$\displaystyle \frac{cosx}{sinx} * cos x dx $
Im so lost that second cos x is really thorwing me off
Nah that's still the best I can come up with.
$\displaystyle \int \frac{1}{\sin(x)} dx = \int \frac{\sin(x)}{\sin^2(x)}dx = \frac{\sin(x)}{1-\cos^2(x)} dx$.
Let $\displaystyle u = \cos(x)$, $\displaystyle du = \sin(x)dx$
Hence integral becomes...
$\displaystyle \int \frac{1}{1-u^2}du = arctanh(u)$
Hello, x5pyd3rx!
$\displaystyle \int \frac{\cos x}{\sin x}\cdot\cos x\, dx $
$\displaystyle \int\frac{\cos^2\!x}{\sin x}\,dx \;\;=\;\;\int\frac{1-\sin^2\!x}{\sin x}\,dx \;\;=\;\;\int\left(\frac{1}{\sin x} - \sin x\right)\,dx$
. . . . . . . .$\displaystyle =\;\int(\csc x - \sin x)\,dx \;\;=\;\;\ln|\csc x - \cot x| + \cos x + C$