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「三角関数の幾つかの2次無理式の積分 (2017年10月31日 [火]」補足その1

[三角関数の幾つかの2次無理式の積分">] での記号の他に、次の記号を導入する ($m,n$ は非負整数)。

\begin{align*}
 &I^{(m,n)} := \int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}(\theta)\Delta(\theta,k)d\theta\\
 &J^{(m,n)} := \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)d\theta}{\Delta(\theta,k)}\\
 &w(x,k) := k^{2}\sin^{2}(x)
\end{align*}

以下、文脈から明らかなときは $\Delta(\theta,k)$ 及び $\Delta(\varphi,k)$ のどちらも $\Delta$ とのみ記するなど、適宜、変数の表示を省略することを、おことわりしておく。

自明な次の等式が成り立つ。

\begin{align*}
 &I^{m,0} = I^{m}_{\mathrm s}, \quad I^{0,n} = I^{n}_{\mathrm c}. \quad I^{(0,0)} = E(\varphi,k)\\
 &J^{m,0} = J^{m}_{\mathrm s}, \quad J^{0,n} = J^{n}_{\mathrm c}, \quad J^{(0,0)} = F(\varphi,k)\\
 &\Delta^{2}\, (:=1-k^{2}\sin^{2}(\theta))\, = k^{\prime2} + k^{2}\cos^{2}(\theta)\\
 &\sin^{2}(\theta) = \frac{1-\Delta^{2}}{k^{2}}, \quad \cos^{2}(\theta) = -\frac{k^{\prime2}-\Delta^{2}}{k^{2}}
\end{align*}

また、$m{\geq}2$ なら、

\begin{align*}
 I^{(m,n)} &= \int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}(\theta){\Delta}d\theta\\
&= \int_{0}^{\varphi}\sin^{m-2}(\theta)(1-\cos^{2}(\theta))\cos^{n}(\theta){\Delta}d\theta\\
&= I^{(m-2,n)} - I^{(m-2,n+2)}\\
 J^{(m,n)} &= \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)}{\Delta}d\theta\\
&= \int_{0}^{\varphi}\frac{\sin^{m-2}(\theta)(1-\cos^{2}(\theta))\cos^{n}(\theta)}{\Delta}d\theta\\
&= J^{(m-2,n)} - J^{(m-2,n+2)}\\
\end{align*}

$n{\geq}2$ なら

\begin{align*}
 I^{(m,n)} &= \int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}(\theta){\Delta}d\theta\\
&= \int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n-2}(\theta)(1-\sin^{2}(\theta)){\Delta}d\theta\\
&= I^{(m,n-2)} - I^{(m+2,n-2)}\\
 J^{(m,n)} &= \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)}{\Delta}d\theta\\
&= \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n-2}(\theta)(1-\sin^{2}(\theta))}{\Delta}d\theta\\
&= J^{(m,n-2)} - J^{(m+2,n-2)}
\end{align*}
が成り立つ。

まとめると、$m{\geq}2$ のとき

\begin{align*}
 &I^{m,n} = I^{(m-2,n)} - I^{(m-2,n+2)}\\
 &J^{m,n} = J^{(m-2,n)} - J^{(m-2,n+2)}
\end{align*}

$n{\geq}2$ のとき

\begin{align*}
 &I^{m,n} = I^{(m,n-2)} - I^{(m+2,n-2)}\\
 &J^{m,n} = J^{(m,n-2)} - J^{(m+2,n-2)}
\end{align*}

また

\begin{align*}
 I^{(m,n)} &= \int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}{\Delta}d\theta\\
&= \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)\Delta^{2}}{\Delta}d\theta\\
&= \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)(1-k^{2}\sin^{2}(\theta))}{\Delta}d\theta\\
&= J^{(m,n)} - k^{2}J^{(m+2,n)}\\
 I^{(m,n)} &= \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)(k^{\prime2}+k^{2}\cos^{2}(\theta))}{\Delta}d\theta\\
&= k^{\prime2}J^{(m,n)} + k^{2}J^{(m,n+2)}\\
\end{align*}
だから、$I$$J$ との間に、次の基本的な関係式が成立する。

\begin{align*}
 I^{(m,n)} &= J^{(m,n)} - k^{2}J^{(m+2,n)}\\
 I^{(m,n)} &= k^{\prime2}J^{(m,n)} + k^{2}J^{(m,n+2)}\\
\end{align*}

当然、次の逆の関係式が成り立つ。 $m{\geq}2$ の時

\begin{align*}
 J^{(m,n)} &= \inverse{k^{2}}\left(J^{(m-2,n)}-I^{(m-2,n)}\right)
\end{align*}

$n{\geq}2$ の時

\begin{align*}
 J^{(m,n)} &= \inverse{k^{2}}\left(-k^{\prime2}J^{(m,n-2)}+I^{(m,n-2)}\right)
\end{align*}

また [三角関数の幾つかの2次無理式の積分] (2017年10月31日[火]) ) で指摘したように

\begin{align*}
 &I^{(1,0)} = I^{1}_{\mathrm s} = -\frac{\cos(\varphi)\Delta(\varphi,k)}{2} + \inverse{2} - \frac{1-k^{2}}{2k}\ln\left(\frac{k\cos(\varphi)+\Delta(\varphi,k)}{k+1}\right)\\
 &\qquad\qquad\qquad      = -\frac{\cos(\varphi)\Delta(\varphi,k)}{2} + \inverse{2} + \frac{1-k^{2}}{2k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\\
 &I^{(2,0)} = I^{2}_{\mathrm s} = -\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3} + \frac{2k^{2}-1}{3k^{2}}E(\varphi,k) + \frac{1-k^{2}}{3k^{2}}F(\varphi,k)\\
 &I^{(0,1)} = I^{1}_{\mathrm c} = \frac{\sin(\varphi)\Delta(\varphi,k)}{2} + \inverse{2k}{\arcsin(k\sin(\varphi))}\\
 &I^{(0,2)} = I^{2}_{\mathrm c} = \frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3} + \frac{1+k^{2}}{3k^{2}}E(\varphi,k) - \frac{1-k^{2}}{3k^{2}}F(\varphi,k)\\
 &J^{(1,0)} = J^{1}_{\mathrm s} = \inverse{k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right) = -\inverse{k}\ln\left(\frac{k\cos(\varphi)+\Delta(\varphi,k)}{k+1}\right)\\
 &J^{(2,0)} = J^{2}_{\mathrm s} = \inverse{k^{2}}\left(F(\varphi,k)-E(\varphi,k)\right)\\
 &J^{(0,1)} = J^{1}_{\mathrm c} = \inverse{k}\arcsin(k\sin(\varphi))\\
 &J^{(0,2)} = J^{2}_{\mathrm c} = \inverse{k^{2}}E(\varphi,k) - \frac{1-k^{2}}{k^{2}}F(\varphi,k)\\
\end{align*}
である。

