$1/\cos x$ integration Question: The aim of this question is to integrate $\frac{1}{\cos x}$
a. Write an integral as $\int \frac{\cos x}{\cos^2(x)}dx$ [DONE]
b. Use $u= \sin x$ [DONE]
c. Use partial fraction to integrate [DONE]
d. Multiply a fraction top and bottom by $1+\sin x$ to simplify it.
My answer to part c is $\frac{\ln(1+\sin x)}{2} - \frac{\ln(1-\sin x)}{2} +c$ but I could not simplify it by multiplyting $1 + \sin x$. Is there a way I could do it?

From your answer in part (c) 
$ A =\frac{1}{2} \ln(1+\sin x ) - \frac{1}{2}\ln (1-\sin x) $ 
$  = \ln (\sqrt{1+\sin x}) - \ln (\sqrt{1-\sin x}) $
$  = \ln\left(\frac{\sqrt{1+\sin x }}{\sqrt{1-\sin x}}\right) $
$= \ln\left(\sqrt{\frac{1+\sin x }{1-\sin x} \cdot \frac{1+\sin x}{1+\sin x}}\right) $
$= \ln\left(\frac{1+\sin x }{\sqrt{1-\sin^2 x}}\right)$
$= \ln\left(\frac{1+\sin x }{\cos x}\right)$
$= \ln\left(\sec x + \tan x \right) .$  
