How do you integrate #int 1/sqrt(x^2+4)# by trigonometric substitution?

1 Answer
Jul 25, 2016

#int (d x)/(sqrt (x^2+4))=l n(x/2+sqrt(1+x^2/4))+C#

Explanation:

#int (d x)/(sqrt (x^2+4))=?#

#"let be "u=x/2" ; " d x=2*d u" ; "x^2=4u^2#

#int (2*d u)/(sqrt(4u^2+4))=int(2*d u)/(sqrt(4(u^2+1))#

#int(cancel(2)* d u)/(cancel(2)*sqrt(u^2+1))=int(d u)/(sqrt(u^2+1))#

#"now, substitute "u=tan v" ; "v=arc tan u#

#d u=sec ^2 v* d v#

#int (sec ^2 v*d v)/(sqrt(tan^2 v+1))" ;so "tan^2 v +1=sec ^2 v#

#int(sec^2 v*d v)/(sqrt(sec^2 v))=int (cancel(sec)^2 v*d v)/(cancel(sec) v)=int sec v*d v#

#"expand fraction by " tan v+sec v#

#int sec v*d v*(tan v+sec v)/(tan v+sec v)#

#int (sec v*tan v + sec ^2 v)/(tan v + sec v) *d v#

#k=tan v+ sec v#

#d k=(sec v* tan v+sec ^2 v) *d v#

#int ( d k)/k=l n k+C#

#"undo substitution "k=tan v +sec v#

#int (d x)/(sqrt (x^2+4))= l n(tan v+sec v)+C#

#sec v=sqrt(1+tan ^2 v)=sqrt (1+u^2)#

#int (d x)/(sqrt (x^2+4))=l n(u+sqrt(1+u^2))+C#

#"undo substitution " u=x/2#

#int (d x)/(sqrt (x^2+4))=l n(x/2+sqrt(1+x^2/4))+C#