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[三角関数の幾つかの2次無理式の積分]補足その2

$m,n$ を非負整数とし、$k,k^{\prime}$$0<k,k^{\prime}<1,\;k^{2}+k^{\prime2}=1$ を満たすとし、さらに、


\begin{align*}
 \Delta(x,k) :&= \sqrt{1-k^{2}\sin^{2}(x)}\\
&= \sqrt{1-k^{2}+k^{2}\cos^{2}(x)} = \sqrt{k^{\prime2}+k^{2}\cos^{2}(x)}
\end{align*}
と定義する時、$I^{(m,n)}$ 及び $J^{(m,n)}$
\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)}{\Delta(\theta,k)}d\theta\\
\end{align*}
を意味するものとする (以下、文脈から明らかな場合は $\Delta(x,k)$ の変数を明示しないで $\Delta$ とのみ書くことがある。また、以下実際には、記法 $k^{\prime2}$ ではなく $1-k^{2}$ が用いられている)。

[三角関数の幾つかの2次無理式の積分: nouse (2017年10月31日 (火))] 及び [「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: nouse (2017年11月30日 (木))] により、現段階では $0{\leq}m+n{\leq}3$$I^{(m,n)}$, $J^{(m,n)}$、そして $I^{(2,2)}$ 及び $J^{(2,2)}$ の表式が得られている。本稿では、それ以降の $I^{(m,n)}$ 及び $J^{(m,n)}$ を、階数 ($=m+n$) に従う漸化式により求めていくことにする。従って、以下では $m+n>3$ が満たされていることを前提とする。

その準備として確認しておくが、[補足その1] の初めの方で既に導いてある関係式より
$m{\geq}2$ の時

\begin{align*}
 &I^{(m,0)} = I^{(m-2,0)} - I^{(m-2,2)}\\
 &J^{(m,0)} = J^{(m-2,0)} - J^{(m-2,2)}
\end{align*}
$n{\geq}2$ の時
\begin{align*}
 &I^{(0,n)} = I^{(0,n-2)} - I^{(2,n-2)}\\
 &J^{(0,n)} = J^{(0,n-2)} - J^{(2,n-2)}
\end{align*}
が成り立つことに注意すると、$I^{(m-2,0)}$, $J^{(m-2,0)}$, $I^{(0,n-2)}$, $J^{(0,n-2)}$ の漸化式と、 $I^{(m-2,2)}$, $J^{(m-2,2)}$, $I^{(2,n-2)}$, $J^{(2,n-2)}$ の漸化式とが得られるなら、$I^{(m,0)}$,$J^{(m,0)}$, $I^{(0,n)}$, $J^{(0,n)}$ の漸化式も得られることが分かる。そして $m+n>3$ が前提となっているから、以下の $I^{(m,n)}$ 及び $J^{(m,n)}$ の計算では $m,n>0$ の場合に限定することにする。

すると

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

まず、$m=1$ 又は $n=1$ の時、上記の最後の式中の $\sin^{m-2}(\theta)$ 又は $\cos^{n-2}(\theta)$ を含む項は不適切になるが、それぞれ、係数 $m-1=0$ 又は $n-1=0$ がかかっており、その不適切性を無視してよいことに注意。

さて $m>1$ なら

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

従って

\begin{align*}
 \frac{m+n+1}{3}I^{(m,n)} &= -\inverse{3k^{2}}\sin^{m-1}(\varphi)\cos^{n-1}(\varphi)\Delta(\varphi,k)^{3}\\
&\qquad + \frac{m-1}{3k^{2}}I^{(m-2,n)} - \frac{n-1}{3k^{2}}(1-k^{2})I^{(m,n-2)}\\
\end{align*}

つまり

\begin{align*}
 I^{(m,n)} &= \inverse{({m+n+1})k^{2}}\Big\{-\sin^{m-1}(\varphi)\cos^{n-1}(\varphi)\Delta(\varphi,k)^{3}\\
&\qquad + (m-1)I^{(m-2,n)} - (n-1)(1-k^{2})I^{(m,n-2)}\Big\}\\
\end{align*}