以下、前記記事で求めなかった $I^{(m,n)}$ 及び $J^{(m,n)}$ を計算していこう。


まず、前記記事と同様な手法で、$I^{(1,1)}$ を計算すると、次のようになる。

\begin{align*}
 I^{(1,1)} &= \int_{0}^{\varphi}\Delta(\theta,k)\sin(\theta)\cos(\theta)d\theta\\
&= \int_{0}^{\varphi}\sqrt{1-k^{2}\sin^{2}(\theta)}\sin(\theta)\cos(\theta)d\theta\\
&= \inverse{2k^{2}}\int_{0}^{k^{2}\sin^{2}(\varphi)}\sqrt{1-w}dw\\
&= \inverse{2k^{2}}\Big[-\frac{2}{3}(1-w)^{3/2}\Big]_{0}^{k^{2}\sin^{2}(\varphi)}\\
&= \inverse{3k^{2}}\left(1-(1-k^{2}\sin^{2}(\varphi))^{3/2}\right)\\
&= \inverse{3k^{2}}(1-(\Delta(\varphi,k))^{3})
\end{align*}


実は、これには、簡単な別解がある。つまり、$(\Delta(\theta,k))^{3}$$\theta$ で微分すると、

\[
 \odiff{(\Delta^{3})}{\theta} = 3\Delta^{2}\left(-\frac{k^{2}\sin(\theta)\cos(\theta)}{\Delta}\right) = -3k^{2}\Delta\sin(\theta)\cos(\theta)
\]
だから
\begin{align*}
 I^{(1,1)} &= \int_{0}^{\varphi}\Delta(\theta,k)\sin(\theta)\cos(\theta)d\theta\\
&= -\inverse{3k^{2}}\int_{0}^{\varphi}\odiff{(\Delta^{3})}{\theta}d\theta\\
&= -\inverse{3k^{2}}\left\{(\Delta(\varphi,k))^{3}-(\Delta(0,k))^{3}\right\}\\
&= \inverse{3k^{2}}\left\{1-(\Delta(\varphi,k))^{3}\right\}\\
\end{align*}


$J^{(1,1)}$ では次のようになる。

\begin{align*}
J^{(1,1)} &= \int_{0}^{\varphi}\frac{\sin(\theta)\cos(\theta)d\theta}{\Delta(\theta,k)} = \int_{0}^{\varphi}\frac{\sin(\theta)\cos(\theta)d\theta}{\sqrt{1-k^{2}\sin^{2}(\theta)}}\\
&= \inverse{2k^{2}}\int_{0}^{k^{2}\sin^{2}(\varphi)}\frac{dw}{\sqrt{1-w}}\\
&= \inverse{k^{2}}\Big[-\sqrt{1-w}\Big]_{0}^{k^{2}\sin^{2}(\varphi)}\\
&= \inverse{k^{2}}(1-\Delta(\varphi,k))
\end{align*}


これも

\[
 \odiff{\Delta}{\theta} = -\frac{k^{2}\sin(\theta)\cos(\theta)}{\Delta}
\]
を使えば、別解
\begin{align*}
 J^{(1,1)} &= \int_{0}^{\varphi}\frac{\sin(\theta)\cos(\theta)d\theta}{\Delta(\theta,k)}\\
&= -\inverse{k^{2}}\int_{0}^{\varphi}\odiff{\Delta}{\theta}d\theta\\
&= \inverse{k^{2}}(1-\Delta(\varphi,k))
\end{align*}
が得られる。


\begin{align*}
 J^{(2,1)} &=\int_{0}^{\varphi}\frac{\sin^{2}(\theta)\cos(\theta)d\theta}{\Delta(\theta,k)}\\
&= \int_{0}^{\varphi}\frac{(1-(\Delta(\theta,k))^{2})\cos(\theta)d\theta}{k^{2}\Delta(\theta,k)}\\
&= \inverse{k^{2}}\left\{J^{(0,1)}-I^{(0.1)}\right\}\\
&= -\inverse{2k^{2}}\sin(\varphi)\Delta(\varphi,k) + \inverse{2k^{3}}\arcsin(k\sin(\varphi))\\
\end{align*}


別解:

\begin{align*}
 J^{(2,1)} &=\int_{0}^{\varphi}\frac{\sin^{2}(\theta)\cos(\theta)d\theta}{\Delta(\theta,k)}\\
&= -\inverse{k^{2}}\int_{0}^{\varphi}\sin(\theta)\left(\odiff{\Delta}{\theta}\right)d\theta\\
&= -\inverse{k^{2}}\left\{\Big[\sin(\theta)\Delta\Big]_{0}^{\varphi} - \int_{0}^{\varphi}\cos(\theta){\Delta}d\theta\right\}\\
&= -\inverse{k^{2}}\left\{\sin(\varphi)\Delta(\varphi,k) - I^{(0,1)}\right\}\\
&= -\inverse{k^{2}}\left\{\sin(\varphi)\Delta(\varphi,k) - \frac{\sin(\varphi)\Delta(\varphi,k)}{2} - \inverse{2k}{\arcsin(k\sin(\varphi))}\right\}\\
&= -\inverse{2k^{2}}\sin(\varphi)\Delta(\varphi,k) + \inverse{2k^{3}}\arcsin(k\sin(\varphi))\\
\end{align*}


\begin{align*}
 J^{(1,2)} &=\int_{0}^{\varphi}\frac{\sin(\theta)\cos^{2}(\theta)d\theta}{\Delta(\theta,k)}\\
&= \int_{0}^{\varphi}\frac{\sin(\theta)(-k^{\prime2}+(\Delta(\theta,k))^{2})d\theta}{k^{2}\Delta(\theta,k)}\\
&= \inverse{k^{2}}\left\{-k^{\prime2}J^{(1,0)}+I^{(1,0)}\right\}\\
&= \frac{k^{2}-1}{k^{2}}\left\{\inverse{k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\right\}\\
&\qquad +\inverse{k^{2}}\left\{
-\frac{\cos(\varphi)\Delta(\varphi,k)}{2} + \inverse{2} + \frac{1-k^{2}}{2k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\right\}\\
&= -\frac{\cos(\varphi)\Delta(\varphi,k)}{2k^{2}} + \inverse{2k^{2}} - 
\frac{1-k^{2}}{2k^{3}}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)
\end{align*}


