Question

Integral of 1/(x^2+1)^2 from 0 to infinity

Answer 1

have you tried plugging in
\(\Large x = \tan u\)

Answer 2

dx = ...?

also change the limits,
when x= tan u=0, u =...?
when x= tan u = \(\infty \), u = ..?

Answer 3

yeah i tried that and i dont think i got the write answer

Answer 4

because you would get tanu=x and dx= sec^2

Answer 5

thus it would be (secx)^2/((tanx)^2)+1)^2

Answer 6

yea and you know whats
\(1+ \tan^2 x=..??\)

remember your trig identities :)

Answer 7

yeah sec squared

Answer 8

and it would be secx squared squared

Answer 9

then you'll need, 1/ sec x = cos x :)
go further, and let me know if u get stusk

Answer 10

**stuck

Answer 11

did you get it ? :)

Answer 12

no i kinda got stuck

Answer 13

i know im close and i know the answer but its not working out exactly

Answer 14

very good! :)

Answer 15

and then of course theta is equal to arctanx. and when you set up the triangle with theta equal to arctanx, then sin of 2 theta should be 2x/sqrtx^2+1 right

Answer 16

or

Answer 17

while doing substitution, haven't you found out the limits for theta?

Answer 18

when x= 0, theta =...?
when x= infinity, theta =...?

Answer 19

when x=0, then tan of theta would be zero which means theta would equal 0

Answer 20

thats correct

Answer 21

tan of theta never actually equals infinity at a point right? it has vertical asymptotes at odd numbers of pi/2

Answer 22

you're right.

so we can say that when theta is pi/2,
tan theta is +infinity, can't we?

Answer 23

ok yeah

Answer 24

and then just evaluate in terms of the bounds being iin terms of theta?

Answer 25

tan theta has a very very large value at theta = pi/2
which we "named" as infinity :)

Answer 26

so theta/2 + sin2theta/4 evaluated from o to pi/2

Answer 27

so then it should just be pi/4

Answer 28

\(\huge \checkmark \)

Answer 29

is that ok? i know it all makes sense but from a teachers perspective is it considered "slacking" maybe to not substitute theta for what we calculated it for in the beginning and then evaluating it in terms of the original bounds?

Answer 30

any method is fine,
whether you substitute back for x and then evaluate from 0 to infinity
or whatever we did, both method are absolutely and completely correct.

Answer 31

Ok. good enough for me. Thank you so much

Answer 32

welcome ^_^

Answer 33

One quick question

Answer 34

oh, please ask :)