Which method to choose to calculate the integral of f(x)=sqrt(16-x^2)?

Before choosing any method of calculation, you have to
multiply and divide the square root with itself.

Int
[sqrt(16-x^2)*sqrt(16-x^2)]dx/sqrt(16-x^2)

Int
(16-x^2)dx/sqrt(16-x^2) = Int 16/sqrt(16-x^2) + Int
(-x^2)/sqrt(16-x^2)

Int 16/sqrt(16-x^2) = 16 arcsin (x/4) +
C

Int (-x^2)/sqrt(16-x^2) we'll solve it by parts method,
choosing f=x and g'(x)=sqrt(16-x^2)dx

Let's see
why:

[sqrt(16-x^2)]' =
[1/sqrt(16-x^2)]*(16-x^2)'

[sqrt(16-x^2)]' = -2x/sqrt
(16-x^2), which is almost what we have in Integral
x*[-x/sqrt(16-x^2)]dx.

The method of integration by parts
is:

Int f*g'=f*g-Int
f'*g

So,

Integral
x*[-x/sqrt(16-x^2)]dx=xsqrt(16-x^2)-Int sqrt
(16-x^2)dx

Integral sqrt(16-x^2)dx=16 arcsin
(x/4)+xsqrt(16-x^2)-Int sqrt (16-x^2)

2Int (16-x^2)dx=16
arcsin (x/4)+xsqrt(16-x^2)

Int
(16-x^2)dx=8arcsin (x/4)+[xsqrt(16-x^2)/2] + C