Method Integral What method is necessary to calculate the integral of f(x) = sqrt(16-x^2) ? Solution First, you\'ll have to multiply and divide the square root with itself, to make more easy to choose the proper method of integration. 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.

1. Method Integral
What method is necessary to calculate the integral of f(x) = sqrt(16-x^2) ?
Solution
First, you'll have to multiply and divide the square root with itself, to make more easy to choose
the proper method of integration.
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