I am having a hard time following Equation 2.6 of Taib, Bachok Bin. "Boundary integral method applied to cavitation bubble dynamics." (1985). The equation is on the middle of page 8 of the document or page 17 of the linked pdf file.
$$\int_0^{2\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta=4\int_0^{\frac{\pi}2}\frac1{(1-k^2\cos^2(\alpha))^\frac12}\,\mathrm{d}\alpha\tag{1}$$
$$4\int_0^{\frac{\pi}2}\frac1{(1-k^2\cos^2(\alpha))^\frac12}\,\mathrm{d}\alpha=4\int_0^{\frac{\pi}2}\frac1{(1-k^2\sin^2(\alpha))^\frac12}\,\mathrm{d}\alpha\tag{2}$$
I understand how to arrive at the left hand side of equation $(1)$ above, however, I do not see how to convert to $\alpha$ and from $\cos^2$ to $\sin^2$ in that way.
First question: What is the substitution from $\theta$ to $\alpha$?
I see that the positive periodic nature of $\cos^2$ and $\sin^2$ means that I can break the integral into two parts that are equal.
$$\int_0^{2\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta=$$ $$\int_0^{\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta\,+\int_\pi^{2\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta\tag{3}$$
$$\int_0^{\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta\,=\int_{n\pi}^{(n+1)\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta\tag{4}$$
$$\int_0^{2\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta=2\int_0^{\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta\tag{5}$$
\begin{align} \alpha=\frac{\theta}{2} \\ \frac{\mathrm{d}\alpha}{\mathrm{d}\theta}=\frac{1}{2} \end{align}
$$2\int_0^{\pi}\frac1{(1-k^2\cos^2(\frac{\theta}2))^\frac12}\,\mathrm{d}\theta=$$ $$2\int_0^{\pi}2\frac1{(1-k^2\cos^2(\alpha))^\frac12}\,\mathrm{d}\alpha=4\int_0^{\frac{\pi}2}\frac1{(1-k^2\cos^2(\alpha))^\frac12}\,\mathrm{d}\alpha\tag{6}$$
Second question: How is equation $(2)$ obtained?
When I convert from $\cos^2$ to $\sin^2$, I would imagine it as the following:
$$\cos^2(\theta)+\sin^2(\theta)=1\tag{7}$$
$$\cos^2(\theta)=1-\sin^2(\theta)\tag{8}$$
Substituting $(8)$ into $(2)$:
$$4\int_0^{\frac{\pi}2}\frac1{(1-k^2\cos^2(\alpha))^\frac12}\,\mathrm{d}\alpha=4\int_0^{\frac{\pi}2}\frac1{(1-k^2(1-\sin^2(\alpha)))^\frac12}\,\mathrm{d}\alpha\tag{9}$$
In order to obtain the first identity you can split the integral into two (not four) equal parts: $$ \int \limits_0^\pi \frac{\mathrm{d}\theta}{1-k^2 \cos^2\left(\frac{\theta}{2}\right)} = \int \limits_\pi^{2\pi} \frac{\mathrm{d}\theta}{1-k^2 \cos^2\left(\frac{\theta}{2}\right)} \, .$$ This implies $$\int \limits_0^{2\pi} \frac{\mathrm{d}\theta}{1-k^2 \cos^2\left(\frac{\theta}{2}\right)} = 2 \int \limits_0^\pi \frac{\mathrm{d}\theta}{1-k^2 \cos^2\left(\frac{\theta}{2}\right)}$$ and now the substitution $\alpha = \frac{\theta}{2}$ yields equation $(1)$ .
In the second step the substitution $\beta = \frac{\pi}{2} - \alpha$ is used before renaming $\beta$ to $\alpha$ again. It leaves the interval of integration unchanged but converts the cosine into a sine.