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$\int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}(\theta){\sqrt{1-k^{2}\sin^{2}(\theta)}}d\theta$ 及び $\int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)}{\sqrt{1-k^{2}\sin^{2}(\theta)}}d\theta$ の計算

以下は、これまで本ブログの [三角関数の幾つかの2次無理式の積分] (2017年10月31日 [火])、[「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] (2017年11月30日[木])、及び [[三角関数の幾つかの2次無理式の積分]補足その2] (2017年12月31日[日]) において計算した結果を、まとめたものである。表式に若干手を入れたが、内容に変化はない。

まず、主要な記号の定義を示しておく。

\begin{align*}
 &0<k<1, 0\leq \varphi \leq \frac{\pi}{2}\\
 &\Delta_{\theta} := \sqrt{1-k^{2}\sin^{2}(\theta)} = \sqrt{1-k^{2}+k^{2}\cos^{2}(\theta)}\\
 &\Delta_{\varphi} := \sqrt{1-k^{2}\sin^{2}(\varphi)} = \sqrt{1-k^{2}+k^{2}\cos^{2}(\varphi)}\\
 &E(\varphi,k) := \int_{0}^{\varphi}\Delta_{\theta}d\theta, \quad F(\varphi,k) := \int_{0}^{\varphi}\frac{d\theta}{\Delta_{\theta}}\\
 &I^{(m,n)} := \int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}(\theta){\Delta_{\theta}}d\theta\\
 &J^{(m,n)} := \int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)}{\Delta_{\theta}}d\theta\\
\end{align*}

言うまでもなかろうが $F(\varphi,k)$ 及び $E(\varphi,k)$ は、それぞれ「第一種の楕円積分」及び「第二種の楕円積分」である。そして、次の関係式は自明だろう。

\begin{align*}
 &E(\varphi,k) = I^{(0,0)} = I^{(2,0)}+I^{(0,2)}\\
 &F(\varphi,k) = J^{(0,0)} = J^{(2,0)}+J^{(0,2)}\\
\end{align*}

漸化式の纏め。

$m{\geq}2$ の時

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

$n{\geq}2$ の時

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

$m,n \geq 2$ の時

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

$m \geq 4, n>0$ の時

\begin{align*}
 &\int_{0}^{\varphi}\sin^{m}(\theta)\cos^{n}(\theta){\Delta_{\theta}}d\theta = \inverse{(m+n+1)k^{2}}\Big\{\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta_{\varphi}^{3}\\
 &\qquad\qquad\qquad\qquad\qquad + (m+n-2 + mk^{2})I^{(m-2,n)} - (m-3)I^{(m-4,n)}\Big\}\\
 &\int_{0}^{\varphi}\frac{\sin^{m}(\theta)\cos^{n}(\theta)}{\Delta_{\theta}}d\theta = - \inverse{(m+n-1)k^{2}}\Big\{-\sin^{m-3}(\varphi)\cos^{n+1}(\varphi)\Delta_{\varphi}\\
 &\qquad\qquad\qquad\qquad\qquad + (m-3)J^{(m-4,n)} - \left(m+n-2+(m-2)k^{2}\right)J^{(m-2,n)}\Big\}\\
\end{align*}

個別の関係式

\begin{align*}
 &I^{(0,n)}: n=1,2,3\\
 &\int_{0}^{\varphi}\cos(\theta)\Delta_{\theta}d\theta = \frac{\sin(\varphi)\Delta_{\varphi}}{2} + \frac{\arcsin(k\sin(\varphi))}{2k}\\
 &\int_{0}^{\varphi}\cos^{2}(\theta)\Delta_{\theta}d\theta = \frac{\sin(\varphi)\cos(\varphi)\Delta_{\varphi}}{3} + \frac{1+k^{2}}{3k^{2}}E(\varphi,k) - \frac{1-k^{2}}{3k^{2}}F(\varphi,k)\\
 &\int_{0}^{\varphi}\cos^{3}(\theta)\Delta_{\theta}d\theta  = \frac{2k^{2}\cos^{2}(\varphi)+2k^{2}+1}{8k^{2}}\sin(\varphi)\Delta_{\varphi} + \frac{4k^{2}-1}{8k^{3}}{\arcsin(k\sin(\varphi))}\\
\end{align*}

