x=tanθ とおく定積分（1/(1+x²)）（A）

(1) $x=\tan\theta$ とおくと $dx=\dfrac{d\theta}{\cos^2\theta}$，$1+x^2=\dfrac{1}{\cos^2\theta}$．$x:0\to 1$ のとき $\theta:0\to\dfrac{\pi}{4}$ \[ \int_0^1\frac{dx}{1+x^2}=\int_0^{\frac{\pi}{4}}\cos^2\theta\cdot\frac{d\theta}{\cos^2\theta}=\int_0^{\frac{\pi}{4}}d\theta=\frac{\pi}{4} \] (2) $x=\sqrt{3}\tan\theta$ とおくと $dx=\dfrac{\sqrt{3}}{\cos^2\theta}d\theta$，$x^2+3=\dfrac{3}{\cos^2\theta}$．$x:0\to\sqrt{3}$ のとき $\theta:0\to\dfrac{\pi}{4}$ \[ \int_0^{\sqrt{3}}\frac{dx}{x^2+3}=\int_0^{\frac{\pi}{4}}\frac{\cos^2\theta}{3}\cdot\frac{\sqrt{3}}{\cos^2\theta}\,d\theta=\frac{\sqrt{3}}{3}\cdot\frac{\pi}{4}=\frac{\sqrt{3}}{12}\pi \] (3) $x=\tan\theta$ とおくと $dx=\dfrac{d\theta}{\cos^2\theta}$，$(1+x^2)^2=\dfrac{1}{\cos^4\theta}$．$x:0\to 1$ のとき $\theta:0\to\dfrac{\pi}{4}$ \[ \int_0^1\frac{dx}{(1+x^2)^2}=\int_0^{\frac{\pi}{4}}\cos^4\theta\cdot\frac{d\theta}{\cos^2\theta}=\int_0^{\frac{\pi}{4}}\frac{1+\cos 2\theta}{2}\,d\theta=\left[\frac{\theta}{2}+\frac{\sin 2\theta}{4}\right]_0^{\frac{\pi}{4}}=\frac{\pi}{8}+\frac{1}{4} \]