Question

integrate; 27x sec^2 (3x) dx

Answer 1

wolframalpha?

Answer 2

\[27x\frac{tan(3x)}{3}-27(??)+C\]is what i got so far :)

Answer 3

why wolfram? i dont care about an answer, its the process i enjoy

Answer 4

because if u use hashtag evaluate it gives the process, just like mathematica 8

Answer 5

if I can see how to integrate tan(3x) id be doing good :)

Answer 6

cool, ill have to try to remember the hashtag trick then

Answer 7

in the mean time, im practiceing our upcoming int by parts for calcII

Answer 8

maybe u = 27 and dv = x sec^2(3x) ?

Answer 9

i know, that was pointless :)

Answer 10

sec^2 = tan^2 + 1\[\]
27x tan^2(3x) +27x ... but then what to do with the tan^2(3x)\[\]
i cant recall to well

Answer 11

integration by parts

Answer 12

u=27x

Answer 13

ln(cos(3x)) somehow

Answer 14

i got that much :) its that sec^2(3x) part i got a brainurysm on

Answer 15

\[\frac{\tan(3x)}{3}\]

Answer 16

right, then that suits up to ln(cos(3x)) somehow :)

Answer 17

or something akin to it

Answer 18

\[\int\tan(x)dx=\int\frac{sin(x)}{\cos(x)}dx=\int\frac{-1}{u}du=-\ln(|u|)+c=-\ln|cos(x)|+c\]

Answer 19

... yep, thats what I was brain blocked on lol, thnx :)

Answer 20

np