\begin{align*}
 &I^{(1,n)}: n=0,1,2,3\\
 &\int_{0}^{\varphi}\sin(\theta)\Delta_{\theta}d\theta = \frac{1-\cos(\varphi)\Delta_{\varphi}}{2} - \frac{1-k^{2}}{2k}\ln\left\{\frac{k\cos(\varphi)+\Delta_{\varphi}}{k+1}\right\}\\
 &\int_{0}^{\varphi}\sin(\theta)\cos(\theta)\Delta_{\theta}d\theta = \frac{1-\Delta_{\varphi}^{3}}{3k^{2}}\\
 &\int_{0}^{\varphi}\sin(\theta)\cos^{2}(\theta)\Delta_{\theta}d\theta = -\frac{2k^{2}\cos^{2}(\varphi)+1-k^{2}}{8k^{2}}\cos(\varphi)\Delta_{\varphi} + \frac{1+k^{2}}{8k^{2}}\\
 &\qquad\qquad\qquad\qquad\qquad\qquad\qquad - \frac{(1-k^{2})^{2}}{8k^{3}}\ln\left(\frac{\Delta_{\varphi}-k\cos(\varphi)}{1-k}\right)\\
 &\int_{0}^{\varphi}\sin(\theta)\cos^{3}(\theta)\Delta_{\theta}d\theta = -\inverse{15k^{4}}\left\{3\Delta_{\varphi}^{5} - 5(1-k^{2})\Delta_{\varphi}^{3} - 5k^{2} +2\right\}\\
 &\qquad\qquad = -\frac{3k^{4}\sin^{4}(\varphi)-k^{2}(5k^{2}+1)\sin^{2}(\varphi)+5k^{2}-2}{15k^{4}}\Delta_{\varphi} + \frac{5k^{2}-2}{15k^{4}}\\
\end{align*}

\begin{align*}
 &I^{(2,n)}: n=0,1,2,3\\
 &\int_{0}^{\varphi}\sin^{2}(\theta)\Delta_{\theta}d\theta = -\frac{\Delta_{\varphi}\sin(\varphi)\cos(\varphi)}{3} + \frac{2k^{2}-1}{3k^{2}}E(\varphi,k) + \frac{1-k^{2}}{3k^{2}}F(\varphi,k)\\
 &\int_{0}^{\varphi}\sin^{2}(\theta)\cos(\theta)\Delta_{\theta}d\theta = \frac{2k^{2}\sin^{2}(\varphi)-1}{8k^{2}}\sin(\varphi)\Delta_{\varphi} + \frac{\arcsin(k\sin(\varphi))}{8k^{3}}\\
 &\int_{0}^{\varphi}\sin^{2}(\theta)\cos^{2}(\theta)\Delta_{\theta}d\theta = \inverse{15k^{2}}\Big\{(-3k^{3}\cos^{2}(\theta)+2k^{2}-1)\sin(\varphi)\cos(\varphi)\Delta_{\varphi}\\
 &\qquad\qquad\qquad\qquad\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\}\\
 &\int_{0}^{\varphi}\sin^{2}(\theta)\cos^{3}(\theta)\Delta_{\theta}d\theta = \frac{-8k^{4}\sin^{4}(\varphi)+2k^{2}(6k^{2}+1)\sin^{2}(\varphi)-6k^{2}+3}{48k^{4}}{\sin(\varphi)\Delta_{\varphi}}\\
 &\qquad\qquad\qquad\qquad\qquad\qquad\qquad  + \frac{(2k^{2}-1)\arcsin(k\sin(\varphi))}{16k^{5}}
\end{align*}

