Question

help integrating natural log (:

Answer 1

You don't need to integrate it. What's the integral of sec(x)?

Answer 2

ln(sec(x)+tan(x)

Answer 3

Okay, then what is a plausible \(u\)-substitution?

Answer 4

i was thinking about substituting in at ln in the denominator

Answer 5

du/sqrt(4ln(u)) ?

Answer 6

I'd be careful about that.

We know that
\[
\frac{d}{dx}\ln|\tan(x)+\sec(x)|=\sec(x)
\]
Right?

Answer 7

yeah i understand that bit because the tan(x) becomes sec^2 but you pull that out of the parentheses to simplify

Answer 8

All right, well, what happens when we allow \(u=\ln|\tan(x)+\sec(x)|\)? Try integrating it, after doing that substitution.

Answer 9

so just to make sure im on the right track here \[\int\limits_{}^{}25\sec(x)/\sqrt(4\ln(u)\]

correct?

Answer 10

Correct substitution, though you're forgetting to change variables of integration.

Answer 11

so sec(u)

Answer 12

(x25/cos(u))/(2sqrt(ln(u))

Answer 13

Not quite. Do you know how to do a u-substitution?

Answer 14

not with this

Answer 15

Remember that \(du=u'(x)dx\).

Answer 16

im re-integrating this part 25sec(u) part so \[\int\limits_{}^{}25\sec(u)du=25\ln(1/\cos(u)+\tan(u))+C\]

Answer 17

Whoa, wait, be careful, you can't just integrate separate parts.

Answer 18

i thought i could pull it apart and put 1 over the denominator and integrate separately

Answer 19

Not quite.

Answer 20

well im stuck so im trying anything

Answer 21

@LolWolf can I say something?

Answer 22

go ahead oops

Answer 23

Go for it.

Answer 24

To me, let \(u =\sqrt{ln(sec+tan)}\)--> du =\(\dfrac{sec xdx}{2\sqrt{(sec+tan})}\) so that you have 2du = the whole integrand. That's it

Answer 25

sorry, denominator is 2 sqrt (ln (sec +tan)

Answer 26

\[\sqrt{ln(sec+tan)}'=\dfrac {1}{2\sqrt{ln(sec+tan)}}*(ln(sec+tan)' )*(sec+tan)'\]
\[\dfrac {1}{2\sqrt{ln(sec+tan)}}*\dfrac{1}{sec+tan}*(sec+tan)'\]
\[\dfrac {1}{2\sqrt{ln(sec+tan)}}*\dfrac{1}{sec+tan}*sectan+sec^2\]
from the last term, factor sec out, you get sec*(tan+sec) then, cancel out with the denominator
at the end up, you have
\[du=\dfrac {sec }{2\sqrt{ln(sec+tan)}}dx\]

Answer 27

One more thing, I let 25 out of the integral before doing that step

Answer 28

I tried plugging 25 back in and got 25sec(x)/(2sqrt(ln(sec(x)+tan(x))))

Answer 29

my program wouldn't accept it

Answer 30

thanks for the help btw

Answer 31

nope, the integral at the end up is
\[\int 25du= 25u +C = 25\sqrt{ln(secx+tanx)}+C\]

Answer 32

ok thanks. having the right answer helps me backtrack to see how things fit together

Answer 33

lalala, I have back up here http://www.wolframalpha.com/input/?i=int+%2825secx%2F%28sqrt%284ln%28secx%2Btanx%29%29%29dx

Answer 34

haha nice..