特に $m=n{\geq}2$ の場合は、次のようになる。

\begin{align*}
 I^{(m,m)} &= \inverse{(2m+1)k^{2}}\Big\{-\sin^{m-1}(\varphi)\cos^{m-1}(\varphi)\Delta(\varphi,k)^{3}\\
&\qquad + (m-1)I^{(m-2,m)} - (m-1)(1-k^{2})I^{(m.m-2)}\Big\}
\end{align*}

また $m=1$ なら (前記の仮定により $n>2$ であることに注意)

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

従って

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

つまり

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

ここで $\Delta$ の次数を 1 に下げるなら

\begin{align*}
 I^{(1,n)} &=  -\inverse{(n+2)k^{2}}\Big\{(n-1)(1-k^{2})I^{(1,n-2)}\\
&\qquad + \cos^{n-1}(\varphi)(1-k^{2}+k^{2}\cos^{2}(\theta))\Delta(\varphi,k) -1\Big\}\\
&= -\inverse{(n+2)k^{2}}\Big\{(n-1)(1-k^{2})I^{(1,n-2)}\\
&\qquad + \Big((1-k^{2})\cos^{n-1}(\varphi) + k^{2}\cos^{n+1}(\theta)\Big)\Delta(\varphi,k) - 1\Big\}\\
\end{align*}

他方

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

いま $m>1$ ならば

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

従って

\begin{align*}
 (m+n-1)J^{(m,n)} &= -\frac{\sin^{m-1}(\varphi)\cos^{n-1}(\varphi)\Delta(\varphi,k)}{k^{2}}\\
&\qquad + \frac{m-1}{k^{2}}J^{(m-2,n)} - \frac{n-1}{k^{2}}(1-k^{2})J^{(m,n-2)}\\
\end{align*}

つまり

\begin{align*}
 J^{(m,n)} &= \inverse{(m+n-1)k^{2}}\Big\{-\sin^{m-1}(\varphi)\cos^{n-1}(\varphi)\Delta(\varphi,k)\\
&\qquad + (m-1)J^{(m-2,n)} - (n-1)(1-k^{2})J^{(m,n-2)}\Big\}\\
\end{align*}

特に $m=n{\geq}2$ の場合は、次のようになる。

\begin{align*}
 J^{(m,m)} &= \inverse{(2m-1)k^{2}}\Big\{-\sin^{m-1}(\varphi)\cos^{m-1}(\varphi)\Delta(\varphi,k)\\
&\qquad + (m-1)J^{(m-2,m)} - (m-1)(1-k^{2})J^{(m,m-2)}\Big\}\\
\end{align*}

$m=1$ (従って $n>2$) なら

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

従って

\begin{align*}
 nJ^{(1,n)} &= -\inverse{k^{2}}\left(\cos^{n-1}(\varphi)\Delta(\varphi,k)-1\right)\\
&\qquad -\frac{(n-1)(1-k^{2})}{k^{2}}J^{(1,n-2)}
\end{align*}

つまり

\[
 J^{(1,n)} = -\inverse{nk^{2}}\left\{\cos^{n-1}(\varphi)\Delta(\varphi,k)-1 + (n-1)(1-k^{2})J^{(1,n-2)}\right\}
\]

