Verifying an antiderivative found in any integral table If $a > 0$, and $0 < b < c$.
\begin{equation*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
= \frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert .
\end{equation*}
(This is the antiderivative given in any calculus book.) With calculations that I made, I showed that
\begin{align*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
&= \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert
,
\end{align*}
and since the sum of an antiderivative of $1/[b + c\sin(ax)]$ and a constant is another antiderivate of $1/[b + c\sin(ax)]$, and since
\begin{equation*}
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b}
\qquad \text{and} \qquad
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b}
\end{equation*}
are reciprocals of each other,
\begin{align*}
&\int \frac{1}{b + c\sin(ax)} \, {\mathit dx} \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad \qquad \qquad\qquad - \frac{1}{a\sqrt{c^{2} - b^{2}}} \ln\left( \frac{c + \sqrt{c^{2} - b^{2}}}{b} \right) \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\sin(ax) + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Furthermore, we have the following trigonometric identity:
\begin{align*}
&\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)} \\
&\qquad\qquad =\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
} .
\end{align*}
So, the antiderivative given in any calculus book is the same function as the second antiderivative that I obtained:
\begin{align*}
&\frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert \\
&\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Here are my questions.  Is it evident to anyone that the function on the right side of the trigonometric identity simplifies to the function on the left side? (It is surprising that the two sides are equal. The numerators and denominators on each side are "almost" the same: the numerator and denominator on the right side have two more terms than those on the left side.) If it is not evident, can someone give me calculations, starting from integration using the technique of trigonometric substitution, that show that the antiderivative of $1/[b + c\sin(ax)]$ is the function that is found in the integral tables of any calculus book - calculations that avoid all the algebraic manipulations?

For the simplicity let your calculation be $f(x)$ and let the antiderivative given in calculus book be $g(x)$. Assuming that your calculation is correct, then the proper way to express the indefinite integral using your calculation is
$$
\int \frac{1}{b + c\sin(ax)}\ dx=f(x)+C_1,
$$
where $C_1$ is a constant of integration. Similarly, the proper way to express the indefinite integral using table in calculus book is
$$
\int \frac{1}{b + c\sin(ax)}\ dx=g(x)+C_2,
$$
where $C_2$ is also a constant of integration. But you cannot deduce $C_1=C_2$, therefore you cannot equate $f(x)$ with $g(x)$ since $f(x)\neq g(x)$. The problem is you tried to set the constant equal to zero but it doesn't always make sense. Here is the classic example (you may also refer to here), $2\sin x\cos x$ can be integrated in at least three different ways:
\begin{align}
\int 2\sin x\cos x\,dx &=  \sin^2(x) + C = -\cos^2x + 1 + C = -\frac12\cos2x + C\\[12pt]
\int 2\sin x\cos x\,dx &= -\cos^2(x) + C =  \sin^2(x) - 1 + C = -\frac12\cos2x + C\\[12pt]
\int 2\sin x\cos x\,dx &= -\frac12\cos2x + C = \sin^2x + C = -\cos^2x + C
\end{align}
So setting $C$ to zero can still leave a constant.  This means that, for a given function, there is no "simplest antiderivative" and of course $-\frac12\cos2x\neq-\cos^2x$. Note that, the constant of integration is sometimes omitted in lists of integrals for simplicity.

Here is an approach to obtain the given antiderivative in the calculus book. Start with setting $u=ax$, then
$$
\int\frac{1}{b+c\sin ax}\ dx=\frac1a\int\frac{1}{b+c\sin u}\ du.\tag1
$$
Using the Weierstrass substitution by setting $t=\tan\frac u2$, we have
$$
\sin u=\frac{2t}{1+t^2}\quad\text{and}\quad du=\frac{2}{1+t^2}\ dt.
$$
Therefore, $(1)$ becomes
\begin{align}
\int\frac{1}{b+c\sin ax}\ dx&=\frac2a\int\frac{1}{bt^2+2ct+b}\ dt\\
&=\frac2a{\Large\int}\frac{1}{b\left(t+\frac{c+\sqrt{c^2-b^2}}{b}\right)\left(t+\frac{c-\sqrt{c^2-b^2}}{b}\right)}\ dt\\
&=\frac2{ab}\cdot\frac{1}{2\frac{\sqrt{c^2-b^2}}{b}}{\Large\int}\left[\frac{1}{\left(t+\frac{c-\sqrt{c^2-b^2}}{b}\right)}-\frac{1}{\left(t+\frac{c+\sqrt{c^2-b^2}}{b}\right)}\right]\ dt\\
&=\frac1{a\sqrt{c^2-b^2}}\left[\ln{\left|t+\frac{c-\sqrt{c^2-b^2}}{b}\right|}-\ln{\left|t+\frac{c+\sqrt{c^2-b^2}}{b}\right|}\right]\\
&=\frac1{a\sqrt{c^2-b^2}}\ln{\left|\frac{bt+c-\sqrt{c^2-b^2}}{bt+c+\sqrt{c^2-b^2}}\right|}\\
&=\frac1{a\sqrt{c^2-b^2}}\ln{\left|\frac{b\tan\frac{ax}{2}+c-\sqrt{c^2-b^2}}{b\tan\frac{ax}{2}+c+\sqrt{c^2-b^2}}\right|}.
\end{align}
Notice that
$$
\tan\frac{ax}{2}=\frac{\sin ax}{1+\cos ax}.
$$
It can easily be proven by using double angle formula.
$$\color{red}{\text{to be continued...}}$$
