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

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the integral sqrt(16-x^2) dx

Let I = Integral sqrt(16-x^2) dx.

Put x =4sint. Then dx = 4costdt

I = integral sqrt(16-16sin^2t)4cost dt

=4*4integral sqrt(1-sin^2t) cost dt

=16integral sqrt (cos^2) *cost dt

=16 integral  cos^2 t dt

=16 integral [(1+cos2t)/2]dt

=16*t/2 + (16/2) (sin2t)/2 + dt

=8t+16sintcost +C

=8t+16sint(1-sin^2 t) + C. But sint = x/4 and so cost = 1-sqrt(x^2/16)

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

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



giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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


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

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