さて $I^{(m,n)}$ の表式

\begin{align*}
 I^{(m,n)} &= \inverse{({m+n+1})k^{2}}\Big\{-\sin^{m-1}(\varphi)\cos^{n-1}(\varphi)\Delta(\varphi,k)^{3}\\
&\qquad + (m-1)I^{(m-2,n)} - (n-1)(1-k^{2})I^{(m,n-2)}\Big\}\\
\end{align*}
に戻って、右辺の $I$ の階数を若干引き下げる変形を考えると (ただし $m{\geq}4$ する)

\begin{align*}
 I^{(m,n)} &= I^{(m-2,n)} - I^{(m-2,n+2)}\\
&= I^{(m-2,n)}\\
&\qquad - \inverse{(m+n+1)k^{2}}\Big\{
-\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta(\varphi,k)^{3}\\
&\qquad\qquad\qquad + (m-3)I^{(m-4,n+2)}\\
&\qquad\qquad\qquad - (n+1)(1-k^{2})I^{(m-2,n)}\Big\}\\
&= I^{(m-2,n)}\\
&\qquad - \inverse{(m+n+1)k^{2}}\Big\{
-\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta(\varphi,k)^{3}\\
&\qquad\qquad\qquad + (m-3)(I^{(m-4,n)}-I^{(m-2,n)})\\
&\qquad\qquad\qquad - (n+1)(1-k^{2})I^{(m-2,n)}\Big\}\\
&= \inverse{(m+n+1)k^{2}}\Big\{\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta(\varphi,k)^{3}\\
&\qquad\qquad\qquad + (m+n-2 + mk^{2})I^{(m-2,n)}\\
&\qquad\qquad\qquad - (m-3)I^{(m-4,n)}\Big\}\\
\end{align*}

$J^{(m,n)}$ の表式

\begin{align*}
 J^{(m,n)} &= \inverse{(m+n-1)k^{2}}\Big\{-\sin^{m-1}(\varphi)\cos^{n-1}(\varphi)\Delta(\varphi,k)\\
&\qquad + (m-1)J^{(m-2,n)} - (n-1)(1-k^{2})J^{(m,n-2)}\Big\}\\
\end{align*}
に就いても同様に計算すると (やはり $m{\geq}4$ とする)

\begin{align*}
 J^{(m,n)} &= J^{(m-2,n)} - J^{(m-2,n+2)}\\
&= J^{(m-2,n)}\\
&\qquad - \inverse{(m+n-1)k^{2}}\Big\{-\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta(\varphi,k)\\
&\qquad + (m-3)J^{(m-4,n+2)} - (n+1)(1-k^{2})J^{(m-2,n)}\Big\}\\
&= J^{(m-2,n)}\\
&\qquad - \inverse{(m+n-1)k^{2}}\Big\{-\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta(\varphi,k)\\
&\qquad + (m-3)\left(J^{(m-4,n)}-J^{(m-2,n)}\right)\\
&\qquad - (n+1)(1-k^{2})J^{(m-2,n)}\Big\}\\
&= - \inverse{(m+n-1)k^{2}}\Big\{-\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta(\varphi,k)\\
&\qquad + (m-3)J^{(m-4,n)} - \left(m+n-2+(m-2)k^{2}\right)J^{(m-2,n)}\Big\}
\end{align*}

