calculus finding indefinite integral: integral cos^4 (3x) dx

[krys]

I need help solving this problem or what I am doing wrong here is my work

integral cos^4 (3x) dx

integral cos^n u du = cos^n-1 u sin u/n +n-1/n integral cos^n-2 u du

I got

1/3 integral cos^3(3x)sin(3x)/4 + 3/4 integral cos^2 (3x) du

 

[stapel]

integral cos^4 (3x) dx

integral cos^n u du = cos^n-1 u sin u/n +n-1/n integral cos^n-2 u du

Your formatting is confusing, especially with respect to grouping symbols. What you have posted in the second line above means something like this:

. . . . .$\displaystyle \int\, \cos^n(u)\, du\, =\, \cos^n\, -\, 1u\sin\left(\dfrac{u}{n}\right)\, +\, n\, -\, \dfrac{1}{n}\, \int\, \cos^n\, -\, 2udu$

But I'm fairly certain that this isn't what you meant...?

When you reply with clarification, please include a clear listing of all of your steps leading to your answer. Thank you!

 

[krys]

integral cos^4(3x) dx

integral cos^n(u)=((cos^n-1(u)sin(u)/(n) +(n-1)/(n) integral cos^(n-2)(u)) du

1/3 integal (cos^3(3x)sin(3x)/(4)+(3/4)integral cos^2 (3x)) du


how is this?

 

[stapel]

integral cos^4(3x) dx

This is what you're trying to find, right?

integral cos^n(u)=((cos^n-1(u)sin(u)/(n) +(n-1)/(n) integral cos^(n-2)(u)) du

I'm going to guess that a "du" is missing from the left-hand side and that, by "cos^n-1(u)sin(u)/(n)", you mean "[ cos^(n-1)(u) * sin(u)] / n", so the above means the following:

. . . . .$\displaystyle \int\, \cos^n(u)\, du\, =\, \dfrac{\cos^{n-1}(u)\, \sin(u)}{n}\, +\, \dfrac{n\, -\, 1}{n}\, \int\, \cos^{n-2}(u)\, du$

Is this correct? Also, is this a formula you were given to apply, or does this mean something else?

1/3 integal (cos^3(3x)sin(3x)/(4)+(3/4)integral cos^2 (3x)) du

I'm not sure what this is...? Are you saying that this is your result, after you plugged "4" in for "n" in the formula (?) in the second line? Are you asking if you did the plug-n-chug correctly, or are you proposing this as being your final answer?

Thank you!

 

[krys]

Yes that is what I meant.
and yes that is my final answer

 

[stapel]

Yes that is what I meant.

You were asked, in many more than one place, what you meant. For which is the above your reply?

and yes that is my final answer

Where have you ever seen the answer to an "integrate the following" exercise be something that included an integral? (Hint: Never. Having an integral in your "answer" means that you're not done.)