- #1

Zack K

- 166

- 6

## Homework Statement

Integrate: $$\int \frac{dx}{x^2\sqrt{4-x^2}}dx$$

## Homework Equations

## The Attempt at a Solution

I got to the final solution ##\int \frac{dx}{x^2\sqrt{4-x^2}}dx=-\frac{1}{4}cot(arcsin(\frac{1}{2}x))##. But It's the method where you transform that to the solution ##-\frac{1}{4}cot(arcsin(\frac{1}{2}x))=-\frac{\sqrt{4-x^2}}{x}+C## that confuses me. I understand that you get that by seeing that on a right triangle, ##cot=\frac{adjacent}{opposite}##, but how do you know the value of the opposite, adjacent and hypotenuse?

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