積分２ - bate's blog

$\int \frac{1}{1+x^2} \, dx = \tan^{-1}x+C$
Cは積分定数
$x=\tan \, t, \hspace{5pt} dx=\frac{1}{\cos^{2} t} \hspace{2pt} dt, \hspace{5pt} t=\tan^{-1}x$
$\int \frac{1}{1+x^2} \, dx = \int \frac{1}{1+\tan^{2}t} (\frac{1}{\cos^{2}t}\,dt)$
$\hspace{45pt}(1+\tan^{2}t=\frac{1}{\cos^{2}t})$
$\hspace{45pt} = \int \frac{\cos^{2}t}{\cos^{2}t} \, dt$
$\hspace{45pt} = \int 1 \, dt$
$\hspace{45pt} = t + C$
$\hspace{45pt} = \tan^{-1}x+C$