Question

integrate question

Answer 1

\[\cosh 2x = \frac{ 1= \tanh^2x }{ 1-\tanh^2x }\]\[t=tanhx\] integrate\[\int\limits sech2x dx\]

Answer 2

I assume I am not welcome here?

Answer 3

\[sech 2x = \frac{ 1 }{ \cosh 2x }\]\[=\frac{ 1 }{ \frac{ 1 +\tanh^2x }{ 1-\tanh^2x }}\]

Answer 4

@inkyvoyd why are you saying that?

Answer 5

because last time I used wolfram alpha show steps :)

Answer 6

still it doesnt explained why are you saying that?

Answer 7

because I spoil your fun :)

Answer 8

also this is a calculus 2 question

Answer 9

ignore the question mark at the end

Answer 10

well in my place, calculus is calculus, doesnt matter 1 or 2, so im not used to the label

Answer 11

anyway you going to help me figure it out??

Answer 12

uhh lemme go ask wolfram alpha lol

Answer 13

mind you it entirely ignores your hint

Answer 14

humm i need to used the hint, i prefer to used definition actually

Answer 15

@ayeshaafzal221 please help

Answer 16

after subs it became
\[\frac{ 1 }{ \frac{ 1+t^2 }{ 1-t^2 } }\] \[= \frac{ 1-t^2 }{ 1+t^2 }\]

Answer 17

yes thats right what is it that ur not understanding?

Answer 18

i dont know what is the next step

Answer 19

have you worked out dt yet? dt = d(tanh x) ....

then use \( sech ^{2} x = 1 - \tanh^{2} x \)

it should come out easy after that.

Answer 20

@IrishBoy123 i dont know how to work the dt
can you show the first step?

Answer 21

from wiki
\( \frac{d}{dx}\tanh x = 1 - \tanh^2 x = \operatorname{sech}^2 x = 1/\cosh^2 x \,\)

so t = tanh x

dt = ??

PS i have not verified that idntity you are using, so i assume that that's good...

PPS i got an annoying email yesterday from the Stasi so i am having to be less helpful than i would like :-))

Answer 22

is okay, i think i get the idea, thanks

Answer 23

time at hand, let's get at work now.
\[\int\limits_{}^{}sech2xdx=\int\limits_{}^{}\frac{ 1-\tan^2hx }{1+\tan ^2hx }dx\\ now\ use \ \sec^2hx=1-\tan^2hx\ and \ put\ tanhx=t\\ \implies\ \sec^2hxdx=dt\\ \therefore\ \int\limits_{}^{}sech2xdx=\ \int\limits_{}^{} \frac{ dt }{ 1+t^2 }\\= \tan^{-1} t +K\\=\tan^{-1} tanhx +K\]

Answer 24

@inkyvoyd: would you please stay on topic (which does not involve your feeling unwelcome). Thanks.

Answer 25

@mathmale , would you please stay on topic by deleting things off topic, and sending me a message instead of further derailing the thread?

oh wait sorry you're the mod here yah I"ll stay on topic oops

Answer 26

That'd be highly advisable. ;)

Answer 27

@eninone thank you for help me to confirm my work, and thanks to those who help :)

Answer 28

+1 @inkyvoyd