I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.
How did we go from:
$$\int\frac{\csc^2 x – \csc x \cot x}{\csc x – \cot x}\,dx%$$
to
$$\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$$
Thank you for your time!
$$
\int \csc^3 x\,dx = \int\csc^2x \csc x\,dx$$
To integrate by parts, let $dv = \csc^2x$ and $u=\csc x$. Then $v=-\cot x$ and $du = -\cot x \csc x \,dx$.
Integrating by parts, we have:
$$\begin{align*}
\int\csc^2 x \csc x \,dx &= -\cot x \csc x – \int(-\cot x)(-\cot x\csc x\,dx)\\
&= -\cot x \csc x – \int \cot^2 x \csc x\,dx\\
&= -\cot x\csc x – \int(\csc^2x – 1)\csc x\,dx &\text{(since }\cot^2 x = \csc^2-1\text{)}\\
&= -\cot x \csc x – \int(\csc^3 x – \csc x)\,dx\\
&= -\cot x\csc x – \int\csc^3 x\,dx + \int \csc x\,dx
\end{align*}$$
From
$$\int \csc^3 x\,dx = -\cot x\csc x – \int\csc^3 x\,dx + \int \csc x\,dx$$
we obtain
$$\begin{align*}
\int\csc^3x\,dx + \int\csc^3 x\,dx &= -\cot x \csc x + \int\csc x\,dx\\
2\int\csc^3 x\,dx &= -\cot x\csc x + \int\csc x\,dx\\
\int\csc^3x\,dx &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\
&=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x – \cot x)}{\csc x – \cot x}\,dx\\
&= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x – \csc x\cot x}{\csc x – \cot x}\,dx\\
&= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\
&= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x – \cot x|+ C
\end{align*}$$
Best Answer
Okay, your actual question is about integrating $\csc x$ (the rest doesn't matter).
You are fine with $$\int \csc x\,dx = \int\frac{\csc x(\csc x -\cot x)}{\csc x - \cot x}\,dx = \int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx.$$
The next step is just a basic substitution. Let $w = \csc x - \cot x$. Then $dw = (-\csc x\cot x + \csc^2x) \,dx$. This happens to be the numerator of the integral we have, while the denominator is $w$. So $$\int\frac{\csc^2x - \csc x\cot x}{\csc x- \cot x}\,dx = \int\frac{dw}{w}.$$
But instead of doing the substution explicitly, they wrote that the numerator, $(\csc^2x - \csc x\cot x)\,dx$ was the differential of the denominator, $\csc x - \cot x$.