Not following an integral solution

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SUMMARY

The discussion centers on the misunderstanding of integral solutions related to the function ln(cosx) and its derivative. The user initially questioned the necessity of dividing by sinx to cancel out the derivative's -sinx, but later retracted the question upon realizing the need to substitute both u and du back into the equation. This highlights the importance of correctly applying substitution methods in integral calculus.

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SamRoss
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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?

9
 
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SamRoss said:
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?

9
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).
 

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