Question

When taking the integral of something, do you use the chain rule??
Ex: integral of secudu = ln(secu + tanu) +C <--- would you take the prime of u and multiply it to that equation?

Answer 1

I don't quite understand what you're asking.
Are you asking if you should multiply your answer by secu*tanu?
(Because that's the derivative of secu)

Answer 2

yes!!! :P

Answer 3

If you were taking a derivative that would be a good idea :o
....
But you're integrating.

Answer 4

no

Answer 5

well, it's more like a reverse of chain rule.

Answer 6

so to get the integral, you would NOT use the chain rule but you would just plug in the u into that equation? correct?

Answer 7

You don't use chain rule directly because it's for derivative. However, the application of chain rule is used when deriving the integral of secx

Answer 8

Kind of continuing off of what sour said,
you start with some messy looking integral,
it requires some type of "reverse chain rule".
That's what the u-substitution takes care of.

The rest is just integrating normally.
Are you asking about this specific problem?
Like do you want to know where that weird looking log is coming from?

Answer 9

Ohhhh okay I understand now :) And sure, where does the ln come from?

Answer 10

You probably got this from a table or something right? :)
It's a bit of a weird substitution to solve this integral.
Not something that you would probably think of, off the top of your head.

Answer 11

you multiply the top and bottom by secu+tanu,\[\Large\bf\sf \int\limits \sec u\frac{\sec u+\tan u}{\sec u+\tan u}\;du\quad=\quad \int\limits \frac{\sec u \tan u+\sec^2u}{\sec u + \tan u}\;du\]Then you do a substitution,\[\Large\bf\sf m=\sec u+\tan u\]Numerator ends up being your dm,\[\Large\bf\sf \int\limits \frac{dm}{m}\quad=\quad \ln|m|+c\]And then undo the substitution :U
Bit of a weird one huh?

Answer 12

That is an odd one lol :P But it makes sense! Thanks!