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

what is the derivative of your answer then? it will not give you your original problem.

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

Saladsamurai

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

Saladsamurai

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
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