Not following an integral solution

[SamRoss]

In the image below, why is the third line not \frac {ln(cosx)} {sinx}+c ? Wouldn't dividing by sinx be necessary to cancel out the extra -sinx that you get when taking the derivative of ln(cosx)? Also, wouldn't the negatives cancel?

 

[SamRoss]

In the image below, why is the third line not \frac {ln(cosx)} {sinx}+c ? Wouldn't dividing by sinx be necessary to cancel out the extra -sinx that you get when taking the derivative of ln(cosx)? Also, wouldn't the negatives cancel?

I retract my question. I was imagining plugging cosx back in for u and imagining the second line was simply the integral of 1/cosx. Of course I was forgetting that I would also have to plug back in for du also which would in fact give me the sinx back (not to mention being a ridiculous thing to do because it just brings us back where we started).