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Given the following function we find its derivative:

$$a) fracddxleft[arcsin(sin(x))right] = frac1sqrt1 - sin^2(x)cos(x) = fraccos(x)sqrtcos^2(x) = fraccos(x)$$

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:

$$b) int fracdxsqrt(x - a)(b - x)$$

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:

$$int fracdxsqrt(x - a)(b - x) = frac2(b - a)sin(u)cos(u)sqrt(b - a)sin^2(u)(b - a)cos^2(u)du = frac2(b - a)b - aint fracsin(u)cos(u)sin(u)cos(u)du$$

which is then trivial. It is important that here $sqrtsin^2(x)cos^2(x) = sin(x)cos(x)$ 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 $sqrtcos^2(x)sin^2(x) = sin(x)cos(x)$, while it should be like in (a).

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 $left[0,fracpi2right]$. 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 $sqrtsin^2 u cos^2 u=sin u cos u$, without absolute values.

Note that in the second integral, if you are using real numbers, then $ale xle b$ or $ble xle a$ (otherwise the argument of the square root is negative). The the substitution you are making implies $0le ule pi/2$, so both $sin$ and $cos$ have positive values.