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

[SpicyPepper]

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

Homework Statement

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

Homework Equations

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

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

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.

 

[LCKurtz]

Looks good to me except you need + C.

 

[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|

 

[Mentallic]

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|

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, let's 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.

 

[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.

 

[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

 

[Bohrok]

The first link has your answer with the absolute values and the second doesn't. I'll take a closer look later, unless someone else can explain what's up.
http://www.wolframalpha.com/input/?i=-ln|csc(x)+cotx|+-+ln|csc(x)-cotx|

http://www.wolframalpha.com/input/?i=-ln|csc(x)+cotx|+-+ln(csc(x)-cotx)

 

[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.

 

[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 :).