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anonymous
 5 years ago
integrate sin^43x dx
anonymous
 5 years ago
integrate sin^43x dx

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \sin^{43}(x) dx=  \cos^{43}(x)\] honestly, this is a weird question lol, so I'm not sure? ._.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sstar, i don't know the answer but i'm almost sure that's not correct ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL >_< I know, it looks weird

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, is it possible to let u = 43? I've never solved such thing lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0meh... no...this method is funny lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sstarica the question is\[\int\limits \sin ^{4}3x dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, you have to substitute (cos x)^2 = 1  (sin x)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh that makes everything a lot better.... oneprince USE PARENTHESIS!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no wait a min, isn't the sin to the power of 43?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0was it to the power of 43? or to the power of 2? lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ugh...sub 3x for u and then use the half angle formula to get cosine and sine together, integrate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sstar are you writing it out? caz i don't really want to...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0>_< nvm, proceed chia, I misunderstood the question lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oneprince, did you read what i wrote? is that enough?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AMISTRE! you can do this! ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my 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
 5 years ago
Best ResponseYou've already chosen the best response.0[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
 5 years ago
Best ResponseYou've already chosen the best response.0it's not really missing something, you can let u = 3x and solve normally like chia said

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if only I remember lol >_<

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Dx(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
 5 years ago
Best ResponseYou've already chosen the best response.0lets multiply your original problem by "1"; or rather a convenient for of "1" such as 15/15..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did you get cos^5?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then we can pull out the bottom "15" and integrate the rest up like normal

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin^4 comes from cos^5; or at least a version of it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do this : \[\int\limits (1\cos^2(3x))^2 dx = \int\limits(12\cos^2u+\cos^2u)du\] and simply solve, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't it simpler that way, lol?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} \frac{5(3)}{5(3)} \sin^4 (3x) dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre, i think you're doing it wrong.. and i feel bad caz i'm not actually doing it...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{15} \int\limits_{} 5(3) \sin^4 (3x) dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use the half angle formula to integrate sin or cos with a power you MUST have sin * cos

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, amistre, that's definitely wrong TT

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why 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
 5 years ago
Best ResponseYou've already chosen the best response.0derive it back again, its right.... it should be right lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using 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
 5 years ago
Best ResponseYou've already chosen the best response.0^_^ there, I hope it's right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sstar is completely right though i have a question where's the person who ASKED this question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL! watching us, hey oneprince did you understand what we did? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah _ but you got it sstar i have integrating cosine sins TT so messy...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[I=\frac{1}{3}(\frac{1}{32} (12 u8 \sin(2 u)+\sin(4 u))+c)\]where u = 3x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's the final answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sub. u=3x, then du = 3dx > dx = du/3. Then you have 1/3*int[(sin^4(u) du]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use a reduction formula on sin^4(u)...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no need for reduction formula ^_^ + i got you lol :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what happens when we derive:\[ \frac{\cos^5 (3x)}{15} + C\] ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why do you want to derive it? that's not the answer lol? ._.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0He wants to see if it matches the integrand.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol.... if its not the answer, then what does it derive to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there's only one way to find out, T.R.Y ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give me a sec, i'll do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I won't even try since my page is lagging BADLY

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyhow, I'm off to review for my DM test tomorrow, good luck all ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0225 cos(3x)*sin(3x)^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Omg, I gave you the answer ^^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait i lied! i did * 15 instead of / 15 give me another min :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh i see why you put that 15 there it's cos(3x)*sin(3x)^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Aren't you trying to find the integral of sin^4(3x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah but amistre is trying to show that he's right but i showed that he's wrong ^^

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol...not trying to show that im right, just trying to figure out what I did wrong :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:x sorry i didn't mean it like that but do you get it now?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I see where my error cropped up ;)
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