Evaluation of $\int\sqrt{1+2\sin(\theta)^2\cos(\theta)^2}\mathrm{d}\theta~~\text{where}~~\theta\in\left[0,{\pi\over4}\right]$ $$\begin{align}
I&:=\int \sqrt{1+2\sin(\theta)^2\cos(\theta)^2} \mathrm{d} \theta ~~~\text{where}~~~~\theta\in\left[0,{\pi\over4}\right]\\
\cos(\theta)^2&=1-\sin(\theta)^2\\
I&=\int\sqrt{1+2\sin(\theta)^2 \left(1-\sin(\theta)^2 \right) } \mathrm{d} \theta\\
&= \underbrace{\int\sqrt{1+2\sin(\theta)^2-2\sin(\theta)^4} \mathrm{d} \theta}_{\text{Seems nothing can be done from this form} } \\\sin(2\theta)&=2\sin(\theta)\cos(\theta)\\
&\iff {\sin(2\theta) \over2  } = \sin(\theta)\cos(\theta)
\\ {\sin(2\theta)^2 \over 4 }&= \sin(\theta)^2\cos(\theta)^2
\\I&=\int \sqrt{1+2 \left({\sin(2\theta)^2 \over 4 } \right)  } \mathrm{d} \theta\\&=\int \sqrt{1+{\sin(2\theta)^2 \over 2 }} \mathrm{d} \theta\\&=\int \sqrt{ {2+\sin(2\theta)^2 \over 2 }} \mathrm{d} \theta\\&= {1 \over \sqrt{2}}\underbrace{\int\sqrt{2+\sin(2\theta)^2} \mathrm{d} \theta}_{\text{It is confusing for me} }  
\end{align}$$
How can I integrate the originally given integral?
 A: $$I=\int_0^{\frac \pi 4}\sqrt{1+\frac{1}{2} \sin ^2(2 \theta )}\,d\theta=\frac{1}{2}\int_0^{\frac \pi 2} \sqrt{1+\frac{1}{2}\sin ^2(x)}\,dx=\frac{1}{2}E\left(-\frac{1}{2}\right)$$ where appears the  complete elliptic integral of the second kind.
If it had been
$$J=\int_0^{t}\sqrt{1+\frac{1}{2} \sin ^2(2 \theta )}\,d\theta=\frac{1}{2}\int_0^{2t} \sqrt{1+\frac{1}{2}\sin ^2(x)}\,dx=\frac{1}{2} E\left(2 t\left|-\frac{1}{2}\right.\right)$$where appears the  incomplete elliptic integral of the second kind.
Have a look here for definitions and properties.
In the worst case, write
$$\sqrt{1+\frac{1}{2}\sin ^2(x)}=\frac 1 {\sqrt \pi}\sum_{n=0}^\infty (-1)^{n+1}\frac{ \Gamma \left(n-\frac{1}{2}\right) }{ 2^{n+1}\,\Gamma(n+1)}\sin ^{2 n}(x)$$ and using
$$\int_0^{\frac \pi 2}\sin ^{2 n}(x)\,dx=\frac{\sqrt{\pi } \Gamma \left(n+\frac{1}{2}\right)}{2 \Gamma (n+1)}$$
$$\int_0^{\frac \pi 2} \sqrt{1+\frac{1}{2}\sin ^2(x)}\,dx=\sum_{n=0}^\infty (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}$$
If
$$a_n=\frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}\implies \frac{a_{n+1}}{a_n}=\frac{4 n^2-1}{8 (n+1)^2}= \frac{1}{2}-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ and since it is alternating it will converge quite fast.
Summing from $n=0$ to $n=5$ would give $1.75185$ while $E\left(-\frac{1}{2}\right)=1.75177$.
If this has to be computed, write
$$\sum_{n=0}^\infty (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}=$$ $$\sum_{n=0}^p (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}+\sum_{n=p+1}^\infty (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}$$ and tou want to know $p$ such that
$$R_p=\frac{ \Gamma \left(p+\frac{1}{2}\right) \Gamma \left(p+\frac{3}{2}\right)}{2^{p+3}\,\Gamma (p+2)^2}\leq \epsilon$$ Taking logarithms and using Stirling approximation, this is almost
$$2^{p+3} p^2 \geq \frac 1 \epsilon \implies p \geq \frac{2}{\log (2)}\,W\left(\frac{\log (2)}{4 \sqrt{2\epsilon } }\right)$$ where $W(.)$ is Lambert function.
Using $\epsilon=10^{-16}$, $p \geq 39.5403$ while the exact solution is $p=39.4815$. So, coding, no more comparison tests.