ここで、一般の漸化式の $(m,n)$ に、幾つかの具体的な値を代入してみよう

$(m,n)=(3,1)$

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

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

$(m,n)=(1,3)$

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

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

$(m,n)=(2,3)$

\begin{align*}
 I^{(2,3)} &= \inverse{6k^{2}}\left\{-\sin(\varphi)\cos^{2}(\varphi)\Delta^{3} + I^{(0,3)}-2(1-k^{2})I^{(2,1)}\right\}\\
&= \inverse{6k^{2}}\Big\{-\sin(\varphi)\cos^{2}(\varphi)\Delta^{3}\\
&\qquad + \frac{\sin(\varphi)\Delta}{8k^{2}}(2k^{2}\cos^{2}(\varphi)+2k^{2}+1) + \frac{4k^{2}-1}{8k^{3}}{\arcsin(k\sin(\varphi))}\\
&\qquad - 2(1-k^{2})\Big(\inverse{8k^{2}}\Delta(2k^{2}\sin^{2}(\varphi)-1)\sin(\varphi) + \inverse{8k^{3}}{\arcsin(k\sin(\varphi))}\Big)\Big\}\\
&= \inverse{6k^{2}}\Big\{\frac{\sin(\varphi)\Delta}{8k^{2}}\Big(-8k^{2}\cos^{2}(\varphi)\Delta^2\\
&\qquad + ((2k^{2}\cos^{2}(\varphi)+2k^{2}+1) - 2(1-k^{2})(2k^{2}\sin^{2}(\varphi)-1)\Big)\\
&\qquad + \frac{\arcsin(k\sin(\varphi))}{8k^{3}}\Big(4k^{2}-1-2(1-k^{2})\Big)\Big\}\\
&= \frac{-8k^{4}\sin^{4}(\varphi)+2k^{2}(6k^{2}+1)\sin^{2}(\varphi)-6k^{2}+3}{48k^{4}}{\sin(\varphi)\Delta}\\
&\qquad  + \frac{(2k^{2}-1)\arcsin(k\sin(\varphi))}{16k^{5}}
\end{align*}

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

$(m,n)=(3,2)$

\begin{align*}
 I^{(3,2)} &= \inverse{6k^{2}}\left\{-\sin^{2}(\varphi)\cos(\varphi)\Delta^{3} + 2I^{(1,2)} - (1-k^{2})I^{(3,0)}\right\}\\
&= \inverse{6k^{2}}\Bigg\{-\sin^{2}(\varphi)\cos(\varphi)\Delta^{3}\\
&\qquad\qquad + 2\Bigg(-\frac{\cos(\varphi){\Delta}}{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{k\cos(\varphi)+{\Delta}}{k+1}\right)\Bigg)\\
&\qquad\qquad - (1-k^{2})\Bigg(-\frac{2k^{2}\sin^{2}(\varphi)+3k^{2}-1}{8k^{2}}\cos(\varphi){\Delta} + \frac{3k^{2}-1}{8k^{2}}\\
&\qquad\qquad\qquad\qquad\qquad + \frac{3k^{4}-2k^{2}-1}{8k^{3}}\ln\left(\frac{k\cos(\varphi)+{\Delta}}{k+1}\right)\Bigg)\Bigg\}\\
&= \inverse{6k^{2}}\Bigg\{\frac{\cos(\varphi){\Delta}}{8k^{2}}(8k^{4}\sin^{4}(\varphi) - 2k^{2}(k^{2}+1)\sin^{2}(\varphi) - 3k^{4} + 2k^{2} - 3)\\
&\qquad +\frac{3k^{4}-2k^{2}+3}{8k^{2}} + \frac{3(1-k^{2})^{2}(1+k^{2})}{8k^{3}}\ln\left(\frac{k\cos(\varphi)+{\Delta}}{k+1}\right)\Bigg\}\\
\end{align*}

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

$(m,n)=(3,3)$
$(3,3)$ に就いては、$\Delta$ を軸にして式を纏める。そのため、式を遡って

\begin{align*}
 I^{(1,3)} &= -\inverse{15k^{4}}\left\{2(1-k^{2})(1-\Delta^3)+3k^{2}\cos^{2}(\varphi)\Delta^{3} - 3k^{2}\right\}\\
&\qquad = -\inverse{15k^{4}}\left\{2(1-k^{2})(1-\Delta^3)-3(1-k^{2}-\Delta^{2})\Delta^{3} - 3k^{2}\right\}\\
&\qquad = -\inverse{15k^{4}}\left\{3\Delta^{5} - 5(1-k^{2})\Delta^{3} - 5k^{2} +2\right\}\\
 I^{(3,1)} &= \inverse{15k^{4}}\left\{-(3k^{2}\sin^{2}(\varphi)+2)\Delta^{3}+2\right\}\\
&\qquad = \inverse{15k^{4}}\left\{-(-3\Delta^{2}+5)\Delta^{3}+2\right\}\\
&\qquad = \inverse{15k^{4}}\left\{3\Delta^{5}-5\Delta^{3}+2\right\}\\
\end{align*}
であることに注意する。

