逆三角関数と逆双曲線関数の冪乗積分漸化式

by nomura · 2020年5月15日

逆三角関数の冪乗積分漸化式

(1)

\[ \int\sin^{\bullet,n}xdx=x\sin^{\bullet,n}x+n\sqrt{1-x^{2}}\sin^{\bullet,n-1}x-n(n-1)\int\sin^{\bullet,n-2}xdx \]

(2)

\[ \int\sin^{\bullet,n}xdx=\frac{1}{n+1}\sqrt{1-x^{2}}\sin^{\bullet,n+1}x+\frac{1}{(n+1)(n+2)}x\sin^{\bullet,n+2}x-\frac{1}{(n+1)(n+2)}\int\sin^{\bullet,n+2}xdx \]

(3)

\[ \int\cos^{\bullet,n}xdx=x\cos^{\bullet,n}x-n\sqrt{1-x^{2}}\cos^{\bullet,n-1}x-n(n-1)\int\cos^{\bullet,n-2}xdx \]

(4)

\[ \int\cos^{\bullet,n}xdx=-\frac{1}{n+1}\sqrt{1-x^{2}}\cos^{\bullet,n+1}x+\frac{1}{(n+1)(n+2)}x\cos^{\bullet,n+2}x-\frac{1}{(n+1)(n+2)}\int\cos^{\bullet,n+2}xdx \]

(1)

\begin{align*} \int\sin^{\bullet,n}xdx & =x\sin^{\bullet,n}x-n\int\frac{x}{\sqrt{1-x^{2}}}\sin^{\bullet,n-1}xdx\\ & =x\sin^{\bullet,n}x+n\sqrt{1-x^{2}}\sin^{\bullet,n-1}x-n(n-1)\int\sin^{\bullet,n-2}xdx \end{align*}

(2)

\begin{align*} \int\sin^{\bullet,n}xdx & =\int\sqrt{1-x^{2}}\sin^{\bullet,n}x(\sin^{\bullet}x)'dx\\ & =\frac{1}{n+1}\sqrt{1-x^{2}}\sin^{\bullet,n+1}x+\frac{1}{n+1}\int x\sin^{\bullet,n+1}x\frac{1}{\sqrt{1-x^{2}}}dx\\ & =\frac{1}{n+1}\sqrt{1-x^{2}}\sin^{\bullet,n+1}x+\frac{1}{(n+1)(n+2)}x\sin^{\bullet,n+2}x-\frac{1}{(n+1)(n+2)}\int\sin^{\bullet,n+2}xdx \end{align*}

(3)

\begin{align*} \int\cos^{\bullet,n}xdx & =x\cos^{\bullet,n}x+n\int\frac{x}{\sqrt{1-x^{2}}}\cos^{\bullet,n-1}xdx\\ & =x\cos^{\bullet,n}x-n\sqrt{1-x^{2}}\cos^{\bullet,n-1}x-n(n-1)\int\cos^{\bullet,n-2}xdx \end{align*}

(4)

\begin{align*} \int\cos^{\bullet,n}xdx & =-\int\sqrt{1-x^{2}}\cos^{\bullet,n}x(\cos^{\bullet}x)'dx\\ & =-\frac{1}{n+1}\sqrt{1-x^{2}}\cos^{\bullet,n+1}x+\frac{1}{n+1}\int x\cos^{\bullet,n+1}x\frac{-1}{\sqrt{1-x^{2}}}dx\\ & =-\frac{1}{n+1}\sqrt{1-x^{2}}\cos^{\bullet,n+1}x+\frac{1}{(n+1)(n+2)}x\cos^{\bullet,n+2}x-\frac{1}{(n+1)(n+2)}\int\cos^{\bullet,n+2}xdx \end{align*}

逆双曲線関数の冪乗積分漸化式

(1)

\[ \int\sinh^{\bullet,n}xdx=x\sinh^{\bullet,n}x-n\sqrt{x^{2}+1}\sinh^{\bullet,n-1}x+n(n-1)\int\sinh^{\bullet,n-2}xdx \]

(2)

\[ \int\sinh^{\bullet,n}xdx=\frac{1}{n+1}\sqrt{x^{2}+1}\sinh^{\bullet,n+1}x-\frac{1}{(n+1)(n+2)}x\sinh^{\bullet,n+2}x+\frac{1}{(n+1)(n+2)}\int\sinh^{\bullet,n+2}xdx \]

(3)

\[ \int\cosh^{\bullet,n}xdx==x\cosh^{\bullet,n}x-n\sqrt{x^{2}-1}\cosh^{\bullet,n-1}x+n(n-1)\int\cosh^{\bullet,n-2}xdx \]

(4)

\[ \int\cosh^{\bullet,n}xdx=\frac{1}{n+1}\sqrt{x^{2}-1}\cosh^{\bullet,n+1}x-\frac{1}{(n+1)(n+2)}x\cosh^{\bullet,n+2}x+\frac{1}{(n+1)(n+2)}\int\cosh^{\bullet,n+2}xdx \]

(1)

\begin{align*} \int\sinh^{\bullet,n}xdx & =x\sinh^{\bullet,n}x-n\int\frac{x}{\sqrt{x^{2}+1}}\sinh^{\bullet,n-1}xdx\\ & =x\sinh^{\bullet,n}x-n\sqrt{x^{2}+1}\sinh^{\bullet,n-1}x+n(n-1)\int\sinh^{\bullet,n-2}xdx \end{align*}

(2)

\begin{align*} \int\sinh^{\bullet,n}xdx & =\int\sqrt{x^{2}+1}\sinh^{\bullet,n}x(\sinh^{\bullet}x)'dx\\ & =\frac{1}{n+1}\sqrt{x^{2}+1}\sinh^{\bullet,n+1}x-\frac{1}{n+1}\int x\sinh^{\bullet,n+1}x\frac{1}{\sqrt{x^{2}+1}}dx\\ & =\frac{1}{n+1}\sqrt{x^{2}+1}\sinh^{\bullet,n+1}x-\frac{1}{(n+1)(n+2)}x\sinh^{\bullet,n+2}x+\frac{1}{(n+1)(n+2)}\int\sinh^{\bullet,n+2}xdx \end{align*}

(3)

\begin{align*} \int\cosh^{\bullet,n}xdx & =x\cosh^{\bullet,n}x-n\int\frac{x}{\sqrt{x^{2}-1}}\cosh^{\bullet,n-1}xdx\\ & =x\cosh^{\bullet,n}x-n\sqrt{x^{2}-1}\cosh^{\bullet,n-1}x+n(n-1)\int\cosh^{\bullet,n-2}xdx \end{align*}

(4)

\begin{align*} \int\cosh^{\bullet,n}xdx & =\int\sqrt{x^{2}-1}\cosh^{\bullet,n}x(\cosh^{\bullet}x)'dx\\ & =\frac{1}{n+1}\sqrt{x^{2}-1}\cosh^{\bullet,n+1}x-\frac{1}{n+1}\int x\cosh^{\bullet,n+1}x\frac{1}{\sqrt{x^{2}-1}}dx\\ & =\frac{1}{n+1}\sqrt{x^{2}-1}\cosh^{\bullet,n+1}x-\frac{1}{(n+1)(n+2)}x\cosh^{\bullet,n+2}x+\frac{1}{(n+1)(n+2)}\int\cosh^{\bullet,n+2}xdx \end{align*}