# Trig integration (cos(x)+sin(x)) / sin(2x)

1. Oct 3, 2009

### SpicyPepper

Prepping for my test, and I can't seem to solve this problem.

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

$$\int \frac{cos(x) + sin(x)}{sin(2x)}dx$$

2. Relevant equations

Not sure if it led me astray, but I used the trig. property:

$$sin(2x) = 2sin(x)cos(x)$$

3. The attempt at a solution

$$= \int \frac{cos(x) + sin(x)}{2sin(x)cos(x)}dx$$

$$= 1/2 \int(csc(x) + sec(x))dx$$

$$= 1/2ln|csc(x) - cot(x)| + 1/2ln|sec(x) + tan(x)| + C$$

$$= 1/2 ln|csc(x)sec(x) + sec(x) - csc(x) - 1| + C$$

I feel like this is wrong, but not sure where I'm messing up. Maybe my trig. property isn't the way to go.

2. Oct 3, 2009

### LCKurtz

Looks good to me except you need + C.

3. Oct 3, 2009

### Bohrok

I would keep the answer in the form of the line before the last one. There's one little mistake in that line, though.
∫cscx dx = -ln|cscx + cotx| + C, not ln|cscx - cotx|

4. Oct 3, 2009

### Mentallic

Uhh...

$$LHS=ln\left(\frac{1}{cscx+cotx}\right)$$

$$=ln\left(\frac{1}{\frac{1}{sinx}+\frac{cosx}{sinx}}\right)$$

$$=ln\left(\frac{sinx}{1+cosx}\right)$$

$$RHS=ln\left(\frac{1}{sinx}-\frac{cosx}{sinx}\right)$$

$$=ln\left(\frac{1-cosx}{sinx}\right)$$

Now, lets assume:$$ln\left(\frac{sinx}{1+cosx}\right)=ln\left(\frac{1-cosx}{sinx}\right)$$

$$ln\left(\frac{sinx}{1+cosx}\right)-ln\left(\frac{1-cosx}{sinx}\right)=0$$

$$ln\left(\frac{(\frac{sinx}{1+cosx})}{(\frac{1-cosx}{sinx})}\right)=0$$

$$ln\left(\frac{sin^2x}{1-cos^2x}\right)=0$$

$$ln\left(\frac{1-cos^2x}{1-cos^2x}\right)=0$$

$$ln(1)=0$$

Therefore, they are equivalent.

5. Oct 3, 2009

### Bohrok

Graphically, it looks like they are the same if you ignore the absolute values. LHS - RHS looks like 0 on wolframalpha, but with absolute values, the graph of their difference is not 0.

6. Oct 4, 2009

### Mentallic

I re-checked my work and didn't seem to find any errors or possibilities of extraneous solutions etc. so I graphed it as well, but on graphmatica. It seems that with or without absolute values, both functions are equal. If you believe you haven't made a mistake, please post a picture of what you see on wolfram

7. Oct 4, 2009

8. Oct 4, 2009

### Bohrok

Ok, there is no difference between our integrals of cscx. I didn't recognize the other form immediately (I never worked with ∫cscx dx by itself either) so I thought it was wrong. Looking at the table of integrals in my calc book, I see that same integral, ln|cscx - cotx| and I did find it by looking on the internet, but apparently that form is not as common as the other one.

I learn something new almost every day on this site.

9. Oct 4, 2009

### SpicyPepper

cool discussion guys. I just worked the integral of csc(x) on my own, and I came up with that funky version. Didn't realize it was less common :).