\begin{align*}
 I^{(3,3)} &= \inverse{7k^{2}}\Big\{-\sin^{2}(\varphi)\cos^{2}(\varphi)\Delta^{3} + 2I^{(1,3)} - 2(1-k^{2})I^{(3,1)}\Big\}\\
&= \inverse{7k^{2}}\Bigg\{\Big(\frac{1-\Delta^{2}}{k^{2}}\Big)\Big(\frac{1-k^{2}-\Delta^{2}}{k^{2}}\Big)\Delta^{3}\\
&\qquad - \frac{2}{15k^{4}}\left\{3\Delta^{5} - 5(1-k^{2})\Delta^{3} - 5k^{2} +2\right\}\\
&\qquad - \frac{2(1-k^{2})}{15k^{4}}(3\Delta^{5}-5\Delta^{3}+2)\Bigg\}\\
&= \inverse{105k^{6}}\big\{15(1-\Delta^{2})(1-k^{2}-\Delta^{2})\Delta^{3}\\
&\qquad - 2\left\{3\Delta^{5} - 5(1-k^{2})\Delta^{3} - 5k^{2} +2\right\}\\
&\qquad - 2(1-k^{2})(3\Delta^{5}-5\Delta^{3}+2)\big\}\\
&= \inverse{105k^{6}}\Big\{15\Delta^{7} - 21(2-k^{2})\Delta^{5} + 35(1-k^{2})\Delta^{3} + 14k^{2} - 8\Big\}\\
&= \inverse{k^{6}}\Big\{\frac{\Delta^{7}}{7} - \frac{(2-k^{2})\Delta^{5}}{5} + \frac{(1-k^{2})\Delta^{3}}{3} + \frac{14k^{2}-8}{105}\Big\}
\end{align*}

この場合、漸化式によらない次の別解が存在する。まず

\begin{align*}
 &\sin^{3}(\theta)\cos^{3}(\theta)\Delta\\
&\qquad = \sin(\theta)^{2}\cos^{2}(\theta)\Big(\sin(\theta)\cos(\theta)\Delta\Big)\\
&\qquad = \Big\{\frac{1-\Delta^{2}}{k^{2}}\Big\}\Big\{\frac{-(1-k^{2}-\Delta^{2})}{k^{2}}\Big\}\Big(\sin(\theta)\cos(\theta)\Delta\Big)\\
&\qquad = -\inverse{k^{4}}\Big\{1-k^{2}-(2-k^{2})\Delta^{2}+\Delta^{4}\Big\}\Big(\sin(\theta)\cos(\theta)\Delta\Big)\\
&\qquad = -\inverse{k^{6}}\Big\{(1-k^{2})k^{2}\sin(\theta)\cos(\theta)\Delta\\
&\qquad\qquad\qquad\qquad - (2-k^{2})k^{2}\sin(\theta)\cos(\theta)\Delta^{3}\\
&\qquad\qquad\qquad\qquad + k^{2}\sin(\theta)\cos(\theta)\Delta^{5}\Big\}\\
\end{align*}
に注意する。これより
\begin{align*}
 I^{(3,3)} &= \int_{0}^{\varphi}\sin^{3}(\theta)\cos^{3}(\theta){\Delta}d\theta\\
&= \inverse{k^{6}}\Big[\frac{(1-k^{2})\Delta^{3}}{3} - \frac{(2-k^{2})\Delta^{5}}{5} + \frac{\Delta^{7}}{7}\Big]_{0}^{\varphi}\\
&= \inverse{k^{6}}\Big\{\frac{(1-k^{2})\Delta^{3}}{3} - \frac{(2-k^{2})\Delta^{5}}{5} + \frac{\Delta^{7}}{7}\\
&\qquad\qquad - \Big(\frac{1-k^{2}}{3} - \frac{2-k^{2}}{5} + \frac{1}{7}\Big)\Big\}\\
&= \inverse{k^{6}}\Big\{\frac{\Delta^{7}}{7} - \frac{(2-k^{2})\Delta^{5}}{5} + \frac{(1-k^{2})\Delta^{3}}{3} + \frac{14k^{2}-8}{105}\Big\}
\end{align*}

