逆三角関数と逆双曲線関数の冪乗積分漸化式
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*}
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三角関数と双曲線関数の実部と虚部
\[
\sin z=\sin\left(\Re\left(z\right)\right)\cosh\left(\Im\left(z\right)\right)+i\cos\left(\Re\left(z\right)\right)\sinh\left(\Im\left(z\right)\right)
\]
三角関数と双曲線関数の実部と虚部
\[
\tan z=\frac{\sin\left(2\Re z\right)+i\sinh\left(2\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}
\]
逆三角関数と逆双曲線関数の積分
\[
\int\sin^{\bullet}xdx=x\sin^{\bullet}x+\sqrt{1-x^{2}}
\]
3角関数・双曲線関数の総和
\[
\sum_{k=m_{1}}^{m_{2}}\sin\left(ak+b\right)=\sin^{-1}\left(\frac{a}{2}\right)\sin\left(\left(m_{1}+m_{2}\right)\frac{a}{2}+b\right)\sin\left(\left(1+m_{2}-m_{1}\right)\frac{a}{2}\right)
\]