I want evaluate the follow integrals
$$\displaystyle \int_{0}^{K} \text{dn}^3(u;k)\text{sn}(u;k)^2\;\text{du},\tag{1}$$
and
$$\displaystyle \int_{0}^{K} \text{dn}(u;k)\text{sn}(u;k)^2\text{cn}(u;k)^2\;\text{du},\tag{2}$$
where $\text{sn}$, $\text{dn}$ and $\text{cn}$ are the Jacobi Elliptic snoidal, dnoidal and cnoidal functions, $K:=K(k)$ is the complete elliptic integral of the first kind and number $k \in \left(0,1\right)$ is called the modulus.
I already consulted the reference $[1]$ in search of some formula that helps me, but I found nothing. Do these integrals have an explicit form? Are there any other references I can refer to to help me?
$[1]$ P. F. Byrd. M. D. Friedman. Hand Book of Elliptical Integrals for Engineers and Scientis. Springer-Verlag New York Heidelberg Berlim, $1971$.
Best Answer
By means of the fundamental relations (B&F 121.00) $\newcommand{sn}{\operatorname{sn}}\newcommand{cn}{\operatorname{cn}}\newcommand{dn}{\operatorname{dn}}$ $$\sn^2u+\cn^2u=1$$ $$k^2\sn^2u+\dn^2u=1$$ we can transform the first given integral to $$\int_0^K\dn u(1-k^2\sn^2u)\sn^2u\,du$$ By B&F 364.03 we can rewrite this as a completely rational integral, which is easily evaluated: $$=2\int_0^1\left(\left(\frac{2t}{1+t^2}\right)^2-k^2\left(\frac{2t}{1+t^2}\right)^4\right)\frac1{1+t^2}\,dt=\frac{\pi(4-3k^2)}{16}$$ When we transform the second given integral we get $$\int_0^K\dn u(1-\sn^2u)\sn^2u\,du$$ at which point we realise that this is just a special case of the first given integral with $k^2=1$, so we immediately get the result as $\frac\pi{16}$.