Question

I really need help on trig substitution I get it but when I try to simplify its really hard for me.

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{x^2+4} }\]

Answer 2

can you please show all your steps of simplifying

Answer 3

I managed to simplify it down to:
\[\int\limits_{}^{}\sec \theta\]
where theta equals:\[\arctan(\frac{ x }{ 2 })\]

Answer 4

but the integral of sec theta is really long and becomes HUGE when you back-substitute the xs

Answer 5

i simplified it down as far as possible

Answer 6

What you've done is correct, well I get the same simplification

Answer 7

but then do i substitute arctan(x/2) for theta?

Answer 8

after doing the integral

Answer 9

Then using substitution I would let:
u=arctan(x/2)
then you have: \[\int\limits_{}^{}\sec(u)du\] multiply top and bottom by tan(u)+sec(u) \[=\int\limits_{}^{}\frac{ \sec^2(u)+\tan(u)\sec(u) }{ \tan(u)+\sec(u) }du\]
let p=tan(u)+sec(u)
then dp=(sec^2(u))+tan(u)sec(u) du
\[=\int\limits_{}^{}\frac{ 1 }{ p }dp = log(p) + constant\]
then sub back in that p=tan(u)+sec(u)
and sub your 'u' back in too

Answer 10

ok thx

Answer 11

yeah sorry, the final solution should be in terms of 'x'

Answer 12

i also know that
\[\int\limits_{}^{}\sec \theta=-\log(\cos(θ/2)-\sin(θ/2))+\log(\cos(θ/2)+\sin(θ/2))\]

Answer 13

so now i would sub back in arctan(x/2) for theta right?

Answer 14

and then simplify

Answer 15

i got:
\[1/4 \sqrt{(x^2+4)} x+\sinh^{-1})(\frac{ x }{ 2 })+c\]

Answer 16

something like that

Answer 17

ok thx c ya

Answer 18

the final solution I got was:\[\log(\frac{ 1 }{ 2 }(\sqrt{x^2+4}+x) + constant\]