How to take the integral of this?

1. Nov 28, 2012

hunter77

1. The problem statement, all variables and given/known data

How do you solve for the integral of 1/(16-20cos4x)?

2. Relevant equations

I know the answer but I don't know how to solve it by hand.

3. The attempt at a solution

I tried u-substitution and breaking it into two parts with 1/16 - 1/20cos4x but couldn't get anywhere.

2. Nov 28, 2012

Staff: Mentor

First, you seem to be thinking that a/(b + c) = a/b + a/c. Is 2/(1 + 1) the same as 2/1 + 2/1?

Second, is what you show the actual problem? It looks to be very difficult to me.

3. Nov 28, 2012

hunter77

Oh, my mistake, I see what you mean. But I still can't seem to go anywhere with it.
Yes, that's the problem exactly as it was stated.

4. Nov 28, 2012

Staff: Mentor

And there's a dx in there somewhere?

Does your integral look like this?

$$\int \frac{dx}{16 - 20cos(4x)}$$

5. Nov 28, 2012

hunter77

Yup, that's what it looks like. Sorry about the dx omission.

6. Nov 28, 2012

iRaid

That seems like a really ugly integral...

7. Nov 28, 2012

Dick

There is a general way to do integrals like this if you can't think of any special tricks that will make it easy. Start by the substitution u=4x, then look at http://en.wikipedia.org/wiki/Weierstrass_substitution

8. Nov 28, 2012

Staff: Mentor

I think that the direction to go is to break down the cos(4x) term using the double angle identity for cos(2A) and the basic trig identities. IOW, cos(4x) = cos2(2x) - sin2(2x) = 2cos2(2x) - 1

Possibly you can get the denominator to look like the difference of squares or something to factor, that you can use partial fractions on.

However, this might be a daunting task given the mistake you made in post #1.

9. Nov 28, 2012

lurflurf

recall from elementary trigonometry that

$$\frac{1}{16-20\cos(4x)}=\frac{1}{48} \left( \frac{6 \sec^2(2x)}{3tan(2x)-1} - \frac{6 \sec^2(2x)}{3tan(2x)+1} \right)$$

Last edited: Nov 28, 2012
10. Nov 29, 2012

Ray Vickson

I'm glad you go nowhere this way, because it is WRONG: you cannot write 1/(a-b) as 1/a - 1/b. That violates all known rules of algebra, and you should never, never, never write it.

RGV

11. Nov 29, 2012

Dickfore

Is this
Wolfram Alpha

the answer that you are given? You can register and see a step-by-step solution.

12. Nov 29, 2012

nugraha

$$∫\frac{1 dx}{16-20\cos(4x) }$$, can be easily solved with substituted method :
Let u = 16 – 20cos (4x), so we got:

$$∫\frac{1}{u} dx$$*

By differentiate u we got:
$$\frac{du}{dx} = 80 sin (4x)$$

$$dx = \frac{du}{80 sin (4x)}$$

use this dx for *
$$∫\frac{du}{u80sin(4x)}=\frac{1}{80sin(4x)}∫\frac{du}{u}=\frac{lnu}{80sin(4x)}=\frac{ln(16-20cos(4x))}{80sin(4x)}$$

Last edited: Nov 29, 2012
13. Nov 29, 2012

Curious3141

Firstly, we don't show complete solutions here.

Secondly, your solution is wrong anyway. This step:

∫du/(u 80sin⁡(4x)) = 1/(80 sin⁡(4x)) ∫du/u

is wrong because you cannot bring the variable expression involving sin 4x outside the integral like that. You have to express it in terms of u and then integrate.

14. Nov 29, 2012

lurflurf

^That is not right.

15. Nov 29, 2012

nugraha

[STRIKE]try to see the previous step, we substituted dx with du. therefor our integral not with dx anymore, instead with du. so we can bring any non-u variable expression outside.[/STRIKE]

Last edited: Nov 29, 2012
16. Nov 29, 2012