# Method IntegralWhat method is necessary to calculate the integral of f(x) = sqrt(16-x^2) ?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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Take the equation of the function  sqrt(16-x^2)  and factor 16-x^2 by 16.

sqrt(16-x^2) = sqrt16(1-x^2/16) = 4 sqrt (1 - (x/4)^2)

If (x/4)^2 would be one of the trigonometric functions sine or cosine, you'll get: sqrt (1  - sin^2 a) = +- cos a or  sqrt (1  - cos^2 a) = +- sin a

Put (x/4)^2 = cos ^2 a => cos a = x/4 => -sin a da = dx/4 => da = -dx/4sqrt(1 - (x/4)^2)

int 4 sqrt (1 - (x/4)^2) dx = 4 int sqrt (1 -cos ^2 a)*(-4sin a)da

int 4 sqrt (1 - (x/4)^2) dx = -16 int sin^2 a da

use the formula sin^2 a = (1-cos 2a)/2

-16 int sin^2 a da = -16 int ((1-cos 2a)/2) da = -8int da + 8 int cos 2a da

-16 int sin^2 a da = -8a + 4 sin 2a + c

cos a = x/4 => a = arccos (x/4)

ANSWER: int sqrt(16-x^2) = -8arccos (x/4)+ 4 sin 2arccos (x/4) + c

beckden | High School Teacher | (Level 1) Educator

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Use trigonometric substutuion.  x = 4 sin(u)

This gives u = arcsin(x/4), 4cos(u)  = sqrt(16 - x^2)

or sin(u) = x/4 and cos(u) = sqrt(16-x^2)/4

dx = 4 cos(u) du

16 - x^2 = 16 - (4 sin(u)) = 16 - 16 sin^2(u) = 16(1 - sin^2(u)) = 16cos^2(u)

sqrt(16 - x^2) = 4 cos(u)

so our integral is

integral(sqrt(16 - x^2) dx) = integral(sqrt(16 cos^2(u)) 4 cos(u) du)

= 16 integral(cos^2(u) du) = 16 integral(1/2(cos(2u) + 1))   Half angle formula

= 8 integral(cos(2u) + 1) = 8 (1/2 sin(2u) + u) = 4 (2 sin(u)cos(u) - 1) + 8u + C

= 8 x/4 sqrt(16 - x^2)/4 + 8 arcsin(x/4) + C

= 1/2 x sqrt(16 - x^2) + 8 arcsin(x/4) + C

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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