Question

Calculate the integral of cos^4(3x)dx..Need some help

Answer 1

I got you :)

Answer 2

gonna need a 3 in there somplace ;)

Answer 3

and a -sin ... right?

Answer 4

cos^2 x - cos^2 x sin^2 x

Answer 5

in place of x put 3x

Answer 6

right ... whats the formula for that..

Answer 7

convert cos^2 3x into (1+ cos6x)/2
and cos^2 3x sin^2 3x into 0.5sin^2 (6x)

Answer 8

you just do the same thing thing as a normal cos^2(x) just replace x with 3x?

Answer 9

Rewrite it as:
\[\int\limits (\cos^2(3x))^2 dx\]
You want to use your half angle formula:
\[\cos^2(x)=\frac{1}{2}+\frac{1}{2}\cos(2x)\]
Then:
\[\int\limits (\frac{1}{2}+\frac{1}{2}\cos(6x))^2dx\]
Square it out, then use the identity again. Integrate term by term.
Or use a power reduction formula >.>

Answer 10

1/2 (sin^2 (6x)) = 1/4(1- cos12x)

Answer 11

u can convert like ive done and then integrate

Answer 12

whats teh half angle formula ?

Answer 13

whats teh half angle formula ?

Answer 14

What I used^^

Answer 15

More generally:
\[\cos^2(ax)=\frac{1}{2}+\frac{1}{2}\cos(2ax)\]

Answer 16

still lost

Answer 17

Did you follow what I did above? I can finish working it out if need be.

Answer 18

why did you write as (cos^2(3x))^2 dx?

Answer 19

So that you could see the formula I gave 3 posts up is applicable (sometimes its hard to see with higher powers) but notice that (cos^2(3x))^2=cos^4(3x). Does that make sense?

Answer 20

yes, you converted it just so you can use the double angle formula?

Answer 21

Yes :) So going from what I have above:
\[\int\limits (\frac{1}{2}+\frac{1}{2}\cos(6x))^2dx=\int\limits (\frac{1}{4}+\frac{1}{2}\cos(6x)+\frac{1}{4}\cos^2(6x))dx\]
Do you follow that distribution?

Answer 22

Refresh if the latex looks bad*

Answer 23

Yes i got that far

Answer 24

Okay then use the identity again on the cos^2(6x) giving:
\[\int\limits(\frac{1}{4}+\frac{1}{2}\cos(6x)+\frac{1}{4}(\frac{1}{2}+\frac{1}{2}\cos(12x)))dx\]
Distribute the 1/4. From there do you know how to proceed?

Answer 25

so its the integration of (3/8) +(1/2)cos(6x)+(1/8)cos(12x))dx

Answer 26

Exactly :) What did you get when you did that?

Answer 27

i would split the integration right?

Answer 28

You could. Giving:
\[\int\limits \frac{3}{8} dx+\frac{1}{2} \int\limits \cos(6x)dx+\frac{1}{8} \int\limits \cos(12x)dx\]
If that helps you.

Answer 29

i got (1/8)sin(6x)+ (1/96)sin)(12x)

Answer 30

That's right, almost. Don't forget you have the integral of 3/8 which is (3/8)x. So you need a +(3/8)x too :P

Answer 31

And a +c to be "technically" right xD

Answer 32

oh ya

Answer 33

But good job :) Any time you have powers of cosine and sine and they are both EVEN. You will use that identity to simplify it. The one for sine is the same just off by a negative:
\[\sin^2(ax)=\frac{1}{2}-\frac{1}{2}\sin(2ax)\].
When you start mixing even and odd powers you'll need:
\[\sin^2(x)+\cos^2(x)=1\]
Its dependent on a problem-by-problem basis. Let me know if you get stuck on another one :P

Answer 34

Errrr...that should read sin^2(ax)=1/2-(1/2)COS(2ax). Mistype on my part.

Answer 35

thanks so much .. i definitely will

Answer 36

No problem :) I'll be around for a while

Answer 37

its telling me its incorrect ..

Answer 38

http://www.wolframalpha.com/input/?i=integral+of+cos^4(3x)
It's correct (just distribute the 1/96)

Answer 39

I don't know if it won't take that, me, you, and wolfram got the same thing...

Answer 40

this site is awesome..

Answer 41

It will do any integration for you. Then just click: see steps. However, if you do definite integrals, it won't show you the step by step. So just do indefinite then check your answer, then do the numbers.

Answer 42

Also does any math anything. Or type in something like GDP ireland 2005. It'll give you that. Type in radius of the moon or ANYTHING.

Answer 43

i dont get how they got a final anwer from (3/8)x-(3/32)sin(4x)-(1/8)sin^3(2x)cos(2x)+C ..of (1/64)(24x-8sin(4x)+sin(4x)) +C