別解:

\begin{align*}
 J^{(1,2)} &=\int_{0}^{\varphi}\frac{\sin(\theta)\cos^{2}(\theta)d\theta}{\Delta(\theta,k)}\\
&= -\inverse{k^{2}}\int_{0}^{\varphi}\cos(\theta)\left(\odiff{\Delta}{\theta}\right)d\theta\\
&= -\inverse{k^{2}}\left\{\Big[\cos(\theta)\Delta(\theta,k)\Big]_{0}^{\varphi} + \int_{0}^{\varphi}\sin(\theta)\Delta(\theta,k)d\theta\right\}\\
&= -\inverse{k^{2}}\cos(\varphi)\Delta(\varphi,k) + \inverse{k^{2}} - \inverse{k^{2}}I^{(1,0)}\\
&= -\inverse{k^{2}}\cos(\varphi)\Delta(\varphi,k) + \inverse{k^{2}}\\
&\qquad - \inverse{k^{2}}\left\{-\frac{\cos(\varphi)\Delta(\varphi,k)}{2} + \inverse{2}
+ \frac{1-k^{2}}{2k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\right\}\\
&= -\frac{\cos(\varphi)\Delta(\varphi,k)}{2k^{2}} + \inverse{2k^{2}} - \frac{1-k^{2}}{2k^{3}}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)
\end{align*}


\begin{align*}
 J^{(2,2)} &=\int_{0}^{\varphi}\frac{\sin^{2}(\theta)\cos^{2}(\theta)d\theta}{\Delta(\theta,k)}\\
&= \int_{0}^{\varphi}\frac{(1-(\Delta(\theta,k))^{2}))(-k^{\prime2}+(\Delta(\theta,k))^{2})d\theta}{k^{4}\Delta(\theta,k)}\\
\end{align*}

ここで被積分関数を整理すると

\begin{align*}
 &\frac{(1-(\Delta(\theta,k))^{2}))(-k^{\prime2}+(\Delta(\theta,k))^{2})}{k^{4}\Delta(\theta,k)}\\
&\quad = \inverse{k^{4}}\left\{-\frac{k^{\prime2}}{\Delta(\theta,k)} + (1+k^{\prime2})\Delta(\theta,k) - (\Delta(\theta,k))^{3}\right\}\\
&\quad = \inverse{k^{4}}\left\{-\frac{k^{\prime2}}{\Delta(\theta,k)} + (1+k^{\prime2})\Delta(\theta,k) - (1-k^{2}\sin^{2}(\theta))(\Delta(\theta,k))\right\}\\
&\quad = \inverse{k^{4}}\left\{-\frac{1-k^{2}}{\Delta(\theta,k)} + (1-k^{2})\Delta(\theta,k) + k^{2}\sin^{2}(\theta))(\Delta(\theta,k))\right\}\\
\end{align*}

従って

\begin{align*}
%&= \inverse{k^{4}}\int_{0}^{\varphi}\left\{-\frac{k^{\prime2}}{\Delta(\theta,k)} + (1+k^{\prime2})\Delta(\theta,k) - (\Delta(\theta,k))^{3}\right\}d\theta\\
%&= \inverse{k^{4}}\int_{0}^{\varphi}\left\{-\frac{k^{\prime2}}{\Delta(\theta,k)} + (1+k^{\prime2})\Delta(\theta,k) - (1-k^{2}\sin^{2}(\theta))(\Delta(\theta,k))\right\}d\theta\\
J^{(2,2)} &= \inverse{k^{4}}\left\{-(1-k^{2})F(\varphi,k)+(1-k^{2})E(\varphi,k)+k^{2}I^{(2,0)}\right\}\\
&= -\frac{1-k^{2}}{k^{4}}F(\varphi,k) + \frac{1-k^{2}}{k^{4}}E(\varphi,k)\\
&\qquad\qquad\qquad + \inverse{k^{2}}\left\{-\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3}\right.\\
&\qquad\qquad\qquad\qquad \left.+ \frac{2k^{2}-1}{3k^{2}}E(\varphi,k) + \frac{1-k^{2}}{3k^{2}}F(\varphi,k))\right\}\\
&= -\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3k^{2}} + \frac{2-k^{2}}{3k^{4}}E(\varphi,k) + \frac{2k^{2}-2}{3k^{4}}F(\varphi.k)
\end{align*}


別解:

\begin{align*}
 J^{(2,2)} &=\int_{0}^{\varphi}\frac{\sin^{2}(\theta)\cos^{2}(\theta)d\theta}{\Delta(\theta,k)}\\
&= \int_{0}^{\varphi}\sin(\theta)\cos(\theta)\left(-\inverse{k^{2}}\odiff{\Delta}{\theta}\right)d\theta\\
&= -\inverse{k^{2}}\left\{\Big[\sin(\theta)\cos(\theta)\Delta(\varphi,k)\Big]_{0}^{\varphi} - \int_{0}^{\varphi}(\cos^{2}(\theta)-\sin^{2}(\theta))\Delta(\varphi,k)d\theta\right\}\\
&= -\inverse{k^{2}}\sin(\varphi)\cos(\varphi)\Delta(\varphi,k) + \inverse{k^{2}}\int_{0}^{\varphi}(2\cos^{2}(\theta)-1)\Delta(\varphi,k)d\theta\\
&= -\inverse{k^{2}}\sin(\varphi)\cos(\varphi)\Delta(\varphi,k) + \frac{2}{k^{2}}I^{(0,2)} - \inverse{k^{2}}E(\varphi,k)\\
&= -\inverse{k^{2}}\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + \frac{2}{k^{2}}\left\{\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3} + \frac{1+k^{2}}{3k^{2}}E(\varphi,k) - \frac{1-k^{2}}{3k^{2}}F(\varphi,k)\right\}\\
&\qquad - \inverse{k^{2}}E(\varphi,k)\\
&= -\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3k^{2}} + \frac{2-k^{2}}{3k^{4}}E(\varphi,k) + \frac{2k^{2}-2}{3k^{4}}F(\varphi,k)
\end{align*}


