Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 7

int_0^adx/(a^2+x^2)^(3/2)
Let's first evaluate the indefinite integral by integral substitution,
Let x=atan(u)
dx=asec^2(u)du
intdx/(a^2+x^2)^(3/2)=int(asec^2(u)du)/(a^2+a^2tan^2(u))^(3/2)
=int(asec^2(u))/(a^2(1+tan^2(u)))^(3/2)du
=int(asec^2(u))/((a^2)^(3/2)(1+tan^2(u))^(3/2))du
Use the identity:1+tan^2(x)=sec^2(x)
=int(asec^2(u))/((a^3)(sec^2(u))^(3/2))du
=1/a^2int(sec^2(u))/(sec^3(u))du
=1/a^2int1/sec(u)du
=1/a^2intcos(u)du
=1/a^2(sin(u))
We have used x=atan(u)
tan(u)=x/a
Now let's find sin(u) for triangle with angle theta, opposite side as x and adjacent side as a and hypotenuse as h,
h^2=x^2+a^2
h=sqrt(x^2+a^2)
So, sin(u)=x/sqrt(x^2+a^2)
=1/a^2(x/sqrt(x^2+a^2))
Add a constant C to the solution.
=1/a^2(x/sqrt(x^2+a^2))+C
Now let's evaluate the definite integral,
int_0^a(dx)/(a^2+x^2)^(3/2)=[1/a^2(x/sqrt(x^2+a^2))]_0^a
=[1/a^2(a/sqrt(a^2+a^2))]-[1/a^2(0/sqrt(0^2+a^2))]
=[1/(asqrt(2a^2))]-[0]
=(1/(a^2sqrt(2)))
=1/(sqrt(2)a^2)