Using Trig Substitution in Trig Integration

[Zack K]

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?

 

[Orodruin]

but how do you know the value of the opposite, adjacent and hypotenuse?

The exact values do not matter, just the ratios. You can choose the hypothenuse to have length one. Then you get the opposite because you know what the sine of the angle is. Pythagoras’ theorem then gives you the adjacent.

 

[Ray Vickson]

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?

Well, $\cot(\arcsin(\frac1 2 x)) = \cos(\theta)/\sin(\theta),$ where $\sin \theta = \frac 1 2 x.$

BTW: in LateX, please write $\cot ...$ instead of $cot ...$, and $\arcsin ...$ instead of $arcsin ...$. You do that by typing "\cot" instead of "cot", etc. (Similarly for all the other trig functions, the hyperbolic functions, and things like "ln", "log", "lim", "max", "min" and a host of others).

 

[epenguin]

$dx$ has got in there twice.