\begin{align*}
 I^{(2,1)} &= \int_{0}^{\varphi}\sin^{2}(\theta)\cos(\theta)\Delta(\theta,k)d\theta\\
&= \inverse{k^{2}}\int_{0}^{\varphi}(1-(\Delta(\theta,k))^{2})\cos(\theta)\Delta(\theta,k)d\theta\\
&= \inverse{k^{2}}\left\{\int_{0}^{\varphi}\Delta(\theta,k)\cos(\theta)d\theta - \int_{0}^{\varphi}(\Delta(\theta,k))^{3}\cos(\theta)d\theta\right\}\\
&= \inverse{k^{2}}\int_{0}^{\varphi}\Delta(\theta,k)\cos(\theta)d\theta - \inverse{k^{2}}\int_{0}^{\varphi}(\Delta(\theta,k))^{3}(\sin(\theta))^{\prime}d\theta\\
&=\inverse{k^{2}}I^{(0,1)} -\inverse{k^{2}}\Big[(\Delta(\theta,k))^{3}\sin(\theta)\Big]_{0}^{\varphi}
+\inverse{k^{2}}\int_{0}^{\varphi}((\Delta(\theta,k))^{3})^{\prime}\sin(\theta)d\theta\\
&=\inverse{k^{2}}I^{(0,1)} -\inverse{k^{2}}(\Delta(\varphi,k))^{3}\sin(\varphi)\\
&\qquad\qquad\qquad -\inverse{k^{2}}\int_{0}^{\varphi}\left\{\Big(3(\Delta(\theta,k))^{2}\Big)\left(\frac{k^{2}\sin(\theta)\cos(\theta)}{\Delta(\theta,k)}\right)\sin(\theta)\right\}d\theta\\
&=\inverse{k^{2}}I^{(0,1)} - \inverse{k^{2}}(\Delta(\varphi,k))(1-k^{2}\sin^{2}(\varphi))\sin(\varphi)-3I^{(2,1)}\\
\end{align*}

従って

\begin{align*}
 I^{(2,1)} &=\inverse{4}\left\{\inverse{k^{2}}I^{(0,1)} - \inverse{k^{2}}(\Delta(\varphi,k))(1-k^{2}\sin^{2}(\varphi))\sin(\varphi)\right\}\\
&= \inverse{4k^{2}}\left\{\frac{\sin(\varphi)\Delta(\varphi,k)}{2} + \inverse{2k}{\arcsin(k\sin(\varphi))}\right\}\\
&\qquad\qquad\qquad  - \inverse{4k^{2}}(\Delta(\varphi,k))(1-k^{2}\sin^{2}(\varphi))\sin(\varphi)\\
&= \inverse{8k^{2}}(\Delta(\varphi,k))(2k^{2}\sin^{2}(\varphi)-1)\sin(\varphi) + \inverse{8k^{3}}{\arcsin(k\sin(\varphi))}
\end{align*}


別解:

\begin{align*}
 I^{(2,1)} &= \int_{0}^{\varphi}\sin^{2}(\theta)\cos(\theta)\Delta(\theta,k)d\theta\\
&= -\inverse{3k^{2}}\int_{0}^{\varphi}\sin(\theta)\left(\odiff{(\Delta^{3})}{\theta}\right)d\theta\\
 &= -\inverse{3k^{2}}\left\{\Big[\sin(\theta)\Delta^{3}\Big]_{0}^{\varphi} - \int_{0}^{\varphi}\cos(\theta)\Delta^{3}d\theta\right\}\\
&= -\inverse{3k^{2}}\sin(\varphi)(\Delta(\varphi,k))^{3} + \inverse{3k^{2}}\int_{0}^{\varphi}\cos(\theta)(1-k^{2}\sin^{2}(\theta)){\Delta}d\theta\\
&= -\inverse{3k^{2}}\sin(\varphi)(\Delta(\varphi,k))^{3} + \inverse{3k^{2}}I^{(0,1)} - \inverse{3}I^{(2,1)}\\
\end{align*}

従って

\begin{align*}
 I^{(2,1)} &= \frac{3}{4}\left\{-\inverse{3k^{2}}\sin(\varphi)(\Delta(\varphi,k))^{3} + \inverse{3k^{2}}I^{(0,1)}\right\}\\
&= -\inverse{4k^{2}}\sin(\varphi)(1-k^{2}\sin^{2}(\varphi))(\Delta(\varphi,k))\\
&\qquad\qquad + \inverse{4k^{2}}\left\{ \frac{\sin(\varphi)\Delta(\varphi,k)}{2} + \inverse{2k}{\arcsin(k\sin(\varphi))}\right\}\\
&= \inverse{8k^{2}}(\Delta(\varphi,k))(2k^{2}\sin^{2}(\varphi)-1)\sin(\varphi) + \inverse{8k^{3}}{\arcsin(k\sin(\varphi))}
\end{align*}


