# How do I evaluate int(1-sinx)/cosx dx?

Jan 31, 2015

$= \ln \left(1 + \sin \left(x\right)\right) + c$

Well, this one is a little bit tricky...
I started with the idea that the result must be a logarithm...
So I did a little manipulation to get to a friendlier version of your function by multiplying and dividing by:

$\frac{1 + \sin \left(x\right)}{1 + \sin \left(x\right)}$; so I get:

$\int \frac{1 - \sin \left(x\right)}{\cos} \left(x\right) \cdot \frac{1 + \sin \left(x\right)}{1 + \sin \left(x\right)} \mathrm{dx} =$
$= \int \frac{1 - {\sin}^{2} \left(x\right)}{\cos \left(x\right) \left(1 + \sin \left(x\right)\right)} \mathrm{dx} =$
$= \int \frac{{\cos}^{2} \left(x\right)}{\cos \left(x\right) \left(1 + \sin \left(x\right)\right)} \mathrm{dx} =$
$= \int \frac{\cos \left(x\right)}{\left(1 + \sin \left(x\right)\right)} \mathrm{dx} =$ which is a manipulated version of your original function and we can call it (1). The integral of (1) is indeed a logarithm:

$= \ln \left(1 + \sin \left(x\right)\right) + c$

Which derived gives (1) or your original function!!!!!