For integrals involving the root sqrt(x^2+a^2),√x2+a2, we can use the substitution x=atanthetax=atanθ. Here, a^2=16, a=4, x=4tanthetaa2=16,a=4,x=4tanθ
So, dx=4sec^2thetad thetadx=4sec2θdθ and we can apply the substitution.
int(4sec^2theta)/sqrt(16+16tan^2theta)d theta=int(4sec^2theta)/(sqrt(16(1+tan^2theta)))d theta∫4sec2θ√16+16tan2θdθ=∫4sec2θ√16(1+tan2θ)dθ
=intsec^2theta/sqrt(1+tan^2theta)d theta=∫sec2θ√1+tan2θdθ
Recalling the identity 1+tan^2theta=sec^2theta,1+tan2θ=sec2θ,
=intsec^2theta/sqrt(sec^2theta)d theta=∫sec2θ√sec2θdθ
=intsec^cancel(2)theta/cancel(secthetad) theta
=intsecthetad theta=ln|sectheta+tantheta|+C
This is a fairly common integral, it should be memorized. We need things in terms of x. Recalling that x=4tantheta, tantheta=x/4.
We still need the secant in terms of x. Applying the identity 1+tan^2theta=sec^2theta:
16/16+x^2/16=sec^2theta
sectheta=sqrt(x^2+16)/4
Thus, we have
ln|(sqrt(x^2+16)+x)/4|+C=ln|sqrt(x^2+16)+x|-ln(4)+C
We may absorb the ln4 into the constant of integration. We're left with
intx/sqrt(16+x^2)dx=ln|sqrt(x^2+16)+x|+C