ここで、$\Delta$ の冪乗中、$\sin(\varphi)$ の有理式で置き換えられる部分を、実際に置き換えると

\begin{align*}
 I^{(3,3)} &= \Big(\frac{\sin^{4}(\varphi)}{7k^{2}}-\frac{(7k^{2}-4)\sin^{2}(\varphi)}{35k^{4}}-\frac{14k^{2}-8}{105k^{6}}\Big)(1-k^{2}\sin^{2}(\varphi)){\Delta}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad +\frac{14k^{2}-8}{105k^{6}}
\end{align*}

同様にして

\begin{align*}
 J^{(3,3)} &= \inverse{5k^{2}}\Big\{-\sin^{2}(\varphi)\cos^{2}(\varphi)\Delta + 2J^{(1,3)} - 2(1-k^{2})J^{(3,1)}\Big\}\\
&= \inverse{5k^{2}}\Big\{-\sin^{2}(\varphi)\cos^{2}(\varphi)\Delta\\
&\qquad\qquad + 2\Big\{-\frac{(k^{2}\cos^{2}(\varphi)-2+2k^{2})\Delta}{3k^{4}} + \frac{3k^{2}-2}{3k^{4}}\Big\}\\
&\qquad\qquad - 2(1-k^{2})\Big\{-\frac{(k^{2}\sin^{2}(\varphi)+2)\Delta}{3k^{4}} + \frac{2}{3k^{4}}\Big\}\Big\}\\
&= \inverse{15k^{6}}\Bigg\{-3k^{4}\sin^{2}(\varphi)\cos^{2}(\varphi)\Delta\\
&\qquad\qquad + 2\Big\{-(k^{2}\cos^{2}(\varphi)-2+2k^{2})\Delta + 3k^{2}-2\Big\}\\
&\qquad\qquad - 2(1-k^{2})\Big\{-(k^{2}\sin^{2}(\varphi)+2)\Delta + 2\Big\}\Bigg\}\\
&= \inverse{15k^{6}}\Big\{-\Big(3k^{4}\sin^{2}(\varphi)\cos^{2}(\varphi) + 2k^{2}\cos^{2}(\varphi)\\
&\qquad\qquad\qquad -2(1-k^{2})k^{2}\sin^{2}(\varphi) - 8(1-k^{2})\Big)\Delta +10k^{2} -8\Big\}\\
&= \inverse{15k^{6}}\Big\{-\Big(3(1-\Delta^{2})(-1+k^{2}+\Delta^{2}) + 2(-1+k^{2}+\Delta^{2})\\
&\qquad\qquad\qquad - 2(1-k^{2})(1-\Delta^{2}) - 8(1-k^{2})\Big)\Delta\\
&\qquad\qquad\qquad\qquad +10k^{2} -8\Big\}\\
&= \inverse{15k^{6}}\Big\{\Big(3\Delta^{4} - 5(2-k^{2})\Delta^{2} + 15(1-k^{2})\Big)\Delta + 10k^{2} - 8\Big\}\\
\end{align*}

