How do you find#int x/sqrt(x^2 + 1) dx # using trigonometric substitution?

Question

1395 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-12T19:16:01

#int x/sqrt(x^2+1)*dx=sqrt(x^2+1)+C#

#int x/sqrt(x^2+1)*dx#

After using #x=tany# and #dx=(secy)^2*dy# transforms, this integral became

#int tany/sqrt((tany)^2+1)*(secy)^2*dy#

=#int tany/sqrt((secy)^2)*(secy)^2*dy#

=#int ((secy)^2*tany)/secy*dy#

=#int secy*tany*dy#

=#secy+C#

=#sqrt((tany)^2+1)+C#

=#sqrt(x^2+1)+C#