# Integral question

1. Dec 15, 2007

### frasifrasi

For the integral

cos^(5)(4x), can I just integrate it directly to -1/4*sin^(6)(4x)... or which technique should I use?

2. Dec 15, 2007

### neutrino

Nope you can't do that. To make things easier, you may want to sub u=4x first, and then you have to use a trig identity to make the (cos(u))^5 "easy" to integrate.

Hint: $$\sin^2{x} + \cos^2{x} = 1$$

3. Dec 15, 2007

### rocomath

remember, when you're dealing with powers, you have to take into consideration the "chain rule."

4. Dec 15, 2007

### frasifrasi

But then I would have two 1-sin^2(X) in the integral, is there an easier way?

5. Dec 15, 2007

### arunbg

As neutrino suggested, after obtaining cos^5(u) write it as cos^4(u)*cos(u) = ((1-sin^2(u))^2)*cos(u). Now giving sin(u)=v, can you proceed?

6. Dec 15, 2007

### frasifrasi

Ok, so i get:

(1-v^2)^2 dv

= 1 - 2(v^2) + v^4...

7. Dec 15, 2007

### frasifrasi

Now, I have to integrate the whole thing??????????????????????????????

How would I sub back? --let v = sin(4x)???

8. Dec 15, 2007

### Kurdt

Staff Emeritus
v = sin(u) and u=4x.

You've expanded the brackets in the post above but did you integrate it?

9. Dec 15, 2007

### frasifrasi

Too much work for this question.

10. Dec 15, 2007

### HallsofIvy

Staff Emeritus
If you are only interested in questions that can be done trivially, then I recommend you drop calculus.
Do you really consider it too difficult to integrate
$$\frac{1}{4}\int (v^4- 2v^2+ 1)dv$$?

11. Dec 15, 2007

I second Halls of Ivy. I am not sure what is so difficult here. It is a matter of individually integrating 3 simple terms and then plugging in for v.

In the time it took to write too much work for this question, you could have finished!

Casey

12. Dec 15, 2007

### frasifrasi

In context, my statement is valid...

13. Dec 15, 2007

I have no idea what this means.

Last edited: Dec 15, 2007
14. Dec 15, 2007

### Kurdt

Staff Emeritus
I think frasifrasi means that they only wanted to know how to substitute back for v once the integration was done so it was too much work to put the full integration in the previous post for the relatively simple clarification.

15. Dec 15, 2007

### frasifrasi

Yeah, I was just saying that more work wasn't really necessary, not meaning to complain in any way.

but hey, I have another question...

For the series (-1)^(n)/ln(n), I used the alternating series test to show that it converges, but how do I show that it converges conditionally?

If I look at the abs value 1/ln(x) , I am not sure how to establish that it diverges...

Thank you for the help!

Last edited: Dec 15, 2007