Question

integral of 1/(cos^2x tanx ) dx

Answer 1

what else can we write 1/cos^2 as?

Answer 2

please don't just give the answer whoever is typing

Answer 3

\[\frac{1}{\cos ^{2}x \tan x}=\frac{1}{\cos x \sin x}\]

Answer 4

^there is a better way

Answer 5

=1/cos^2 x .tanx
=1/(sin x . cos x)
=cosecx.secx
integration of this will be ln|tan x |
am i right???

Answer 6

\[u=\tan x\]

Answer 7

yes

Answer 8

thats probably the best way

Answer 9

@turingtest- what other way is there...

Answer 10

\[{1\over\cos^2x\tan x}={\sec^2x\over\tan x}\]\[u=\tan x\]is the best way in my opinion

Answer 11

how do you guys see\[\int \sec x\csc xdx\]as\[\ln|\tan x|+C\]anyway? cuz I don't see it

Answer 12

neither do I know the derivation of that,i had an integral table given to me by one of the users here ,i looked upto to that hence it says that the integral will give ln|tan x|

Answer 13

WolframAlph agrees with @TuringTest on the substitution.

Answer 14

well I prefer integrals I can prove, and I don't know the proof of that
I'm playing with it right now

Answer 15

oh yeah, I promise this problem is probably "supposed" to be done by the sub I posted
but there's more than one way to skin a cat!

Answer 16

i agree...

Answer 17

\[u = \tan x\]\[du = \sec^{2} xdx\]\[\int\limits \sec x \csc x dx=\int\limits \frac{1}{u} du\]

Answer 18

Why do you want to skin a cat anyways?

Answer 19

They are not that bad in general :P

Answer 20

I don't know, satellite always says that, lol

Answer 21

ok I see it now, but it seems a rather roundabout way to do it
how would you know to sub u=tanx right there ?
that way seems more counter-intuitive to me, but that's just my opinion

Answer 22

Turing: If you have to ask, you'll never know. If you know, you need only ask. ;)

Answer 23

true, the best way is sub u=tan x from the beginning

Answer 24

ah-so sensei.... ;)