別解:

\begin{align*}
  J^{(3,3)} &= \int_{0}^{\varphi}\frac{\sin^{3}(\theta)\cos^{3}(\theta)}{\Delta}d\theta\\
&= -\inverse{k^{6}}\int_{0}^{\varphi}\Big\{(1-k^{2})\Big(\frac{k^{2}\sin(\theta)\cos(\theta)}{\Delta}\Big)\\
&\qquad\qquad\qquad\qquad - (2-k^{2})k^{2}\sin(\theta)\cos(\theta)\Delta\\
&\qquad\qquad\qquad\qquad + k^{2}\sin(\theta)\cos(\theta)\Delta^{3}\Big\}d\theta\\
&= \inverse{k^{6}}\Big[(1-k^{2})\Delta - \frac{(2-k^{2})\Delta^{3}}{3} + \frac{\Delta^{5}}{5}\Big]_{0}^{\varphi}\\
&= \inverse{15k^{6}}\Big[15(1-k^{2})\Delta - 5(2-k^{2})\Delta^{3} + 3\Delta^{5}\Big]_{0}^{\varphi}\\
&= \inverse{15k^{6}}\Big\{15(1-k^{2})\Delta - 5(2-k^{2})\Delta^{3} + 3\Delta^{5}\\
&\qquad\qquad - \Big(15(1-k^{2}) - 5(2-k^{2}) + 3\Big)\Big\}\\
&= \inverse{15k^{6}}\Big\{3\Delta^{5} - 5(2-k^{2})\Delta^{3} + 15(1-k^{2})\Delta + 10k^{2} - 8\Big\}\\
%&= \inverse{15k^{6}}\Big\{\Big(3\Delta^{4} - 5(2-k^{2})\Delta^{2} + 15(1-k^{2})\Big)\Delta + 10k^{2} - 8\Big\}\\
\end{align*}

ここでも、$\Delta$ の有理式で置き換えられる部分を、実際に置き換えると

\begin{align*}
 J^{(3,3)} &= \frac{\Delta}{15k^{6}}\Big(3k^{4}\sin^{4}(\varphi)-(5k^{4}-4k^{2})\sin^{2}(\varphi)-10k^{2}+8\Big)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad +\frac{10k^{2}-8}{15k^{6}}\\
\end{align*}

$(m,n)=(4,1)$

\begin{align*}
 I^{(4,1)} &= \inverse{6k^{2}}\Big\{\sin(\varphi)\cos^{2}(\varphi)\Delta^{3} + (3 + 4k^{2})I^{(2,1)} - I^{(0,1)}\Big\}\\
&= \inverse{6k^{2}}\Big\{\sin(\varphi)\cos^{2}(\varphi)\Delta^{3}\\
&\qquad + (3 + 4k^{2})\Big(\inverse{8k^{2}}\Delta(2k^{2}\sin^{2}(\varphi)-1)\sin(\varphi) + \inverse{8k^{3}}{\arcsin(k\sin(\varphi))}\Big)\\
&\qquad - \Big(\frac{\sin(\varphi)\Delta}{2} + \inverse{2k}{\arcsin(k\sin(\varphi))}\Big)\Big\}\\
&= \inverse{6k^{2}}\Big\{\frac{\sin(\varphi)\Delta}{8k^{2}}\Big(8k^{2}\cos^{2}(\varphi)\Delta^{2} + (3+4k^{2})(2k^{2}\sin^{2}(\varphi)-1) - 4k^{2}\Big)\\
&\qquad\qquad + \frac{\arcsin(k\sin(\varphi))}{8k^{3}}\Big((3+4k^{2})-4k^{2}\Big)\Big\}\\
&= \inverse{6k^{2}}\Big\{\frac{\sin(\varphi)\Delta}{8k^{2}}\Big(
8k^{4}\sin^{4}(\varphi) - 2k^{2}\sin^{2}(\varphi) - 3\Big)\\
&\qquad\qquad + \frac{3\arcsin(k\sin(\varphi))}{8k^{3}}\Big\}
\end{align*}

