Calculus: Can someone help me solve this indefinate Integral?

[chiwawa626]

Calculus: Can someone help me solve this indefinate Integral?

integral of: cos(x)^6

 

[Maximus96]

dont you just use the chain rule 6 times? i forget, its been a while since i last took the course and i have no reason to practice this stuff since then.

 

[VioletAura]

That is an easy problem, your really do not help if you know the rule for that sort of problem. Check your math book.

 

[Gibson486]

my bad, not basic...but you need soem work to get it done...but just take it step by step.

 

[chiwawa626]

You can't use the chain rule. This is Trigonometric Integration...

 

[VioletAura]

Didn't get it yet?

answer is
6 sin x + C

 

[Gibson486]

VioletAura
You need to take calc all over again!

the derivative of 6 cos(x) + c is -6 sin(x).

 

[thraxes]

I think it's partial integration (i hate that method).

But that problem of yours is kinda neat though, I'll have a go at it later, wanted to practice that stuff anyway.

EDIT: urrr yeah violet, this is not that easy, it looks it at first but then gets bigger and bigger as you go along.

Mr friendly TI92 says it's a nice long equation that comes out at the end, I'm not quite clear yet on how you actually solve the problem by hand (which is what counts in an exam). As I was saying, I'm guessing its a partial integration since

"sin(x)^6"="sin(x)*sin(x)*sin(x)*sin(x)*sin(x)*sin(x)"

and this is what that method is for.

 

[VioletAura]

Originally posted by: Gibson486
VioletAura
You need to take calc all over again!

the derivative of 6 cos(x) + c is -6 sin(x).

WTF???
I said it was 6 sin(x) + c NOT 6 cos(x) +c

 

[BullsOnParade]

that's not right

There are substitutions for even and odd powers of sin and cos
that make it simple to integrate, lemme find a proper integration
table.

And i dont think partial integration applies and 6sinx + C isn't the parital of that.

try integral(cosx^2) = integral(1/2+1/2cos2x)

or integral((cos(x))^n) = 1/n * (cos(x))^ (n-1)*sin(x)+(n-1)/(n)*integral((cos(x))^(n-2))

 

[chiwawa626]

So far you are all wrong...can someone who knows their calc please help?

 

[Gibson486]

yeah is it, 6 is a constant, you do not take the derivative of a constant unless it is by itself.

edit: whatever, just take change the cos to sin and remove th neg sign. Either way, the integral of it is not 6sinx+c.

 

[sash1]

been a while since I've done these, but IIRC it's:

x/6 + sin(x)/12 + c

`K

 

[Gibson486]

You need integration by parts, or you can do it by hand with teh intergration table, or you can use the p(x) function, which i doubt you learned yet.

 

[yllus]

You mean cos^6(x), right?

I'll write it right now. You may want to use The Integrator to solve it in the meantime.

 

[Legendary]

You have to use uv - int(v*du)
du = cosx
v = cosx

You have to do it a few times.

 

[mAdD INDIAN]

Originally posted by: chiwawa626
Calculus: Can someone help me solve this indefinate Integral?

integral of: cos(x)^6

think of it as (cos^4)(cos^2)

Then break down cos^4 as (cos^2)^2 so then you have cos^2(cos^2)^2 and integrate.....use the double angle formula for cos^2 (i forgot wat it was...1/2 something).

You'll be using the double angle formula quite often.

 

[Gibson486]

the answer will be be pretty big, I mean big as in 3 different terms being added.

 

[yllus]

Sorry, got thrown off, re-writing now...

1. We know that sin^2(x) = (1 - cos(2x) / 2) and sin^2(x) = (1 + cos(2x) / 2). Use the identities to break down cos^6(x) to odd powers of cosine.

2. Save one factor of cos(x) and convert the rest to sin(x) as needed. Use the substitution U = sin(x) and substitute as needed.

3. Integrate under the substitution and finally sub cos(x) back in as needed.

 

[BullsOnParade]

yllus has got the ticket. The obscure formula is



integral((cos(x))^n) = 1/n * (cos(x))^ (n-1)*sin(x)+(n-1)/(n)*integral((cos(x))^(n-2))

 

[Gibson486]

Many ways to do it, but Yllus is teh only one who can give a result. Kudos to him for knowing his calc. Yllus, can you help me with differential equations? You can be my LaPlace Transform Hero

 

[Goosemaster]

Originally posted by: yllus
Sorry, got thrown off, re-writing now...

1. We know that sin^2(x) = (1 - cos(2x) / 2) and sin^2(x) = (1 + cos(2x) / 2). Use the identities to break down cos^6(x) to odd powers of cosine.

2. Save one factor of cos(x) and convert the rest to sin(x) as needed. Use the substitution U = sin(x) and substitute as needed.

3. Integrate under the substitution and finally sub cos(x) back in as needed.

Seriously dude, I was giving up all hope. This problem is EASY. The responses I was reading were SCARY.


All you have to do, as was said, is brake away a cos^2^ x and then convert it. Then multiply them, and do the reverse of the chain rule.