1/1+x^2 型の積分　tan に置きかえる

$$\displaystyle\int_0^{\frac{\pi}{4}} \frac{1}{1+x^2}\space dx$$

$\tan$ に置きかえ１

$$\displaystyle\int_0^{\frac{\pi}{4}} \frac{1}{1+x^2}\space dx$$
$x=\tan t$ とおくと
$\displaystyle dx=\frac{1}{\cos^2 t}\space dt$
$x\space 0\rightarrow \frac{\pi}{4}$
$t\space 0\rightarrow 1$
$\displaystyle =\int_0^1 \frac{1}{1+\tan^2 t}\cdot \frac{1}{\cos^2 t}\space dt$
$\displaystyle =\int_0^1 \frac{1}{\frac{1}{\cos^2 t}}\cdot \frac{1}{\cos^2 t}\space dt$
$\displaystyle =\int_0^1 \cos^2 t\cdot \frac{1}{\cos^2 t}\space dt$
$\displaystyle =\int_0^1 \space dt$

$\displaystyle =\left[t\right]_0^1$
$=1$（答え）

$\tan$ に置きかえ2

$$\displaystyle\int_0^1 \frac{1}{3+x^2}\space dx$$

$x=\sqrt{3}\tan t$ とおくと
$\displaystyle dx=\frac{\sqrt{3}}{\cos^2 t}\space dt$
$x\space 0\rightarrow 1$
$t\space 0\rightarrow \frac{\pi}{6}$
$\displaystyle =\int_0^{\frac{\pi}{6}} \frac{1}{3+3\tan^2 t}\cdot\frac{\sqrt{3}}{\cos^2 t}\space dt$
$\displaystyle =\int_0^{\frac{\pi}{6}} \frac{1}{3(1+\tan^2 t)}\cdot\frac{\sqrt{3}}{\cos^2 t}\space dt$
$\displaystyle =\int_0^{\frac{\pi}{6}} \frac{1}{\frac{3}{\cos^2 t}}\cdot\frac{\sqrt{3}}{\cos^2 t}\space dt$
$\displaystyle =\int_0^{\frac{\pi}{6}} \frac{\cos^2 t}{3}\cdot\frac{\sqrt{3}}{\cos^2 t}\space dt$
$\displaystyle =\int_0^{\frac{\pi}{6}} \frac{\sqrt{3}}{3}\space dt$
$\displaystyle =\left[\frac{\sqrt{3}}{3}t\right]_0^{\frac{\pi}{6}}$
$\displaystyle =\frac{\sqrt{3}}{3}\cdot\frac{\pi}{6}$
$\displaystyle =\frac{\sqrt{3}}{18}\pi$（答え）

$\tan$ に置きかえ3

$$\displaystyle \int_1^4 \frac{dx}{x^2-2x+4}$$

分母を平方完成すると
$x^2-2x+4=(x^2-2x)+4$
$=(x-1)^2-1+4=(x-1)^2+3$
となるので
$\displaystyle =\int_1^4 \frac{dx}{(x-1)^2+3}$

$x-1=\sqrt{3}\tan t$ とおくと
$\displaystyle dx=\frac{\sqrt{3}}{\cos^2 t}\space dt$
$x\space 1\rightarrow 4$
$t\space 0\rightarrow \frac{\pi}{3}$
$\displaystyle \int_0^\frac{\pi}{3} \frac{1}{3+3\tan^2 t}\cdot\frac{\sqrt{3}}{\cos^2 t}\space dt$

$\displaystyle =\frac{\sqrt{3}}{9}\pi$（答え）