Question

can i write tan(^-1)x, (read as tan inverse x) as 1/tanx ?

Answer 1

maybe?

Answer 2

\[\Large \tan^{-1}x \quad\ne\quad \frac{1}{\tan x}\]
It's not an exponent when it's -1, even though it's in the same location.
I usually try to avoid that notation by using:\[\Large \arctan x\]

Answer 3

so how would you integrate \[\tan ^{-1}x\] ?

Answer 4

\[\Large\bf\sf \int\limits \arctan x\;dx\]Ummm we'd have to do integration by parts.\[\Large\bf\sf u=\arctan x, \qquad\qquad dv=dx\]

Answer 5

\[\Large\bf\sf du=?, \qquad\qquad v=?\]

Answer 6

Have you learned about `Integration by Parts` ? :)

Answer 7

uh, yes i have.. do you mean i should write \[\tan ^{-1}x . 1\] and then integrate it by parts?

Answer 8

or there some other method?

Answer 9

times 1, is that what that notation is?

Answer 10

yes

Answer 11

then i could take u = tan^-1x and v = 1 and then integrate it... but would that work?

Answer 12

or will i get stuck somewhere?

Answer 13

Ok so I guess you learned parts the sloppy way hehe: without using the differential dx.

That's fine, we can work with that.

So our parts will be:
\[\Large\bf\sf u=\tan^{-1}x,\qquad\qquad v'=1\]So then we need to find our:\[\Large\bf\sf u'=?,\qquad\qquad v=?\]In order to do the whole setup, right?

Answer 14

You remember the derivative of tan^(-1)x?

The integral you end up with after doing parts will work out really nicely.
It'll be a simple U-substitution.

Answer 15

so is the integration like this \[xtan ^{-1}x - \frac{ 1}{ 2 }\log_{e}|1+x ^{2}| + c \] ?

Answer 16

Mmm yes that looks right! good job.

Answer 17

oh dude!!! this means i didnt get this right in my test this morning. i wrote tan inverse as 1/tan which is cot and then integrated it because it seemed so simple!!... anyway, thanks for your help :)

Answer 18

oh boy :3