Integrate #int x/sqrt(1+x^2) dx# by Trigonometric substitution
Solution:
Let #x=tan theta#
Let #x^2=tan^2 theta#
Let #dx=sec^2 theta* d theta#
#int x/sqrt(1+x^2) dx=int (tan theta*sec^2 theta* d theta)/sqrt(1+tan^2 theta)=int (tan theta*sec^2 theta* d theta)/sqrt(sec^2 theta)#
#int x/sqrt(1+x^2) dx=int (tan theta*sec^2 theta* d theta)/sec theta#
#int x/sqrt(1+x^2) dx=int tan theta*sec theta* d theta#
#int x/sqrt(1+x^2) dx=int (sin theta/cos theta*1/cos theta)* d theta#
#int x/sqrt(1+x^2) dx=int sin theta/cos^2 theta d theta#
#int x/sqrt(1+x^2) dx=int cos^(-2) theta*sin theta* d theta#
#int x/sqrt(1+x^2) dx=-int cos^(-2) theta*(-sin theta)* d theta#
#int x/sqrt(1+x^2) dx=- (cos theta)^(-2+1)/(-2+1)#
#int x/sqrt(1+x^2) dx=(cos theta)^(-1)=sec theta#
Our Right Triangle with acute angle #theta#
#tan theta=x/1#
#sec theta=sqrt(1+x^2)/1#
#int x/sqrt(1+x^2) dx=(cos theta)^(-1)=sec theta=sqrt(1+x^2)/1#
Therefore
#int x/sqrt(1+x^2) dx=sqrt(1+x^2)+C#
God bless....I hope the explanation is useful.