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

オマケ:
$(m,n)=(4,0)$

\begin{align*}
 I^{(4,0)} &= I^{(2,0)}-I^{(2,2)}\\
&= -\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)\\
&\qquad -\inverse{30k^{4}}\Big\{\Big(2k^{2}(-3k^{2}\cos^{2}(\theta)+2k^{2}-1)\Big)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad\qquad + 4(k^{4}-k^{2}+1)E(\varphi,k) - 2(1-k^{2})(2-k^{2})F(\varphi,k)\Big\}\\
&= \inverse{30k^{4}}\Big\{-\Big(10k^{4}+2k^{2}(-3k^{2}\cos^{2}(\theta)+2k^{2}-1)\Big)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + \Big(10k^{2}(2k^{2}-1)-4(k^{4}-k^{2}+1)\Big)E(\varphi,k)\\
&\qquad + \Big(10k^{2}(1-k^{2})+2(1-k^{2})(2-k^{2})\Big)F(\varphi,k)\Big\}\\
&= \inverse{30k^{4}}\Big\{-2k^{2}(5k^{2}-3k^{2}\cos^{2}(\theta)+2k^{2}-1)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + 2\Big(5k^{2}(2k^{2}-1)-2(k^{4}-k^{2}+1)\Big)E(\varphi,k)\\
&\qquad + 2(1-k^{2})(5k^{2}+2-k^{2})F(\varphi,k)\Big\}\\
&= \inverse{30k^{4}}\Big\{-2k^{2}(3k^{2}\sin^{2}(\theta)+4k^{2}-1)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + 2(8k^{4}-3k^{2}-2)E(\varphi,k) + 4(1-k^{2})(2k^{2}+1)F(\varphi,k)\Big\}\\
&= -\frac{3k^{2}\sin^{2}(\theta)+4k^{2}-1}{15k^{2}}\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + \frac{8k^{4}-3k^{2}-2}{15k^{4}}E(\varphi,k) + \frac{2(1-k^{2})(2k^{2}+1)}{15k^{4}}F(\varphi,k)\\
\end{align*}

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

$(m,n)=(0,4)$

\begin{align*}
 I^{(0,4)} &= I^{(0,2)}-I^{(2,2)}\\
&= \Big(\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)\Big)\\
&\qquad -\inverse{30k^{4}}\Big\{\Big(2k^{2}(-3k^{2}\cos^{2}(\theta)+2k^{2}-1)\Big)\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad\qquad + 4(k^{4}-k^{2}+1)E(\varphi,k) - 2(1-k^{2})(2-k^{2})F(\varphi,k)\Big\}\\
&= \frac{\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)}{30k^{4}}\Big\{10k^{4}-2k^{2}\big(-3k^{2}\cos^{2}(\theta)+2k^{2}-1)\big)\Big\}\\
&\qquad + \frac{E(\varphi,k)}{30k^{4}}\Big\{10k^{2}(1+k^{2})-4(k^{4}-k^{2}+1)\Big\}\\
&\qquad - \frac{F(\varphi,k)}{30k^{4}}\Big\{10k^{2}(1-k^{2})-2(1-k^{2})(2-k^{2})\Big\}\\
&= \frac{\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)}{15k^{4}}\big(5k^{2}-(-3k^{2}\cos^{2}(\varphi)+2k^{2}-1)\big)\\
&\qquad + \frac{E(\varphi,k)}{15k^{4}}\big(5k^{2}(1+k^{2})-2(k^{4}-k^{2}+1))\big)\\
&\qquad - \frac{F(\varphi,k)}{15k^{4}}(1-k^{2})\big(5k^{2}-(2-k^{2})\big)\\
&= \frac{3k^{2}\cos^{2}(\varphi)+3k^{2}+1}{15k^{2}}\sin(\varphi)\cos(\varphi)\Delta(\varphi,k)\\
&\qquad + \frac{3k^{4}+7k^{2}-2}{15k^{4}}E(\varphi,k) + \frac{2(1-k^{2})(1-3k^{2})}{15k^{4}}F(\varphi,k)
\end{align*}

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

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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 | トップページ | $\int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}(\theta){\sqrt{1-k^{2}\sin^{2}(\theta)}}d\theta$ 及び $\int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)}{\sqrt{1-k^{2}\sin^{2}(\theta)}}d\theta$ の計算 »

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