1/(1+x^2)の[0, t]での定積分がarctan(t)となる話

問

original: https://www.youtube.com/watch?v=PvL2RDp1H4Y

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中心を点$P$に持つ円と直線$g$との交点$Q$を求める．

$x$座標を求める． $$\begin{array}{rcl}{x}^2+{(y+1)}^2& =& 1\\ x^2+{\left(\frac{x}{t}-1+1\right)}^2& =& 1\cdots g:y=\frac{x}{t}-1\mathrm{を}\mathrm{代}\mathrm{入}\\ x^2+{\left(\frac{x}{t}\right)}^2& =& 1\\ x^2(1+\frac{1}{t^2})& =& 1\\ x^2\left(\frac{t^2+1}{t^2}\right)& =& 1\\ x^2& =& \frac{t^2}{t^2+1}\\ x& =& \sqrt{\frac{t^2}{t^2+1}}\cdots x\ge 0\\ & =& \frac{t}{\sqrt{t^2+1}}\end{array}$$ $y$座標は$g$を用いて求める． $$\begin{array}{rcl}y& =& \frac{x}{t}-1\\ & =& \frac{\frac{t}{\sqrt{t^2+1}}}{t}-1\\ & =& \frac{1}{\sqrt{t^2+1}}-1\end{array}$$

面積$A_1$(=四角形$ABCO$)を求める．

$$\begin{array}{rcl}{A}_1& =& \overline{OA}\cdot \overline{AB}\cdots \mathrm{長}\mathrm{方}\mathrm{形}\mathrm{の}\mathrm{面}\mathrm{積}\\ & =& t\cdot \frac{1}{2}\frac{1}{1+t^2}\\ & =& \frac{t}{2(1+t^2)}\end{array}$$

面積$A_2$(=凾数$f$と線分$\overline{CB}$及び$y$軸で囲まれる領域(図の$x<0$の領域の着色は間違い))を求める．

$$\begin{array}{rcl}2y& =& \frac{1}{1+x^2}\\ 1+x^2& =& \frac{1}{2y}\\ x^2& =& \frac{1}{2y}-1\\ x^2& =& \frac{1}{2y}-1\\ & =& \frac{1-2y}{2y}\\ x& =& \sqrt{\frac{1-2y}{2y}}\end{array}$$ $$\begin{array}{rcl}{A}_2& =& {\int }_{\frac{1}{2(1+t^2)}}^{\frac{1}{2}}x\mathrm{d}y\\ & =& {\int }_{\frac{1}{2(1+t^2)}}^{\frac{1}{2}}\sqrt{\frac{1-2y}{2y}}\mathrm{d}y\\ & =& {\int }_{\sqrt{\frac{1}{(1+t^2)}}-1}^0\sqrt{\frac{1-{(u+1)}^2}{\cancel{2y}}}\cancel{\sqrt{2y}}\mathrm{d}u\\ & & \cdots u=\sqrt{2y}-1,\frac{\mathrm{d}u}{\mathrm{d}y}=\frac{1}{2}\frac{1}{\sqrt{2y}}\cdot 2=\frac{1}{\sqrt{2y}},\mathrm{d}y=\sqrt{2y}\mathrm{d}u\\ & & \cdots \frac{1}{2}\to \sqrt{2\cdot \frac{1}{2}}-1=0\\ & & \cdots \frac{1}{2(1+t^2)}\to \sqrt{2\cdot \frac{1}{2(1+t^2)}}-1=\frac{1}{\sqrt{1+t^2}}-1\\ & =& {\int }_{\frac{1}{\sqrt{1+t^2}}-1}^0\sqrt{1-{(u+1)}^2}\mathrm{d}u\end{array}$$

面積$B_1$(=三角形$PQR$)を求める．

$$\begin{array}{rcl}{B}_1& =& \frac{1}{2}\overline{QR}\cdot \overline{PR}\cdots \mathrm{三}\mathrm{角}\mathrm{形}\mathrm{の}\mathrm{面}\mathrm{積}\\ & =& \frac{1}{2}\cdot \frac{t}{\sqrt{t^2+1}}\cdot \{1+(\frac{1}{\sqrt{t^2+1}}-1)\}\\ & =& \frac{1}{2}\cdot \frac{t}{\sqrt{t^2+1}}\cdot \frac{1}{\sqrt{t^2+1}}\\ & =& \frac{t}{2(1+t^2)}\end{array}$$

面積$B_2$(=円$x^2+(y-1{)}^2=1$と線分$\overline{QR}$及び$y$軸で囲まれる領域)を求める．

$$\begin{array}{rcl}{x}^2+{(y+1)}^2& =& 1\\ x^2& =& 1-{(y+1)}^2\\ x& =& \sqrt{1-{(y+1)}^2}\end{array}$$ $$\begin{array}{rcl}{B}_2& =& {\int }_{\frac{1}{\sqrt{t^2+1}}-1}^{0}x\mathrm{d}y\\ & =& {\int }_{\frac{1}{\sqrt{t^2+1}}-1}^0\sqrt{1-{(y+1)}^2}\mathrm{d}y\end{array}$$

扇PQRの面積を求める．

$$\begin{array}{rclrc}\mathrm{扇}PQR& =& \mathrm{半}\mathrm{径}1\mathrm{の}\mathrm{面}\mathrm{積}\cdot \mathrm{比}\mathrm{率}=(1\cdot 1\cdot \pi )\cdot \frac{\theta }{2\pi }& =& \frac{\theta }{2}\end{array}$$

$\theta$と$t$の関係式を求める．

$$\begin{array}{rcl}\tan \left(\theta \right)& =& \frac{t-0}{0-(-1)}=\frac{t}{1}=t\\ {\tan }^{-1}\left(t\right)& =& \theta \end{array}$$

扇PQRの面積を介して．

$$\begin{array}{rcl}\mathrm{扇}PQR& =& \frac{\theta }{2}\\ & =& \frac{{\tan }^{-1}\left(t\right)}{2}\\ & =& B_1+B_2\\ & =& A_1+A_2\cdots B_1=A_1,B_2=A_2\\ & =& {\int }_0^t\frac{1}{2}\frac{1}{1+x^2}\mathrm{d}x=\mathrm{凾}\mathrm{数}f\mathrm{の}\mathrm{区}\mathrm{間}[0,t]\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}\mathrm{定}\mathrm{積}\mathrm{分}\end{array}$$ $$\begin{array}{rcl}{\int }_0^t\frac{1}{1+x^2}\mathrm{d}x& =& {\tan }^{-1}\left(t\right)\end{array}$$