How do you evaluate #\int _ { 0} ^ { 1} \sqrt { 2+ x ^ { 2} } dx#?

Question

1472 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-07T14:39:51

#int_0^1 sqrt(2+x^2)*dx=sqrt3/2#+#ln((sqrt6+sqrt2)/2)#

#int_0^1 sqrt(2+x^2)*dx#

After using #x=sqrt2tanu# and #dx=sqrt2(secu)^2*du# transforms , this integral became

#I=int_0^arctan(sqrt2/2) 2(secu)^3*du#

=#int_0^arctan(sqrt2/2) 2(secu)^2*secu*du#

=#[2tanu*secu]_0^arctan(sqrt2/2)#-#int_0^arctan(sqrt2/2) 2tanu*secu*tanu*du#

=#2*(sqrt2/2*sqrt6/2-0*1)#-#int_0^arctan(sqrt2/2) 2secu*(tanu)^2*du#

=#sqrt3#-#int_0^arctan(sqrt2/2) 2secu*[(secu)^2-1]*du#

=#sqrt3#-#int_0^arctan(sqrt2/2) 2(secu)^3*du#+#int_0^arctan(sqrt2/2) 2secu*du#

=#sqrt3#-#I#+#2[ln(secu+tanu)]_0^arctan(sqrt2/2)#

#2I=sqrt3#+#2[ln(secu+tanu)]_0^arctan(sqrt2/2)#

#2I=sqrt3#+#2ln((sqrt6+sqrt2)/2)#

#I=sqrt3/2#+#ln((sqrt6+sqrt2)/2)#