積分漸化式

積分（３）からの続き

三角関数の積分の漸化式が教科書に載っていたことをいまごろ思い出した。これならば双曲線関数にも簡単にあてはめられるはずだ。

$I_n = \int \sin^n x \ dx =  (-\cos x) \sin^{n-1} x - \int  (-\cos x) (n-1) \sin^{n-2} x \cos x\ dx$

$\quad = (-\cos x) \sin^{n-1} x + (n-1) \int (1-\sin^2 x) \sin^{n-2} x \ dx$

$\quad = (-\cos x) \sin^{n-1} x + (n-1) (I_{n-2} - I_n )$

$\therefore I_n = -\frac{1}{n} \cos x \sin^{n-1} x + \frac{n-1}{n} I_{n-2} \quad (n \ge 2) $

$I_n = \int \cos^n x \ dx =  (\sin x) \cos^{n-1} x - \int  (\sin x) (n-1) \cos^{n-2} x (-\sin x) \ dx$

$\quad = (\sin x) \cos^{n-1} x + (n-1) \int (1-\cos^2 x) \cos^{n-2} x \ dx$

$\quad = (\sin x) \cos^{n-1} x + (n-1) (I_{n-2} - I_n )$

$\therefore I_n = \frac{1}{n} \sin x \cos^{n-1} x + \frac{n-1}{n} I_{n-2} \quad (n \ge 2) $

$I_n = \int \tan^n x \ dx =   \int  (\frac{1}{\cos^2 x} -1) \tan^{n-2} x \ dx$

$\quad = \int (\tan x)' \tan^{n-2} x\ dx -  \int \tan^{n-2} x \ dx$

$\therefore I_n = \frac{1}{n-1}\tan^{n-1} x - I_{n-2}  \quad (n \ge 2) $

$I_n = \int \sinh^n x \ dx =  (\cosh x) \sinh^{n-1} x - \int  (\cosh x) (n-1) \sinh^{n-2} x \cosh x\ dx$

$\quad = \cosh x\ \sinh^{n-1} x - (n-1) \int (1+\sinh^2 x) \sinh^{n-2} x \ dx$

$\quad = \cosh x\ \sinh^{n-1} x - (n-1) (I_{n-2} + I_n )$

$\therefore I_n = \frac{1}{n} \cosh x\ \sinh^{n-1} x - \frac{n-1}{n} I_{n-2} \quad (n \ge 2) $

$I_n = \int \cosh^n x \ dx =  (\sinh x) \cosh^{n-1} x - \int  (\sinh x) (n-1) \cosh^{n-2} x \sinh x \ dx$

$\quad = \sinh x \cosh^{n-1} x - (n-1) \int (\cosh^2 x -1) \cosh^{n-2} x \ dx$

$\quad = \sinh x \cosh^{n-1} x + (n-1) (I_{n-2} - I_n )$

$\therefore I_n = \frac{1}{n} \sinh x \cosh^{n-1} x + \frac{n-1}{n} I_{n-2} \quad (n \ge 2) $

$I_n = \int \tanh^n x \ dx =   \int  (1-\frac{1}{\cosh^2 x} ) \tanh^{n-2} x \ dx$

$\quad = -\int (\tanh x)' \tanh^{n-2} x\ dx + \int \tanh^{n-2} x \ dx$

$\therefore I_n = -\frac{1}{n-1}\tanh^{n-1} x + I_{n-2}  \quad (n \ge 2) $