\begin{align*}
 I^{(1,2)} &= \int_{0}^{\varphi}\sin(\theta)\cos^{2}(\theta)\Delta(\theta,k)d\theta\\
&= \inverse{k^{2}}\int_{0}^{\varphi}\sin(\theta)(-k^{\prime2}+(\Delta(\theta,k))^{2})\Delta(\theta,k)d\theta\\
&= \inverse{k^{2}}\left\{-k^{\prime2}\int_{0}^{\varphi}\sin(\theta)\Delta(\theta,k)d\theta + \int_{0}^{\varphi}\sin(\theta)(\Delta(\theta,k))^{3}d\theta\right\}\\
&= -\frac{k^{\prime2}}{k^{2}}\int_{0}^{\varphi}\sin(\theta)\Delta(\theta,k)d\theta
 + \inverse{k^{2}}\int_{0}^{\varphi}(-\cos(\theta))^{\prime}(\Delta(\theta,k))^{3}d\theta\\
&=-\frac{k^{\prime2}}{k^{2}}I^{(1,0)} +\inverse{k^{2}}\Big[-\cos(\theta)(\Delta(\theta,k))^{3}\Big]_{0}^{\varphi}+\inverse{k^{2}}\int_{0}^{\varphi}\cos(\theta)\left(\odiff{(\Delta^{3})}{\theta}\right)d\theta\\
&=-\frac{k^{\prime2}}{k^{2}}I^{(1,0)} -\inverse{k^{2}}\left\{\cos(\varphi)(\Delta(\varphi,k))^{3}-1\right\}\\
&\qquad\qquad\qquad\qquad -\inverse{k^{2}}\int_{0}^{\varphi}\cos(\theta)(3(\Delta(\theta,k))^{2})\left\{\frac{k^{2}\sin(\varphi)\cos(\phi)}{\Delta(\theta,k)}\right\}d\theta\\
&=-\frac{k^{\prime2}}{k^{2}}I^{(1,0)} - \inverse{k^{2}}\cos(\varphi)(1-k^{2}\sin^{2}(\varphi))(\Delta(\varphi,k)) + \inverse{k^{2}} -3I^{(1,2)}\\
\end{align*}

従って

\begin{align*}
 I^{(1,2)} &= \inverse{4}\left\{-\frac{k^{\prime2}}{k^{2}}I^{(1,0)} - \inverse{k^{2}}\cos(\varphi)(1-k^{2}\sin^{2}(\varphi))(\Delta(\varphi,k)) + \inverse{k^{2}}\right\}\\
&= \inverse{4}\left\{-\frac{1-k^{2}}{k^{2}}
\left\{-\frac{\cos(\varphi)\Delta(\varphi,k)}{2} + \inverse{2} + \frac{1-k^{2}}{2k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\right\}\right.\\
&\qquad\qquad\qquad\qquad \left. - \inverse{k^{2}}\cos(\varphi)(1-k^{2}\sin^{2}(\varphi))(\Delta(\varphi,k)) + \inverse{k^{2}}\right\}\\
&= \inverse{4}\left\{\frac{\cos(\varphi)\Delta(\varphi,k)}{2k^{2}}(2k^{2}\sin^{2}(\varphi)-1-k^{2}) + \frac{1+k^{2}}{2k^{2}}\right.\\
&\qquad\qquad\qquad\qquad \left.- \frac{(1-k^{2})^{2}}{2k^{3}}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\right\}\\
&= \frac{\cos(\varphi)\Delta(\varphi,k)}{8k^{2}}(2k^{2}\sin^{2}(\varphi)-1-k^{2}) + \frac{1+k^{2}}{8k^{2}}\\
&\qquad\qquad\qquad\qquad - \frac{(1-k^{2})^{2}}{8k^{3}}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\\
&= -\frac{\cos(\varphi)\Delta(\varphi,k)}{8k^{2}}(2k^{2}\cos^{2}(\varphi)+1-k^{2}) + \frac{1+k^{2}}{8k^{2}}\\
&\qquad\qquad\qquad\qquad - \frac{(1-k^{2})^{2}}{8k^{3}}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\\
%&= -\frac{\cos(\varphi)\Delta(\varphi,k)}{8k^{2}}(2k^{2}\cos^{2}(\varphi)+1-k^{2}) + \frac{1+k^{2}}{8k^{2}}\\
%&\qquad\qquad\qquad\qquad - \frac{(1-k^{2})^{2}}{8k^{3}}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\\
\end{align*}


別解:

\begin{align*}
 I^{(1,2)} &= \int_{0}^{\varphi}\sin(\theta)\cos^{2}(\theta)\Delta(\theta,k)d\theta\\
&= -\inverse{3k^{2}}\int_{0}^{\varphi}\cos(\theta)\left(\odiff{(\Delta^{3})}{\theta}\right)d\theta\\
&= -\inverse{3k^{2}}\left\{\Big[\cos(\theta)(\Delta^{3})\Big]_{0}^{\varphi} + \int_{0}^{\varphi}\sin(\theta)(\Delta^{3})d\theta\right\}\\
&= -\inverse{3k^{2}}\left\{\cos(\varphi)(\Delta(\varphi,k)^{3})-1\right\} - \inverse{3k^{2}} \int_{0}^{\varphi}\sin(\theta)(k^{\prime2}+k^{2}\cos^{2}(\theta)){\Delta}d\theta\\
&= -\inverse{3k^{2}}\cos(\varphi)(k^{\prime2}+k^{2}\cos^{2}(\varphi))\Delta(\varphi,k) + \inverse{3k^{2}} - \frac{k^{\prime2}}{3k^{2}}I^{(1,0)} - \inverse{3}I^{(1,2)}
\end{align*}

従って

\begin{align*}
 I^{(1,2)} &= \frac{3}{4}\left\{-\inverse{3k^{2}}\cos(\varphi)(k^{\prime2}+k^{2}\cos^{2}(\varphi))\Delta(\varphi,k) + \inverse{3k^{2}} - \frac{k^{\prime2}}{3k^{2}}I^{(1,0)}\right\}\\
&= \frac{3}{4}\left\{-\inverse{3k^{2}}\cos(\varphi)(k^{\prime2}+k^{2}\cos^{2}(\varphi))\Delta(\varphi,k) + \inverse{3k^{2}}\right.\\
& \qquad\qquad - \left.\frac{k^{\prime2}}{3k^{2}}\left\{-\frac{\cos(\varphi)\Delta(\varphi,k)}{2} + \inverse{2} + \frac{1-k^{2}}{2k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\right\}\right\}\\
&= \frac{3}{4}\left\{-\inverse{3k^{2}}\cos(\varphi)\left(k^{\prime2}+k^{2}\cos^{2}(\varphi)-\inverse{2}k^{\prime2}\right)\Delta(\varphi,k) + \inverse{3k^{2}}\left(1-\frac{k^{\prime2}}{2}\right)\right.\\
&\qquad\qquad - \left.\frac{k^{\prime2}}{3k^{2}}\cdot\frac{1-k^{2}}{2k}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)
\right\}\\
&= -\frac{\cos(\varphi)\Delta(\varphi,k)}{8k^{2}}(2k^{2}\cos^{2}(\varphi)+1-k^{2}) + \frac{1+k^{2}}{8k^{2}}\\
&\qquad\qquad\qquad\qquad - \frac{(1-k^{2})^{2}}{8k^{3}}\ln\left(\frac{\Delta(\varphi,k)-k\cos(\varphi)}{1-k}\right)\\
\end{align*}


