What is the function y if dy/dx=square root(1-x^2)/x^2?

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beckden | High School Teacher | (Level 1) Educator

Posted on

`(dy)/(dx) = sqrt(1-x^2)/x^2`

We need to integrate to get the function

`y = int sqrt(1-x^2)/x^2 dx`

 

`x = cos(u) `

`dx = -sin(u) du`

`y = int sqrt(1-cos^2(u))/(cos^2(u)) (-sin(u)) du=-int (sin^2(u))/(cos^2(u)) du=-int tan^2(u) du`

`y = -int (sec^2(u)-1)du = u - tan(u) + C = arccos(x) - sqrt(1-x^2)/x + C`

So the answer is

`y = arccos(x) - sqrt(1-x^2)/x + C`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll have to integrate dy to get the primitive funtion Y.

dy = sqrt(1-x^2)dx/x^2

We'll integrate both sides:

`int` dy = `int` sqrt(1-x^2)dx/x^2

Let x = sin t => dx = cos t dt

t = arcsin x

`int` sqrt(1-x^2)dx/x^2 = `int` sqrt[1-(sin t)^2]*cos t dt/(sin t)^2

We'll apply Pythagorean identity to numerator:

`int` [sqrt (cos t)^2]*cos t dt/(sin t)^2 = `int` (cos t)^2dt/(sin t)^2

`int` (cos t)^2dt/(sin t)^2 = `int` [1-(sin t)^2]dt/(sin t)^2

`int` [1-(sin t)^2]dt/(sin t)^2 =`int` ` ` dt/(sin t)^2 - `int` dt

`int` [1-(sin t)^2]dt/(sin t)^2 = -cot t - t + C

`int` sqrt(1-x^2)dx/x^2 = - cot (arcsin x) - arcsin x + C

The requested primitive function Y is: Y = - cot (arcsin x) - arcsin x + C.

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