# Using Trignometric Integration

1. Aug 12, 2009

### ntox101

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

I am having problems with this integral. I cannot find a similar example in the book(Calculus 8th Edition) or an odd numbered problem. Here it is

Integral [ 1 / csc [x] - 1 ] dx

2. Relevant equations

None to my knowledge.

3. The attempt at a solution

I have tried multiplying by the conjugate to form Integral [ csc[x] + 1 / cot^2 x ] dx . But for some reason I think I am doing something incorrect. I have also tried converting into terms of cosine and sine. No luck. :(

2. Aug 12, 2009

### jpreed

This is more of a trig problem than a calculus problem.

$$\int\frac{1}{\csc x -1}$$

$$=\int\frac{1}{\frac{1}{\cos x}-1}$$

$$=\int\frac{1}{\frac{1-\cos x}{\cos x}}$$

$$=\int\frac{\cos x}{1 - \cos x}$$

$$=\int\frac{(\cos x)(1 + \cos x)}{(1 - \cos x)(1 + \cos x)}$$

$$=\int\frac{\cos x + \cos^2 x}{\sin^2 x}$$

You should be able to do the rest. Hope this helps.

3. Aug 13, 2009

### Staff: Mentor

The way you have written the integrand is ambiguous. jpreed has made an assumption about what you meant, which might not be what the original problem is. Is the denominator of your integrand just csc(x) or is it csc(x) - 1? If it's the latter you must put parentheses around it to prevent ambiguity.

IOW, is this the integral?
$$\int (\frac{1}{csc(x)} - 1) dx$$
(This is how most would interpret what you have written.)

Or is it this, which is how jpreed interpreted it?
$$\int \frac{1}{csc(x) - 1} dx$$

4. Aug 13, 2009

### ntox101

Mark44,

jpreed's notation is correct according to this worksheet. I ended up splitting the integrals into two separate ones and using U-Substitution on the first set and just taking the integral of cot2x which is provided by an integration table in the back of my text.

Thanks for your help, both of you.

Jon

5. Aug 13, 2009

### Staff: Mentor

This is a case of two wrongs making a right. You miswrote the integral that was in your worksheet, and jpreed translated it as what he thought you really meant.

My point is that if you write a rational expression such as a/b + c, you might mean this to be a/(b + c), but most would read this as (a/b) + (c).

Use parentheses!

6. Aug 13, 2009

### ntox101

Yeah, I noticed that, hence when you try something similar in a graphing calculator/matlab/mathmatica it gives wrong answers. I will in the future, thanks.

7. Aug 13, 2009

### Staff: Mentor

The graphing calculator/matlab/Mathematica are giving correct answers to what you entered, which doesn't happen to be the same as what you intended.

8. Aug 13, 2009

### jpreed

To be clear, I realized the ambiguity in the integral as he had it written. If you take the time to read the entire original post he mentions multiplying the conjugate at the bottom.

I untangled the ambiguity from what he had written there and then worked out the rest.