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I am currently trying to get a general expression of the following integral, I spaned many questions with the above tags and found nothing close to it:

$$ I_n = \int_0^{n \pi} \sin^2(x) \sqrt{1+\alpha^2 \cos^2(x)} \mathrm d x $$ where $\alpha^2 > 0 $ and $n \in \mathbb{Z}$.

I thought of changes of variable $u =\cos(x)$ or $u = \cos^2(x)$, but after a try they don't lead somewhere interesting. I linearized the $sin^2(x)$ factor, but again I end up with two integrals:

$$ I_n = \frac{1}{2}\int_0^{n \pi}\sqrt{1+\alpha^2 \cos^2(x)} \mathrm d x - \frac{1}{2}\int_0^{n \pi} \cos(2x)\sqrt{1+\alpha^2 \cos^2(x)} \mathrm d x $$ and they don't seem easier.

Has anyone any ideas regarding a change of variable worth a try? Is there a way to obtain a closed form, if it exists one?

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    $\begingroup$ A closed form can be found in terms of elliptic integrals. Wolfram alpha can handle this job. $\endgroup$ – David H May 8 '14 at 13:06
  • $\begingroup$ There is a solution for the antiderivative (found by a CAS). Is this for homework ? $\endgroup$ – Claude Leibovici May 8 '14 at 13:06
  • $\begingroup$ @ClaudeLeibovici I came across this integral while trying to calculate kinetic energy in some systyem. I was doing the calculations at home but this isn't homework I guess. $\endgroup$ – ChocoPouce May 9 '14 at 0:02
  • $\begingroup$ @ChocoPouce. What is you range for $\alpha$ ? I am sure I can develop a very accurate approximation. Let me know and if you wish, ring me at 06 84 62 89 48. Cheers. $\endgroup$ – Claude Leibovici May 9 '14 at 11:25
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Hint: Define $I_n(\alpha)$ for each $\alpha\in\mathbb{R}\setminus\{0\}$ and $n\in\mathbb{Z}$ by the definite integral,

$$I_n(\alpha):=\int_{0}^{n\pi}\sin^2{x}\,\sqrt{1+\alpha^2\cos^2{x}}\,dx.$$

Note that the integrand $f(x;\alpha)=\sin^2{x}\,\sqrt{1+\alpha^2\cos^2{x}}$ is a periodic function of $x$ with period $\pi$, hence

$$I_n(\alpha)=\int_{0}^{n\pi}\sin^2{x}\,\sqrt{1+\alpha^2\cos^2{x}}\,dx\\ =n\int_{0}^{\pi}\sin^2{x}\,\sqrt{1+\alpha^2\cos^2{x}}\,dx\\ =n\,I_1(\alpha).$$

So it suffices to just consider the integral $I(\alpha):=I_1(\alpha)=\int_{0}^{\pi}\sin^2{x}\,\sqrt{1+\alpha^2\cos^2{x}}\,dx$.

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Using Mathematica, I found that the value is $$ I_n = n \frac{2 \sqrt{\alpha ^2+1} \left(\left(\alpha ^2-1\right) E\left(\frac{1}{1+\frac{1}{\alpha ^2}}\right)+K\left(\frac{1}{1+\frac{1}{\alpha ^2}}\right)\right)}{3 \alpha ^2} $$ Here $E$ is Mathematica's EllipticE and $K$ is EllipticK. These functions are not elementary but can be efficiently evaluated to any desired precision.

To remove ambiguity regarding definitions, $E(m) = E_2(\sqrt m)$ and $K(m) = E_1(\sqrt m)$, where $E_j$ is the (ordinary) complete elliptic integral of the $j$'th kind.

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Using another CAS, I obtained $$I_1=\frac{\left(\alpha ^2-1\right) E\left(-\alpha ^2\right)+\left(\alpha ^2+1\right) K\left(-\alpha ^2\right)+\sqrt{\alpha ^2+1} \left(\left(\alpha ^2-1\right) E\left(\frac{\alpha ^2}{\alpha ^2+1}\right)+K\left(\frac{\alpha ^2}{\alpha ^2+1}\right)\right)}{3 \alpha ^2}$$ I have not been able to simplify as much as user111187 did. As already mentioned by other answers, $I_n=n I_1$.

For small values of $\alpha$, we can establish that$$I_1 \simeq\frac{\pi }{2}+\frac{\pi \alpha ^2}{16}-\frac{\pi \alpha ^4}{128}+\frac{5 \pi \alpha ^6}{2048}-\frac{35 \pi \alpha ^8}{32768}+\frac{147 \pi \alpha ^{10}}{262144}-\frac{693 \pi \alpha ^{12}}{2097152}+O\left(\alpha ^{14}\right)$$ while for very large values of $\alpha$, $$I_1 \simeq \frac{2 \alpha }{3}+\frac{-\log \left(\frac{1}{\alpha }\right)-\frac{1}{2}+2 \log (2)}{\alpha }+O\left(\left(\frac{1}{\alpha }\right)^3\right)$$

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