Question

On problem set 9, Integration techniques, exercise 5A-3 (f), shouldn't the answer have a |x|*sqr(xˆ2-aˆ2) on the denominator instead of x*sqr(xˆ2-aˆ2)? I'm asking because the domain of the arcsin() function is (-1<=x<=1), hence the negative branch should be reflected on the answer. Also, I believe that the answer should express the restriction on x<>0, because the original function is not defined at this point.

Answer 1

The derivative is
\[ \frac{d}{dx} \sin^{-1} u = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \]
in this case with \( u= ax^{-1}\)
and \( \frac{du}{dx} = -ax^{-2} \)
we have
\[ \frac{d}{dx} \sin^{-1}\left( \frac{a}{x} \right) = \frac{1}{\sqrt{1- \frac{a^2}{x^2}}}\cdot -\frac{a}{x^2} \]
notice we have only x^2 terms:
\[ = \frac{-a}{x^2\sqrt{1- \frac{a^2}{x^2}}} \]
you can also "bring" one x into the square root:
\[\frac{-a}{x\sqrt{x^2- a^2}} \]

I suppose you can make note of when this expression is real, but it is self evident we want x^2> a^2

Answer 2

though you are right, we should write |x| if we "bring" one of the x's inside the root.

Answer 3

I'm still confused:

Considering that arcsin (a/x), the original function, is defined whenever x<=-a or x>=a.

and that the answer can be stated in two forms:

(1) \[ = \frac{-a}{x^2\sqrt{1- \frac{a^2}{x^2}}} \]

and

(2) \[\frac{-a}{x\sqrt{x^2- a^2}} \],

From (1) is clear that x can take negative values, because as long as x <= -a, the square root will be real.

However, without the absolute value on (2), then, for x <= -2, (1) and (2) will give different answers.

Answer 4

@phi, thank you for this exchange. It really helped clear my doubts!