\begin{align*}
 I^{(2,2)} &=\int_{0}^{\varphi}\sin^{2}(\theta)\cos^{2}(\theta){\Delta}d\theta = \inverse{4}\int_{0}^{\varphi}(\sin(2\theta))^{2}{\Delta}d\theta\\
&= \inverse{4}\int_{0}^{\varphi}\left(\frac{1-\cos(4\theta)}{2}\right){\Delta}d\theta\\
&= \inverse{8}\left\{\int_{0}^{\varphi}{\Delta}d\theta - \int_{0}^{\varphi}\cos(4\theta){\Delta}d\theta\right\}\\
&= \inverse{8}\left\{E(\varphi,k) - \inverse{4}\int_{0}^{\varphi}(\sin(4\theta))^{\prime}{\Delta}d\theta\right\}\\
&= \inverse{8}E(\varphi,k) - \inverse{32}\left\{\Big[\sin(4\theta)\Delta\Big]_{0}^{\varphi} + k^{2}\int_{0}^{\varphi}\sin(4\theta)\frac{\sin(\theta)\cos(\theta)}{\Delta}d\theta\right\}\\
&= \inverse{8}E(\varphi,k) - \inverse{32}\sin(4\varphi)\Delta(\varphi,k)\\
&\qquad - \frac{k^{2}}{8}\int_{0}^{\varphi}\left\{\sin(\theta)\cos(\theta)\left\{\cos^{2}(\theta)-\sin^{2}(\theta)\right\}\frac{\sin(\theta)\cos(\theta)}{\Delta}\right\}d\theta\\
&= \inverse{8}E(\varphi,k) - \inverse{8}\sin(\varphi)\cos(\varphi)\left\{\cos^{2}(\varphi)-\sin^{2}(\varphi)\right\}\Delta(\varphi,k)\\
&\qquad - \frac{k^{2}}{8}\int_{0}^{\varphi}\left\{\frac{\sin^{2}(\theta)\cos^{2}(\theta)\left\{\cos^{2}(\theta)-\sin^{2}(\theta)\right\}}{\Delta}\right\}d\theta\\
\end{align*}

しかるに

\begin{align*}
 \cos^{2}(\theta)-\sin^{2}(\theta) &= \frac{k^{2}-1+\Delta^{2}}{k^{2}} - \frac{1-\Delta^{2}}{k^{2}}\\
&= \frac{-2+k^{2}+2\Delta^{2}}{k^{2}}
\end{align*}

だから

\[
 \frac{\cos^{2}(\theta)-\sin^{2}(\theta)}{\Delta} = \frac{-2+k^{2}+2\Delta^{2}}{k^{2}\Delta} = \frac{-2+k^{2}}{k^{2}}\inverse{\Delta} + \frac{2}{k^{2}}\Delta
\]
なので

\begin{align*}
 I^{(2,2)} &= \inverse{8}E(\varphi,k) - \inverse{8}\sin(\varphi)\cos(\varphi)\left\{\cos^{2}(\varphi)-\sin^{2}(\varphi\right\}\Delta(\varphi,k)\\
&\qquad - \frac{k^{2}}{8}\int_{0}^{\varphi}\sin^{2}(\theta)\cos^{2}(\theta)\left\{\frac{-2+k^{2}}{k^{2}}\inverse{\Delta} + \frac{2}{k^{2}}\Delta\right\}d\theta\\
&= \inverse{8}E(\varphi,k) - \inverse{8}\sin(\varphi)\cos(\varphi)\left\{\cos^{2}(\varphi)-\sin^{2}(\varphi)\right\}\Delta(\varphi,k)\\
&\qquad - \frac{-2+k^{2}}{8}J^{(2,2)} - \inverse{4}I^{(2,2)}
\end{align*}

結局

\begin{align*}
 I^{(2,2)} &= \frac{4}{5}
\left\{\inverse{8}E(\varphi,k) - \inverse{8}\sin(\varphi)\cos(\varphi)\left\{\cos^{2}(\varphi)-\sin^{2}(\varphi)\right\}\Delta(\varphi,k)\right.\\
&\qquad - \left.\frac{-2+k^{2}}{8}\left(-\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3k^{2}} + \frac{2-k^{2}}{3k^{4}}E(\varphi,k) + \frac{2k^{2}-2}{3k^{4}}F(\varphi,k)\right)\right\}\\
&= \inverse{10}\left\{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)\left(-(\cos^{2}(\theta)-\sin^{2}(\theta))+\frac{-2+k^{2}}{3k^{2}}\right)\right.\\
&\qquad + \left.\left(1+\frac{(2-k^{2})^{2}}{3k^{4}}\right)E(\varphi,k) -\frac{(2k^{2}-2)(-2+k^{2})}{3k^{4}}F(\varphi,k)\right\}\\
&= \inverse{30k^{4}}\Big\{\Big(-(\cos^{2}(\theta)-\sin^{2}(\theta))(3k^{4})+(-2+k^{2})(k^{2})\Big)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + \Big(3k^{4}+(2-k^{2})^{2}\Big)E(\varphi,k) - 2(1-k^{2})(2-k^{2})F(\varphi,k)\Big\}\\
&= \inverse{30k^{4}}\Big\{\Big(2k^{2}(-3k^{2}\cos^{2}(\theta)+2k^{2}-1)\Big)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + 4(k^{4}-k^{2}+1)E(\varphi,k) - 2(1-k^{2})(2-k^{2})F(\varphi,k)\Big\}\\
\end{align*}

