Trigonometric substitution usually is needed when the function contains a term of #x^2+-a^2# or #a^2+-x^2#.
Let's convert it to such a form by substituting #u=e^x# and #du=e^xdx#:
#int\ 1/sqrt(u^2+36)\ du#
Now, we need to decide which trigonometric function to substitute in this integral. The options are #u={(asin(theta)),(atan(theta)),(asec(theta)):}#, where #a# is some constant. It should be apparent by trial substitution that only #u=6tan(theta)# will simplify the integral.
Thus, substitute #u=6tan(theta)# and #du=6sec^2(theta)d theta# to get
#6int\ sec^2(theta)/sqrt(36tan^2(theta)+36)\ d theta#
Pull out the #36# from the denominator (which cancels the #6# on the outside):
#int\ sec^2(theta)/sqrt(tan^2(theta)+1)\ d theta#
Now, the reason we could substitute #x=atan(theta)# in the first place was so that we could take advantage of the identity #tan^2(theta)+1=sec^2(theta)#.
This simplifies nicely to
#int\ sec(theta)\ d theta#
Now, to finish this problem, you can recall the integral of the secant function as #ln|sec(theta)+tan(theta)|+C# (which is probably given to you for tests that have formula booklets). Then, substitute back #u=6tan(theta)# and #u=e^x# to get the final answer of
#ln|1/6(sqrt(u^2+36)+u)|+C#
#=ln|1/6(sqrt(e^(2x)+36)+e^x)|+C#
#=ln|sqrt(e^(2x)+36)+e^x|-6+C#
#=ln|sqrt(e^(2x)+36)+e^x|+C#
(Note how in the final line we absorbed the #6# into the arbitrary constant #C#.)
#--------------------#
Here I will give you a quick derivation for the integral of #sec(theta)#.
The trick here is to multiply it by #(sec(theta)+tan(theta))/(sec(theta)+tan(theta))#:
#int\ sec(theta)*(sec(theta)+tan(theta))/(sec(theta)+tan(theta))\ d theta#
#=int\ (sec(theta)tan(theta)+sec^2(theta))/(sec(theta)+tan(theta))\ d theta#
This may seem at first like needlessly complicating the problem, but in fact now you can simplify this problem by substituting #u=sec(theta)+tan(theta)# and #du=(sec(theta)tan(theta)+sec^2(theta))d theta#:
#int\ (du)/u#
#=ln|u|+C#
#=ln|sec(theta)+tan(theta)|+C#
