How to evaluate $\int_0^{\pi/2}(\sin^2\theta+a)^{-3/2}\,d\theta$, where $a>0$

Question

How to evaluate the following integral? $$\int_0^{\pi/2}(\sin^2\theta+a)^{-3/2}\,d\theta,$$where $a>0$

Answer 1

$$\int_0^{\frac{\pi}{2}}(\sin^2\theta+a)^{-\frac{3}{2}}\,d\theta$$

$$\int_0^{\frac{\pi}{2}} \frac{1}{(\sin^2\theta+a)\sqrt{\sin^2\theta+a}} \,d\theta$$

$$\int_0^{\frac{\pi}{2}} \frac{1}{a(\frac{1}{a}\sin^2\theta+1)\sqrt{a\left( \frac{1}{a}\sin^2\theta+1\right) }} \,d\theta$$

$$\frac{1}{a\sqrt{a}}\int_0^{\frac{\pi}{2}} \frac{1}{(\frac{1}{a}\sin^2\theta+1)\sqrt{\frac{1}{a}\sin^2\theta+1}} \,d\theta$$

$$\frac{1}{a\sqrt{a}}\int_0^{\frac{\pi}{2}} \frac{1}{(1+a^{-1}\sin^2\theta)\sqrt{1+a^{-1}\sin^2\theta}} \,d\theta$$

$$\frac{1}{a\sqrt{a}}\underbrace{\int_0^{\frac{\pi}{2}} \frac{1}{(1 - (-a^{-1})\sin^2\theta)\sqrt{1 - (-a^{-1})\sin^2\theta}} \,d\theta}_{\Pi(-a^{-1}\,|\,-a^{-1})}$$

Also

Solution I $$\int_0^{\frac{\pi}{2}}(\sin^2\theta+a)^{-\frac{3}{2}}\,d\theta = \frac{\Pi\left(-a^{-1}\,|\,-a^{-1}\right)}{a\sqrt{a}}\, $$ For more information

Where $\Pi\left(-a^{-1}\,|\,-a^{-1}\right)$ ist The Complete Elliptic Integral of the Third Kind $$\Pi\left( n\,|\,m\right) = \Pi\left( n;\,\frac{\pi}{2}\,|\,m\right) = \int_0^{\frac{\pi}{2}} \frac{1}{\left( 1 - n\sin^2(t)\right) \sqrt{1 - m\sin^2(t)}} \,dt$$

The integral is also allowed to be expressed in terms of The Complete Elliptic Integral of the Second Kind $E(m)$

$$E(m) = \int_0^{\frac{\pi}{2}}\sqrt{1 - m\sin^2(t)}\, dt$$

There the complete elliptic integral of the second kind satisfies the imaginary modulus identity: $$E(-m)=\sqrt{1+m}\,\,E\left(\frac{m}{1+m}\right)$$ Where \begin{align*} E\left( \frac{m}{m+1}\right) &= \int_0^{\frac{\pi}{2}}\sqrt{1 - \left( \frac{m}{m+1}\right)\sin^2(t)}\, dt\\ &= \frac{1}{\sqrt{m+1}}\int_0^{\frac{\pi}{2}}\sqrt{m + 1 -m\sin^2(t)}\, dt\\ &= \frac{1}{\sqrt{m+1}}\int_0^{\frac{\pi}{2}}\sqrt{1 + m\cos^2(t)}\, dt \end{align*} Changing variables to $\theta = \frac{\pi}{2} - t,\, dt = -d\theta,\implies \cos(t) = \sin\theta$, with $\theta$ going now from $\frac{\pi}{2}$ to $0$, then reversing the path of integration and changing the sign on the integral, therefore $$E\left( \frac{m}{m+1}\right) = \frac{1}{\sqrt{m+1}}\int_0^{\frac{\pi}{2}}\sqrt{1 + m\sin^2\theta }\, d\theta$$
In this way we can write the integral as follows $$\int_0^{\frac{\pi}{2}}(\sin^2\theta+a)^{-\frac{3}{2}}\,d\theta$$ $$\frac{1}{a\sqrt{a}}\int_0^{\frac{\pi}{2}}\frac{1}{(\frac{1}{a}\sin^2\theta+1)\sqrt{\frac{1}{a}\sin^2\theta+1}} \,d\theta$$ Set $k = \frac{1}{a} $ $$\frac{k}{\sqrt{a}}\int_0^{\frac{\pi}{2}}\frac{1}{(k\sin^2\theta+1)\sqrt{k\sin^2\theta+1}} \,d\theta$$ $$\frac{k}{\sqrt{a}}\int_0^{\frac{\pi}{2}}\frac{1}{(k\sin^2\theta+1)^{\frac{3}{2}}}\,d\theta$$