若干変形して

\begin{align*}
 I^{(2,2)} &= \inverse{15k^{2}}\Big\{(-3k^{2}\cos^{2}(\theta)+2k^{2}-1)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + \frac{2(k^{4}-k^{2}+1)}{k^2}E(\varphi,k) - \frac{(1-k^{2})(2-k^{2})}{k^{2}}F(\varphi,k)\Big\}\\
\end{align*}


別解:

\begin{align*}
 I^{(2,2)} &=\int_{0}^{\varphi}\sin^{2}(\theta)\cos^{2}(\theta)\Delta(\theta,k)d\theta\\
&= \int_{0}^{\varphi}\sin(\theta)\cos(\theta)\left\{-\inverse{3k^{2}}\odiff{(\Delta^{3})}{\theta}\right\}d\theta\\
&= -\inverse{3k^{2}}\left\{\Big[\sin(\theta)\cos(\theta)\Delta^{3}\Big]_{0}^{\varphi} - \int_{0}^{\varphi}(\cos^{2}(\theta)-\sin^{2}(\theta)){\Delta^{3}}d\theta\right\}\\
&= -\inverse{3k^{2}}\sin(\varphi)\cos(\varphi)(\Delta(\varphi,k)^{3}) + \inverse{3k^{2}}\int_{0}^{\varphi}(2\cos^{2}(\theta)-1){\Delta^{3}}d\theta\\
&= -\inverse{3k^{2}}\sin(\varphi)\cos(\varphi)(\Delta(\varphi,k)^{3})\\
&\qquad\qquad + \inverse{3k^{2}}\int_{0}^{\varphi}(2\cos^{2}(\theta)-1)(1-k^{2}\sin^{2}(\theta)){\Delta}d\theta\\
&= -\inverse{3k^{2}}\sin(\varphi)\cos(\varphi)(\Delta(\varphi,k)^{3})\\
&\qquad\qquad + \inverse{3k^{2}}\left\{-2k^{2}I^{(2,2)}+k^{2}I^{(2,0)}+2I^{(0,2)}-E(\varphi,k)\right\}\\
\end{align*}

ところが

\begin{align*}
 &k^{2}I^{(2,0)}+2I^{(0,2)}\\
&\quad = k^{2}\left\{-\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3} + \frac{2k^{2}-1}{3k^{2}}E(\varphi,k) + \frac{1-k^{2}}{3k^{2}}F(\varphi,k)\right\}\\
&\qquad\qquad + 2\left\{\frac{\Delta(\varphi,k)\sin(\varphi)\cos(\varphi)}{3} + \frac{1+k^{2}}{3k^{2}}E(\varphi,k) - \frac{1-k^{2}}{3k^{2}}F(\varphi,k)\right\}\\
&\quad = \frac{-k^{2}+2}{3}(\Delta(\varphi,k)\sin(\varphi)\cos(\varphi))\\
&\qquad\qquad + \frac{2k^{4}+k^{2}+2}{3k^{2}}E(\varphi,k) + \frac{-k^{4}+3k^{2}-2}{3k^{2}}F(\varphi,k)
\end{align*}

だから

\begin{align*}
 I^{(2,2)} &= \frac{3}{5}\left\{-\inverse{3k^{2}}\sin(\varphi)\cos(\varphi)(1-k^{2}+k^{2}\cos^{2}(\theta))\Delta(\varphi,k)\right.\\
&\qquad\qquad + \inverse{3k^{2}}\left\{\frac{-k^{2}+2}{3}(\Delta(\varphi,k)\sin(\varphi)\cos(\varphi))\right.\\
&\qquad\qquad\qquad + \left.\left.\frac{2k^{4}+k^{2}+2}{3k^{2}}E(\varphi,k) + \frac{-k^{4}+3k^{2}-2}{3k^{2}}F(\varphi,k)
-E(\varphi,k)\right\}\right\}\\
&= \inverse{15k^{2}}\Big\{(-3k^{2}\cos^{2}(\varphi)+2k^{2}-1)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad\qquad + \frac{2(k^{4}-k^{2}+1)}{k^{2}}E(\varphi,k) - \frac{(1-k^{2})(2-k^{2})}{k^{2}}F(\varphi,k)\Big\}
\end{align*}


\begin{align*}
 J^{(3,0)} &= \int_{0}^{\varphi}\frac{\sin^{3}(\theta)d\theta}{\Delta}\\
&= \int_{0}^{\varphi}\frac{\sin(\theta)}{\Delta}\left(\frac{1-\Delta^{2}}{k^{2}}\right)d\theta\\
&= \inverse{k^{2}}\left\{\int_{0}^{\varphi}\frac{\sin(\theta)d\theta}{\Delta} - \int_{0}^{\varphi}\Delta\sin(\theta)d\theta\right\}\\
&= \inverse{k^{2}}(J^{(1,0)} - I^{(1,0)})\\
&= \inverse{2k^{2}}(\cos(\varphi)\Delta-1) - \frac{1+k^{2}}{2k^{3}}\ln\left(\frac{k\cos(\varphi)+\Delta}{k+1}\right)
\end{align*}


同様に

\begin{align*}
 J^{(0,3)} &= \int_{0}^{\varphi}\frac{\cos^{3}(\theta)d\theta}{\Delta}\\
&= \int_{0}^{\varphi}\frac{\cos(\theta)}{\Delta}\left(\frac{-k^{\prime2}+\Delta^{2}}{k^{2}}\right)d\theta\\
&= \inverse{k^{2}}\left\{-k^{\prime2}\int_{0}^{\varphi}\frac{\cos(\theta)d\theta}{\Delta} + \int_{0}^{\varphi}\Delta\cos(\theta)d\theta\right\}\\
&= \inverse{k^{2}}(-k^{\prime2}J^{(0,1)} + I^{(0,1)})\\
&= \frac{\sin(\varphi)\Delta}{2k^{2}} + \frac{2k^{2}-1}{2k^{3}}\arcsin(k\sin(\varphi))
\end{align*}