\begin{align*}
 &I^{(3,n)}: n=0,1,2,3\\
 &\int_{0}^{\varphi}\sin^{3}(\theta)\Delta_{\theta}d\theta = -\frac{2k^{2}\sin^{2}(\varphi)+3k^{2}-1}{8k^{2}}\cos(\varphi)\Delta_{\varphi} + \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_{\varphi}}{k+1}\right)\\
 &\int_{0}^{\varphi}\sin^{3}(\theta)\cos(\theta)\Delta_{\theta}d\theta = \inverse{15k^{4}}\left\{3\Delta_{\varphi}^{5}-5\Delta_{\varphi}^{3}+2\right\}\\
 &\qquad\qquad\qquad\qquad\qquad = \frac{(3k^{4}\sin^{4}(\varphi)-k^{2}\sin^{2}(\varphi)-2)\Delta_{\varphi}}{15k^{4}}+\frac{2}{15k^{4}}\\
%& = \inverse{15k^{4}}\left\{3\Delta_{\varphi}^{5}-5\Delta_{\varphi}^{3}+2\right\}\\
 &\int_{0}^{\varphi}\sin^{3}(\theta)\cos^{2}(\theta)\Delta_{\theta}d\theta =\\
 &\qquad\qquad \inverse{6k^{2}}\Bigg\{\frac{8k^{4}\sin^{4}(\varphi) - 2k^{2}(k^{2}+1)\sin^{2}(\varphi) - 3k^{4} + 2k^{2} - 3}{8k^{2}}\cos(\varphi){\Delta_{\varphi}}\\
 &\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_{\varphi}}}{k+1}\right)\Bigg\}\\
 &\int_{0}^{\varphi}\sin^{3}(\theta)\cos^{3}(\theta)\Delta_{\theta}d\theta = \inverse{k^{6}}\Big\{\frac{\Delta_{\varphi}^{7}}{7} - \frac{(2-k^{2})\Delta_{\varphi}^{5}}{5} + \frac{(1-k^{2})\Delta_{\varphi}^{3}}{3} + \frac{14k^{2}-8}{105}\Big\}\\
 &\qquad\qquad= \Big(\frac{k^{4}\sin^{4}(\varphi)}{7}-\frac{(7k^{4}-4k^{2})\sin^{2}(\varphi)}{35}-\frac{14k^{2}-8}{105}\Big)(1-k^{2}\sin^{2}(\varphi)){\Delta_{\varphi}}\\
 &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad +\frac{14k^{2}-8}{105}
\end{align*}

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

\begin{align*}
 &J^{(0,n)}: n=1,2,3\\
 &\int_{0}^{\varphi}\frac{\cos(\theta)}{\Delta_{\theta}}d\theta = \inverse{k}\arcsin(k\sin(\varphi))\\
 &\int_{0}^{\varphi}\frac{\cos^{2}(\theta)}{\Delta_{\theta}}d\theta = \frac{E(\varphi,k)}{k^{2}} - \frac{1-k^{2}}{k^{2}}F(\varphi,k)\\
 &\int_{0}^{\varphi}\frac{\cos^{3}(\theta)}{\Delta_{\theta}}d\theta = \frac{\sin(\varphi)\Delta_{\varphi}}{2k^{2}} + \frac{2k^{2}-1}{2k^{3}}\arcsin(k\sin(\varphi))
\end{align*}

\begin{align*}
 &J^{(1,n)}: n=0,1,2,3\\
 &\int_{0}^{\varphi}\frac{\sin(\theta)}{\Delta_{\theta}}d\theta = \inverse{k}\ln\left(\frac{\Delta_{\varphi}-k\cos(\varphi)}{1-k}\right)\\
 &\int_{0}^{\varphi}\frac{\sin(\theta)\cos(\theta)}{\Delta_{\theta}}d\theta = \frac{1-\Delta_{\varphi}}{k^{2}}\\
 &\int_{0}^{\varphi}\frac{\sin(\theta)\cos^{2}(\theta)}{\Delta_{\theta}}d\theta = -\frac{\cos(\varphi)\Delta_{\varphi}}{2k^{2}} + \inverse{2k^{2}} - 
\frac{1-k^{2}}{2k^{3}}\ln\left(\frac{\Delta_{\varphi}-k\cos(\varphi)}{1-k}\right)\\
 &\int_{0}^{\varphi}\frac{\sin(\theta)\cos^{3}(\theta)}{\Delta_{\theta}}d\theta = -\frac{(k^{2}\cos^{2}(\varphi)-2+2k^{2})\Delta_{\varphi}}{3k^{4}} + \frac{3k^{2}-2}{3k^{4}}
\end{align*}

