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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 $(2)$:

$$4\int_0^{\frac{\pi}2}\frac1{(1-k^2\cos^2(\alpha))^\frac12}\,\mathrm{d}\alpha=4\int_0^{\frac{\pi}2}\frac1{(1-k^2(1-\sin^2(\alpha)))^\frac12}\,\mathrm{d}\alpha\tag{9}$$

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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.

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  • $\begingroup$ Yes i did forget all the square roots. oops. So in the original article the second line alpha is a beta? $\endgroup$ – N. Morgan Jul 18 '18 at 10:41
  • $\begingroup$ Exactly. That would be a lot easier to understand if they had not changed the name back to $\alpha$ for some reason. $\endgroup$ – ComplexYetTrivial Jul 18 '18 at 10:51
  • $\begingroup$ And because both the cosine and sine versions of the integral of the function are equal to each other within a range of $n\frac{\pi}{2}$ to $(n+1)\frac{\pi}{2}$ there is no change in the interval of integration when performing the shift to convert between sine and cosine? $\endgroup$ – N. Morgan Jul 18 '18 at 10:56
  • $\begingroup$ Yes. Alternatively you can argue that the $\alpha$-integral starts at $\alpha_1 = 0$ and ends at $\alpha_2=\frac{\pi}{2}$ , so the $\beta$-integral goes from $\beta_1 = \frac{\pi}{2} - \alpha_1 = \frac{\pi}{2}$ to $\beta_2 = \frac{\pi}{2} - \alpha_2 = 0$ . The additional sign from the substitution allows you to interchange the limits and you end up with an integral from $0$ to $\frac{\pi}{2}$ again. $\endgroup$ – ComplexYetTrivial Jul 18 '18 at 11:04
  • $\begingroup$ Sounds good, thank you. $\endgroup$ – N. Morgan Jul 19 '18 at 3:34

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