\begin{align*}
 I^{(3,0)} &= \int_{0}^{\varphi}\sin^{3}(\theta){\Delta}d\theta = \int_{0}^{\varphi}\sin(\theta)(1-\cos^{2}(\theta)){\Delta}d\theta\\
&\qquad = I^{(1,0)} - \int_{0}^{\varphi}\cos(\theta)(\sin(\theta)\cos(\theta)){\Delta}d\theta\\
&\qquad = I^{(1,0)} + \inverse{3k^{2}}\int_{0}^{\varphi}\cos(\theta)\left(\odiff{(\Delta^{3})}{\theta}\right)d\theta\\
&\qquad = I^{(1,0)} + \inverse{3k^{2}}\left\{\Big[\cos(\theta)\Delta^{3}\Big]_{0}^{\varphi} + \int_{0}^{\varphi}\sin(\theta)(1-k^{2}\sin^{2}(\theta)){\Delta}d\theta\right\}\\
&\qquad = I^{(1,0)} + \inverse{3k^{2}}\left(\cos(\varphi)(\Delta(\varphi,k))^{3}-1\right) + \inverse{3k^{2}}I^{(1,0)} - \inverse{3}I^{(3,0)}
\end{align*}

従って

\[
 \frac{4}{3}I^{(3,0)} = \inverse{3k^{2}}\left(\cos(\varphi)(\Delta(\varphi,k))^{3}-1\right) + \frac{3k^{2}+1}{3k^{2}}I^{(1,0)}
\]

つまり

\begin{align*}
 I^{(3,0)} &= \inverse{4k^{2}}\left(\cos(\varphi)(\Delta(\varphi,k))^{3}-1\right) + \frac{3k^{2}+1}{4k^{2}}I^{(1,0)}
\end{align*}


以下、暫くの間、$m+n$ を「階数」と呼んでおくと、これまでの $I^{(m,n)}$ 又は $J^{(m,n)}$ の計算で分かるように、その「取り敢えずの」表式には、より低い階数の $I^{(m,n)}$ 又は $J^{(m,n)}$ が登場する。つまり、$I^{(m,n)}$ 又は $J^{(m,n)}$ の表式は、漸化式として成立している。いままでは、そうした得られた階数が十分低くて扱いやすかったので、 その $I^{(m,n)}$ 及び/又は $J^{(m,n)}$ を更に開いた訣だ。しかし、今後、ヨリ高階数となった場合には、漸化式のまま表示することを視野に入れるべきだろう。ここでも、勿論、このまま、漸化式の形で表示したままてもよいのだが、右辺の「印象」にまだ「余裕」があるので、ひとまずは、両方の形式を見比べると云う意味もあって、右辺における $\Delta$ の次数を 1 まで引き下げ、また、$I^{(1,0)}$ を開くなら

\begin{align*}
 I^{(3,0)} &= \inverse{4k^{2}}\left\{\cos(\varphi)(1-k^{2}\sin^{2}(\varphi))(\Delta(\varphi,k))-1\right\}\\
&\qquad + \frac{3k^{2}+1}{4k^{2}}\left\{
-\frac{\cos(\varphi)\Delta(\varphi,k)}{2} + \inverse{2} - \frac{1-k^{2}}{2k}\ln\left(\frac{k\cos(\varphi)+\Delta(\varphi,k)}{k+1}\right)\right\}\\
&= -\frac{2k^{2}\sin^{2}(\varphi)+3k^{2}-1}{8k^{2}}\cos(\varphi)\Delta(\varphi,k) + \frac{3k^{2}-1}{8k^{2}}\\
&\qquad + \frac{3k^{4}-2k^{2}-1}{8k^{3}}\ln\left(\frac{k\cos(\varphi)+\Delta(\varphi,k)}{k+1}\right)
\end{align*}
となる。


\begin{align*}
 I^{(0,3)}&= \int_{0}^{\varphi}\cos^{3}(\theta){\Delta}d\theta = \int_{0}^{\varphi}\cos(\theta)(1-\sin^{2}(\theta)){\Delta}d\theta\\
&\qquad = I^{(0,1)} - \int_{0}^{\varphi}\sin(\theta)(\sin(\theta)\cos(\theta)){\Delta}d\theta\\
&\qquad = I^{(0,1)} + \inverse{3k^{2}}\int_{0}^{\varphi}\sin(\theta)\left(\odiff{(\Delta^{3})}{\theta}\right)d\theta\\
&\qquad = I^{(0,1)} + \inverse{3k^{2}}\left\{\Big[\sin(\theta)\Delta^{3}\Big]_{0}^{\varphi} - \int_{0}^{\varphi}\cos(\theta)(k^{\prime2}+k^{2}\cos^{2}(\theta)){\Delta}d\theta\right\}\\
&\qquad = I^{(0,1)} + \frac{\sin(\varphi)(\Delta(\varphi,k))^{3}}{3k^{2}} - \frac{k^{\prime2}}{3k^{2}}I^{(0,1)} - \inverse{3}I^{(0,3)}
\end{align*}

従って

\begin{align*}
 \frac{4}{3}I^{(0,3)} = \frac{\sin(\varphi)(\Delta(\varphi,k))^{3}}{3k^{2}} + \frac{4k^{2}-1}{3k^{2}}I^{(0,1)}
\end{align*}

つまり

\begin{align*}
 I^{(0,3)} = \frac{\sin(\varphi)(\Delta(\varphi,k))^{3}}{4k^{2}} + \frac{4k^{2}-1}{4k^{2}}I^{(0,1)}
\end{align*}

ここで、右辺における $\Delta$ の次数を 1 まで引き下げ、また、$I^{(0,1)}$ を開くと

\begin{align*}
 I^{(0,3)} &= \frac{\sin(\varphi)(\Delta(\varphi,k))}{4k^{2}}(1-k^{2}+k^{2}\cos^{2}(\varphi))\\
&\qquad + \frac{4k^{2}-1}{4k^{2}}\left\{
\frac{\sin(\varphi)\Delta(\varphi,k)}{2} + \inverse{2k}{\arcsin(k\sin(\varphi))}\right\}\\
&\qquad = \frac{\sin(\varphi)(\Delta(\varphi,k))}{8k^{2}}(2k^{2}\cos^{2}(\varphi)+2k^{2}+1) + \frac{4k^{2}-1}{8k^{3}}{\arcsin(k\sin(\varphi))}
\end{align*}

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