\begin{align*}
 &J^{(2,n)}: n=0,1,2,3\\
 &\int_{0}^{\varphi}\frac{\sin^{2}(\theta)}{\Delta_{\theta}}d\theta = \frac{F(\varphi,k)-E(\varphi,k)}{k^{2}}\\
 &\int_{0}^{\varphi}\frac{\sin^{2}(\theta)\cos(\theta)}{\Delta_{\theta}}d\theta = -\frac{\sin(\varphi)\Delta_{\varphi}}{2k^{2}} + \frac{\arcsin(k\sin(\varphi))}{2k^{3}}\\
 &\int_{0}^{\varphi}\frac{\sin^{2}(\theta)\cos^{2}(\theta)}{\Delta_{\theta}}d\theta = -\frac{\sin(\varphi)\cos(\varphi)\Delta_{\varphi}}{3k^{2}} + \frac{2-k^{2}}{3k^{4}}E(\varphi,k) + \frac{2k^{2}-2}{3k^{4}}F(\varphi,k)\\
 &\int_{0}^{\varphi}\frac{\sin^{2}(\theta)\cos^{3}(\theta)}{\Delta_{\theta}}d\theta = \inverse{8k^{4}}\Big\{(2k^{2}\sin^{2}(\varphi)-4k^{2}+3){\sin(\varphi)\Delta_{\varphi}} + \frac{4k^{2}-3}{k}\arcsin(k\sin(\varphi))\Big\}\\
\end{align*}

\begin{align*}
 &J^{(3,n)}: n=0,1,2,3\\
 &\int_{0}^{\varphi}\frac{\sin^{3}(\theta)}{\Delta_{\theta}}d\theta = \frac{\cos(\varphi)\Delta-1}{2k^{2}} - \frac{1+k^{2}}{2k^{3}}\ln\left(\frac{k\cos(\varphi)+\Delta_{\varphi}}{k+1}\right)\\
 &\int_{0}^{\varphi}\frac{\sin^{3}(\theta)\cos(\theta)}{\Delta_{\theta}}d\theta = -\frac{(k^{2}\sin^{2}(\varphi)+2)\Delta_{\varphi}}{3k^{4}} + \frac{2}{3k^{4}}\\
 &\int_{0}^{\varphi}\frac{\sin^{3}(\theta)\cos^{2}(\theta)}{\Delta_{\theta}}d\theta = \inverse{4k^{2}}\Bigg\{\frac{\cos(\varphi)\Delta_{\varphi}}{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_{\varphi}}{k+1}\right)\Bigg\}\\
 &\int_{0}^{\varphi}\frac{\sin^{3}(\theta)\cos^{3}(\theta)}{\Delta_{\theta}}d\theta = \inverse{15k^{6}}\Big\{\Big(3\Delta_{\varphi}^{4} - 5(2-k^{2})\Delta_{\varphi}^{2} + 15(1-k^{2})\Big)\Delta + 10k^{2} - 8\Big\}\\
 &\qquad\qquad\qquad = \frac{\Delta_{\varphi}}{15k^{6}}\Big(3k^{4}\sin^{4}(\varphi)-(5k^{4}-4k^{2})\sin^{2}(\varphi)-10k^{2}+8\Big)+\frac{10k^{2}-8}{15k^{6}}
\end{align*}

\begin{align*}
 &J^{(4,1)}\\
 &\int_{0}^{\varphi}\frac{\sin^{4}(\theta)\cos(\theta)}{\Delta_{\theta}}d\theta = \inverse{4k^{2}}\Big\{-\frac{\sin(\varphi)\Delta_{\varphi}}{2k^{2}}\Big(2k^{2}\sin^{2}(\varphi)+3\Big) + \frac{3\arcsin(k\sin(\varphi))}{2k^{3}}\Big\}
\end{align*}

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