Given the following function we find its derivative:
$$a) \frac{d}{dx}\left[\arcsin(\sin(x))\right] = \frac{1}{\sqrt{1 – \sin^2(x)}}\cos(x) = \frac{\cos(x)}{\sqrt{\cos^2(x)}} = \frac{\cos(x)}{|\cos(x)|}$$
It is important to notice that in the last step the square root of the square resulted in the absolute value, which makes sense if we check the graph of $\arcsin(\sin(x))$.
Now, the we try to find another integral:
$$b) \int \frac{dx}{\sqrt{(x – a)(b – x)}}$$
Here we use the substitution
$$ x – a = (b – a)\sin^2(u), dx = 2(b – a)\sin(u)\cos(u)du$$
I do not know why this particular substitution was chosen, except that it is hinted at by Apostol. So, with this substitution it is easy to find:
$$\int \frac{dx}{\sqrt{(x – a)(b – x)}} = \frac{2(b – a)\sin(u)\cos(u)}{\sqrt{(b – a)\sin^2(u)(b – a)\cos^2(u)}}du = \frac{2(b – a)}{|b – a|}\int \frac{\sin(u)\cos(u)}{\sin(u)\cos(u)}du$$
which is then trivial. It is important that here $\sqrt{\sin^2(x)\cos^2(x)} = \sin(x)\cos(x)$ unlike in (a), and $\sin/\cos$ themselves are pulled out, not their absolute values!
What is the difference in (a), and (b)? I do not understand why in (b) we are allowed to pull out non-absolute value $\sqrt{\cos^2(x)\sin^2(x)} = \sin(x)\cos(x)$, while it should be like in (a).
Best Answer
The function $(b-a)\sin^2 u$ is not one-to-one. So in order to make the substitution $x-a=(b-a)\sin^2 u$ you have to restrict the domain of $\sin^2 u$: otherwise, you won't know which value of $u$ corresponds to a given $x$.
The most natural choice of domain where $(b-a)\sin^2 u$ is one-to-one is the interval $\left[0,\frac{\pi}{2}\right]$. Since $(b-a)\sin^2 u$ attains its full range for $u$ in this domain, there is no harm in restricting $u$ to this domain when making the substitution.
But on this interval, $\sin u \cos u$ is nonnegative, because $\sin u$ and $\cos u$ are individually non-negative. So it is genuinely true that $\sqrt{\sin^2 u \cos^2 u}=\sin u \cos u$, without absolute values.