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

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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

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

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

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

= $int secy*tany*dy$

= $secy+C$

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

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

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