# 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