Integration by Trigonometric Substitution
We let #e^x=2*tan theta#
Let #e^(2x)=4*tan^2 theta#
Let #e^x dx=2*sec^2 theta* d theta#
and #dx=(2*sec^2 theta* d theta)/e^x=(2*sec^2 theta* d theta)/(2*tan theta)=(sec^2 theta* d theta)/(tan theta)#
#int e^x/sqrt(e^(2x)+4)dx=int (2*tan theta )/sqrt(4*tan^2 theta+4)(sec^2 theta* d theta)/(tan theta)#
#int e^x/sqrt(e^(2x)+4)dx=int (2*tan theta )/(sqrt(4)sqrt(tan^2 theta+1))(sec^2 theta* d theta)/(tan theta)#
#int e^x/sqrt(e^(2x)+4)dx=int (2*tan theta )/(2*sqrt(sec^2 theta))(sec^2 theta* d theta)/(tan theta)#
#int e^x/sqrt(e^(2x)+4)dx=int (2*tan theta )/(2*sec theta)(sec^2 theta* d theta)/(tan theta)#
#int e^x/sqrt(e^(2x)+4)dx=int (cancel2*canceltan theta )/(cancel2*cancelsec theta)(cancelsec theta* sec theta*d theta)/(canceltan theta)#
#int e^x/sqrt(e^(2x)+4)dx=int sec theta*d theta=ln (sec theta+tan theta)+C_1#
We return to the original variables:
From #e^x=2*tan theta#
#e^x# is Opposite side
#2# is Adjacent side
#sqrt(e^(2x)+4)# is Hypotenuse
#int e^x/sqrt(e^(2x)+4)dx=int sec theta*d theta=ln (sec theta+tan theta)+C#
#int e^x/sqrt(e^(2x)+4)dx=ln (sqrt(e^(2x)+4)/2+e^x/2)+C_1#
#int e^x/sqrt(e^(2x)+4)dx=ln ((sqrt(e^(2x)+4)+e^x)/2)+C_1#
#int e^x/sqrt(e^(2x)+4)dx=ln (sqrt(e^(2x)+4)+e^x)-ln 2+C_1#
#color(blue)(int e^x/sqrt(e^(2x)+4)dx=ln (sqrt(e^(2x)+4)+e^x)+C)#
God bless....I hope the explanation is useful.