$\frac{d^2}{d\theta^2}(\sqrt{1 + a\sin^2\theta})$
$= \frac{-a+2a\sin^2\theta + k^2\sin^4\theta}{(1 + k\sin^2\theta)^{\frac{3}{2}}}$
$= \frac{-a-1+1+2a\sin^2\theta + k^2\sin^4\theta}{(1 + k\sin^2\theta)^{\frac{3}{2}}}$
$= \frac{-(a+1)+ (1 + a\sin^2\theta)^2}{(1 + k\sin^2\theta)^{\frac{3}{2}}}$
$= \frac{a+1}{(1 + k\sin^2\theta)^{\frac{3}{2}}}-\sqrt{1+a\sin^2\theta}$

$$\frac{1}{k + 1}\left[ \frac{k}{\sqrt{a}}\int_0^{\frac{\pi}{2}}\sqrt{1 + k\sin^2\theta} - \frac{d^2}{d\theta^2}(\sqrt{1 + a\sin^2\theta}) \right]\,d\theta$$ $$\frac{1}{k + 1}\left[ \frac{k}{\sqrt{a}}\underbrace{\int_0^{\frac{\pi}{2}}\sqrt{1 + k\sin^2\theta}\, d\theta}_{E(-k)} - \left(\frac{k\sin\theta\cos\theta}{\sqrt{k\sin^2\theta+1}}\bigg\vert_{\frac{\pi}{2}}^{0}\right) \right]$$ $$\frac{1}{k+1}\left[\frac{k}{\sqrt{a}}E(-k) - (0 - 0)\right]$$ $$\frac{k}{(k+1)\sqrt{a}}E(-k)$$ Go to back $k$ $$\frac{1}{\left( \frac{1}{a}+1\right)a^{\frac{3}{2}}}E(-a^{-1})$$ $$\frac{1}{\left( \frac{1}{a}+1\right)a^{\frac{3}{2}}}\sqrt{1+a^{-1}}E\left( \frac{a^{-1}}{1+a^{-1}}\right)$$

Solution II $$\int_0^{\frac{\pi}{2}}(\sin^2\theta+a)^{-\frac{3}{2}}\,d\theta = \frac{1}{\sqrt{\frac{1}{a}+1}\,a^{\frac{3}{2}}}E\left( \frac{1}{a + 1}\right)$$For more Information

Many important applications of these integrals were found at that time.\ The problem of evaluating such integrals was converted into the problem of evaluating only three basic integrals, namely the incomplete elliptic integrals of the first, second, and third kinds— $F(z|m),\, E(z|m)$, and $\Pi(n;z|m)$ (A. M. Legendre). See Introduction to the incomplete elliptic integrals

Here are some interesting links:
1- Definition of Elliptic Integral
2- Introduction to the complete elliptic integrals
3- Compute $\Pi\left(n|m\right)$
4- Elliptic Integrals

Answer 2

With Mathematica notation we have that $\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{a+\sin^2\theta}}$ is an elliptic integral, namely $$ \frac{1}{\sqrt{1+a}}\,K\left(\frac{1}{1+a}\right) $$ hence by differentiating with respect to $a$ we get that $$ \int_{0}^{\pi/2}\frac{d\theta}{(a+\sin^2\theta)^{3/2}} = \frac{1}{a\sqrt{a+1}}\,E\left(\frac{1}{1+a}\right).$$