Finding $\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta$ 
Finding $$\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta$$

Try: $$I=\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta =2\sqrt{2}\int \frac{1}{(3-\cos 2\theta)^{\frac{3}{2}}}d\theta$$
How can I write this in terms of the elliptic integral? Could someone help me to explain it? Thanks

As stated by SmarthBansal in the comments, this can be described as an Incomplete Elliptic Integral of the Third Kind, defined as 
$$\Pi(n;\phi,m)=\int_0^\phi \frac{1}{1-n\sin^2(\theta)}\frac{1}{\sqrt{1-m\sin^2(\theta)}} d\theta$$
So we can write this integral as $\Pi(-1,x|-1)$. However, this is not a very satisfying answer, which I believe is why SmarthBansal mentioned this as a comment. Jack D'Aurizio mentions another solution in terms of the Incomplete Elliptic Integral of the Second Kind, defined as
$$E(\phi, m)=\int_0^\phi \sqrt{1-m\sin^2(\theta)} d\theta$$
These two solutions can be reconciled by an elegant relation, namely, 
$$E(\phi, m) = (1-m) \Pi(m;\phi ,m) + \frac{m \sin(2\phi)}{2\sqrt{1-m\sin^2(\phi)}}$$
Now let's derive that equation! If we start with the left side of the equation we get,
$$
\int_0^\phi \sqrt{1-m\sin^2(\theta)}d\theta  
= \int_0^\phi \frac{(1-m\sin^2(\theta))^2}{(1-m\sin^2(\theta))^{3/2}}d\theta $$$$
=\int_0^\phi \frac{1- 2 m \sin^2(\theta)+m^2 \sin^4 (\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta $$$$
=\int_0^\phi \frac{1-m}{((1-m\sin^2(\theta))^{3/2})}d \theta + m\int_0^\phi \frac{\cos^2(\theta)-\sin^2(\theta)+m\sin^4(\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta$$$$
=(1-m)\Pi(m; \phi, m) + m\int_0^\phi \frac{(\cos^2(\theta)-\sin^2(\theta))(1-m\sin^2(\theta)) + m \sin^2(\theta)\cos^2(\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta$$$$
=(1-m)\Pi(m; \phi, m) + m\int_0^\phi \frac{(\cos(2\theta))(1-m\sin^2(\theta)) + m\sin(2\theta)\sin(\theta)\cos(\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta$$
Now dividing each side by $(1-m\sin^2(\theta))^{1/2}$, we can use the "reverse quotient rule" to get
$$=(1-m)\Pi(m; \phi, m) + \frac{m}{2}\int_0^\phi \frac{2\cos(2\theta)\sqrt{1-m\sin^2(\theta)} - \sin(2\theta)\frac{-2m\sin(\theta)\cos(\theta)}{\sqrt{1-m\sin^2(\theta)}}}{(1-m\sin^2(\theta))}d\theta$$
$$=(1-m) \Pi(m;\phi ,m) + \frac{m}{2} \frac{\sin(2\phi)}{\sqrt{1-m\sin^2(\phi)}}$$
Thus we have our relationship, and solving for $\Pi(m; x, m)$, we get
$$\Pi(m;\phi ,m)= \frac{E(\phi,m)}{1-m} - \frac{m\sin(2\phi)}{2(1-m)\sqrt{1-m\sin^2(\phi)}}$$
And plugging in $m=-1$, we get
$$\int \frac{1}{(1+\sin^2(\theta))^{3/2}}d\theta= \Pi(-1;\phi ,-1)= \frac{E(\phi,-1)}{2} + \frac{\sin(2\phi)}{4\sqrt{1+m\sin^2(\phi)}}$$
