# Integrals of Gradshteyn and Ryzhik: 5.136 – 5.139 – Integrals of Combinations of Jacobi Elliptic Functions (Part 1)

## 5.136.1

\int \mathrm{sn}\,\mathrm{cn} = -\frac{1}{k^2}\mathrm{dn}

## 5.136.2

\int \mathrm{sn}\,\mathrm{dn} = -\mathrm{cn}

## 5.136.3

\int \mathrm{cn}\,\mathrm{dn} = \mathrm{sn}

The three integrals above follow from the derivatives of the Jacobi elliptic functions.

## 5.137.1

\int \frac{\mathrm{sn}}{\mathrm{cn}^2} = \frac{1}{{k^{\prime}}^2} \frac{\mathrm{dn}}{\mathrm{cn}} = \frac{1}{{k^{\prime}}^2} \mathrm{dc}

## 5.137.2

\int \frac{\mathrm{sn}}{\mathrm{dn}^2} = -\frac{1}{{k^{\prime}}^2} \frac{\mathrm{cn}}{\mathrm{dn}}
= -\frac{1}{{k^{\prime}}^2} \mathrm{cd}

## 5.137.3

\int \frac{\mathrm{cn}}{\mathrm{sn}^2} = -\frac{\mathrm{dn}}{\mathrm{sn}} = -\mathrm{ds}

## 5.137.4

\int \frac{\mathrm{cn}}{\mathrm{dn}^2} = \frac{\mathrm{sn}}{\mathrm{dn}} = \mathrm{sd}

## 5.137.5

\int \frac{\mathrm{dn}}{\mathrm{sn}^2} = -\frac{\mathrm{cn}}{\mathrm{sn}} = -\mathrm{cs}

## 5.137.6

\int \frac{\mathrm{dn}}{\mathrm{cn}^2} = \frac{\mathrm{sn}}{\mathrm{cn}} = \mathrm{sc}

The six integrals above follow from the derivatives of the Jacobi elliptic functions.

## 5.138.1

\begin{align}
\int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} &= \int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} \frac{\mathrm{dn}}{\mathrm{dn}} = \int \frac{\mathrm{cn}}{\mathrm{dn}^2} \frac{1}{\mathrm{sd}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sd}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{dn}}\right)
\end{align}
We used the substitution $$w=\mathrm{sd}$$.

## 5.138.2

\begin{align}
\int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} &= \int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}}{\mathrm{cn}^2} \frac{1}{\mathrm{dc}} \\
&= \frac{1}{{k^{\prime}}^2} \int \frac{1}{w} = \frac{1}{{k^{\prime}}^2} \ln(\mathrm{dc}) = \frac{1}{{k^{\prime}}^2} \ln\left(\frac{\mathrm{dn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution $$w=\mathrm{dc}$$.

## 5.138.3

\begin{align}
\int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} &= \int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{sc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sc}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution $$w=\mathrm{sc}$$.

## 5.139.1

\int \frac{\mathrm{cn}\,\mathrm{dn}}{\mathrm{sn}} = \int \frac{1}{w} = \ln(\mathrm{sn})

## 5.139.2

\begin{align}
\int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} &= \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{nc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{nc}) = \ln\left(\frac{1}{\mathrm{cn}}\right)
\end{align}
We used the substitution $$w=\mathrm{nc}$$.

## 5.139.3

\int \frac{\mathrm{cn}\,\mathrm{sn}}{\mathrm{dn}} = -\frac{1}{k^2} \int \frac{1}{w} = -\frac{1}{k^2} \ln(\mathrm{dn})