I can't seem to figure this out.
it's asking me to take the integral of:
These are my options.
1.
2.
3.
4.
Thanks in advance
I've spent a lot of time on it and I could only go so far. I think it's option 4, but I can't see how. Maybe I need to brush up on logarithms. Sorry I wasn't able to go till the end, but I spent 40 minutes on this, and I thought I might as well tell you what I've got, or it would be a wasted 40 minutes.
Hope this helps at least a little
$\displaystyle 33 \int \frac{cos(x) + sin(x)}{sin(2x)}dx$
$\displaystyle =33 \int \frac{cos(x)}{sin(2x)} + \frac{sin(x)}{sin(2x)}dx$
Use trig identity: $\displaystyle sin(2x)=2sin(x)cos(x)$
$\displaystyle =33 \int \frac{cos(x)}{2sin(x)cos(x)} + \frac{sin(x)}{2sin(x)cos(x)}dx$
$\displaystyle =33 \int \frac{1}{2sin(x)} + \frac{1}{2cos(x)}dx$
$\displaystyle =\frac{33}{2} \int \frac{1}{sin(x)} + \frac{1}{cos(x)}dx$
$\displaystyle =\frac{33}{2} \int csc(x) + sec(x) dx$
I've just memorized the integrals of cosecant and secant, but they aren't too impossible to do.
$\displaystyle =\frac{33}{2} [-ln|cot(x)+csc(x)| + ln|tan(x) + sec(x)|]+C$