Trigonometric substitution usually is needed when the function contains a term of x^2+-a^2x2±a2 or a^2+-x^2a2±x2.
Let's convert it to such a form by substituting u=e^xu=ex and du=e^xdxdu=exdx:
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.)
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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
