`int_0^a (dx)/(a^2 + x^2)^(3/2) , a>0` Evaluate the integral

`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)`