Question

Please help, integration.

Answer 1

@phi

Answer 2

@xapproachesinfinity

Answer 3

Hmm nice little u-sub for this one :)
\[\Large\rm u=\log x\]Is that log base 10?

Answer 4

Yes :D it's log base 10

Answer 5

Ok so we need to do a change of base before we can find our du.

Answer 6

\[\Large\rm u=\frac{\ln x}{\ln 10}\qquad\qquad\to\qquad\qquad u=\left(\frac{1}{\ln 10}\right)\ln x\]

Answer 7

Differentiating gives us:\[\Large\rm du=\left(\frac{1}{\ln10}\right)\frac{1}{x}dx\]

Answer 8

\[\Large\rm \int\limits \frac{\sec(\log x)}{x}dx=\int\limits \sec(\log x)\left(\frac{1}{x}dx\right)\]

Answer 9

Oh we'll need to multiply that coefficient to the other side,\[\Large\rm u=\log x\]\[\Large\rm (\ln10)du=\frac{1}{x}dx\]Understand how we're going to plug these in? :o
I know it's a little bit crazy.

Answer 10

"I'm never gonna use this in life..." idk anymore

Answer 11

and I got how you're going to plug that in.
Yeah it's really hard . These are challenge problems my tutor gave so he can have us do a integration competition thingy in this summer class

Answer 12

So you get something like this, yes?\[\Large\rm \ln(10)\int\limits \sec u ~du\]You can find the integral for secant in a table of integrals somewhere.
You can do it by hand, but it's not very intuitive.
It's involves a weird step that you wouldn't just stumble upon on your own.

Answer 13

yes
= ln(10) ln I sec u + tan u I + C

Answer 14

so I just plug u back in and that's it? : D

Answer 15

cool c:
yes, undo your substitution to wrap it up.

Answer 16

Thank you sooooooo much! You're the best :)))