When dealing with integrals involving a root in the form #sqrt(x^2+a^2)# where #a# is a constant, we can make the following trigonometric substitution:
#x=atan(theta)#
Here, #a^2=1, a=sqrt(1)=1, x=tan(theta)#
#dx=sec^2(theta)d theta#
Substituting yields:
#intsec^2(theta)/sqrt(tan^2(theta)+1)d theta#
Recall the identity
#tan^2(theta)+1=sec^2(theta)#. Apply it to what's under the root:
#intsec^2(theta)/sqrt(sec^2(theta))d theta#
#sqrt(sec^2(theta))=|sec(theta)|=sec(theta)#. We'll have to assume secant will remain positive since we're working with an indefinite integral.
#intsec^(cancel(2)1)(theta)/cancel(sec(theta))d theta#
#intsec(theta)d theta#
This is a common integral, I'll show a proof at the end, but it should be memorized:
#intsec(theta)d theta=ln|sec(theta)+tan(theta)|+C#
We must rewrite in terms of #x.# We said #tan(theta)=x,# it was our initial substitution.
To find #sec(theta)#, recall the identity
#1+tan^2(theta)=sec^2(theta)#
Substitute in #x=tan(theta)# meaning that #tan^2(theta)=x^2#
#sec^2(theta)=1+x^2#
Take the root of both sides:
#sec(theta)=sqrt(1+x^2)#
Thus
#intsec(theta)d theta=intdx/sqrt(x^2+1)=ln|sqrt(1+x^2)+x|+C#
Proof of #intsecthetad theta=ln|sec(theta)+tan(theta)|+C:#
#intsecthetad theta=intsectheta (sectheta+tantheta)/(sectheta+tantheta)#
Multiply the integrand by #(sectheta+tantheta)/(sectheta+tantheta).# This is the same as multiplying by #1.#
Make the following substitution:
#u=sectheta+tantheta#
#du=(sec(theta)tan(theta)+sec^2(theta))d theta#
If we multiply out the numerator of our integrand, we have
#intsectheta(sectheta+tantheta)/(sectheta+tantheta)=int(sec^2theta+secthetatantheta)/(sectheta+tantheta)#
So, #du# appears in the numerator and #u# is the denominator. . We now have
#int(du)/u=ln|u|+C#
Rewrite in terms of #theta:#
#intsec(theta)=ln|sec(theta)+tan(theta)|+C#
