integration

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

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

$= \int \frac{\sin^2(x)}{1 - \sin^2(x)} \cos(x) dx$

$= \int \frac{u^2}{1 - u^2} du$ (where $u = \sin(x)$ ).

$= \int \frac{u^2 - 1+1}{1 - u^2} du$

$= \int \left( \frac{u^2 - 1}{1 - u^2} + \frac{1}{1 - u^2} \right) du$

$= \int (-1 + \frac{1}{1 - u^2}) du$

$= -\int du + \int \frac{1}{1 - u^2} du$

$= -u + \int \frac{1}{(1 - u)(1+u)} du$

$= -u + \int \left( \frac{1/2}{(1 - u)} + \frac{1/2}{(1+u)} \right) du$

$= -u + (1/2) \int \frac{du}{(1 - u)} + (1/2) \int \frac{du}{(1+u)}$

$= -u + (1/2) (-\ln|1-u|) + (1/2) (\ln|1+u|) + C$

$= -u + (1/2) \ln\left( \left\lvert \frac{1+u}{1-u} \right\rvert \right) + C$

$= -\sin(x) + (1/2) \ln\left( \left\lvert \frac{1+\sin(x)}{1-\sin(x)} \right\rvert \right) + C$

$= -\sin(x) + (1/2) \ln\left( \left\lvert \frac{(1+\sin(x))^2}{\cos^2(x)} \right\rvert \right) + C$

$= -\sin(x) + (1/2) \ln\left( \left\lvert \frac{1+\sin(x)}{\cos(x)} \right\rvert^2 \right) + C$

$= -\sin(x) + \ln\left( \left\lvert \frac{1+\sin(x)}{\cos(x)} \right\rvert \right) + C$ .

$= -\sin(x) + \ln\left( \frac{1+\sin(x)}{|\cos(x)|} \right) + C$ .

*** Check ***

On a domain where $\cos(x) > 0$ :

$\frac{d}{dx}\left\{ -\sin(x) + \ln\left( \frac{(1+\sin(x))}{\cos(x)} \right) + C \right\}$

$= -(\cos(x)) + \frac{\cos(x)}{(1+\sin(x))} \frac{d}{dx} \left\{ \frac{(1+\sin(x))}{\cos(x)} \right\} + (0)$

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

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

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

$= -\cos(x) + \frac{1}{\cos(x)} = -\frac{\cos^2(x)}{\cos(x)} + \frac{1}{\cos(x)} = \frac{1 - \cos^2(x)}{\cos(x)}$

$= \frac{\sin^2(x)}{\cos(x)}$ .