Please help me solve this integral

Question

Please help me solve this integral \begin{align} \int_{-\pi/2}^{\pi/2}\sqrt{\varepsilon_r-\sin^2\varphi}\cos^2\varphi\ d\varphi. \end{align} Thank you very much!

Answer 1

Here is a hint:

In Elliptic Inegrals of Hancock, your integral appears. It shows that it can be evaluated with elliptic integrals:

$$\int_{-\pi/2}^{\pi/2}\sqrt{\varepsilon_r-\sin^2\varphi}\cos^2\varphi\ d\varphi. = 2\sqrt{\varepsilon_{r}}\int_{0}^{\pi/2}\sqrt{1-\frac{1}{\varepsilon_{r}}\sin^2\varphi}\cos^2\varphi\ d\varphi.$$

Using the notation of elliptic integrals:

$$\int_{0}^{\theta}\sqrt{1-k^2\sin^2\varphi}\cos^2\varphi\ d\varphi = \frac{1}{3}\sqrt{1-k^2\sin^2\theta}\sin\theta\cos\theta + \frac{1+k^2}{3k^2}E(k,\theta) - \frac{k'^2}{3k^2}F(k,\theta)$$

where $k$ is the modulus and $k'$ is the complementary modulus and

$k'= \sqrt{1-k}$

and

$$ F(k,\theta) = \int_{0}^{\theta} \frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}$$

$$ E(k,\theta) = \int_{0}^{\theta} \sqrt{1-k^2\sin^2\phi} d\phi$$

are the incomplete elliptic integrals of the first and second kind, respectively.

If you put $\displaystyle k^2 = \frac{1}{\varepsilon_{r}}$ and $\displaystyle \theta = \frac{\pi}{2}$ we have your integral.