# Calculate some integrals involving Jacobi Elliptic Functions

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, $$1971$$.

• Mathematica V12.1 gives (4-3k^2)Pi/16 for (1) and Pi/16 for (2). Aug 18, 2020 at 1:21
• @Somos Is it possible to do this on Maple? Aug 18, 2020 at 1:51
• And if the integrals are valued from $0$ to $4K (k)$? Aug 18, 2020 at 1:59
• As for Mathematica, try it yourself at WolframCloud. I think it should be possible with Maple but I don't have Maple. Aug 18, 2020 at 13:06
• @Somos I calculated here. And it is according to what you originally said. Aug 18, 2020 at 22:47

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 $$\frac\pi{16}$$.