

oh lol

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

AMISTRE!
you can do this! ^_^

- amistre64

Howdy :)

- anonymous

lol

- anonymous

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? ^_^

- amistre64

[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...

- anonymous

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

- anonymous

if only I remember lol >_<

- amistre64

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

- amistre64

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

- anonymous

where did you get cos^5?

- amistre64

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

- amistre64

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

- anonymous

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?

- anonymous

where u = 3x ? :)

- anonymous

right? .-.

- anonymous

isn't it simpler that way, lol?

- amistre64

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

- anonymous

why?

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

use the half angle formula
to integrate sin or cos with a power
you MUST have sin * cos

- anonymous

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

- anonymous

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?

- amistre64

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

- anonymous

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)

- anonymous

star is right

- anonymous

^_^ there, I hope it's right

- anonymous

as simple as that :)

- anonymous

sstar is completely right
though i have a question
where's the person who ASKED this question

- anonymous

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

- anonymous

wait...he left ._.

- anonymous

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

- anonymous

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

- anonymous

o_o....no wait O_O!

- anonymous

that's the final answer?

- anonymous

Sub. u=3x, then du = 3dx --> dx = du/3. Then you have
1/3*int[(sin^4(u) du]

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

He wants to see if it matches the integrand.

- anonymous

oh

- amistre64

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

- anonymous

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

- anonymous

give me a sec, i'll do it

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Omg, I gave you the answer ^^

- anonymous

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

- anonymous

what answer ._.

- anonymous

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

- anonymous

cheers :D

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

I see where my error cropped up ;)

- anonymous

lols kewls

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