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．
$$\begin{eqnarray}
x^2+\left(y+1\right)^2&=&1
\\x^2+\left(\color{red}{\frac{x}{t}-1}\color{black}{}+1\right)^2&=&1\;\cdots\;g:y=\frac{x}{t}-1を代入
\\x^2+\left(\frac{x}{t}\right)^2&=&1
\\x^2\left(1+\frac{1}{t^2}\right)&=&1
\\x^2\left(\frac{t^2+1}{t^2}\right)&=&1
\\x^2&=&\frac{t^2}{t^2+1}
\\x&=&\sqrt{\frac{t^2}{t^2+1}}\;\cdots\;x\ge0
\\&=&\frac{t}{\sqrt{t^2+1}}
\end{eqnarray}$$
\(y\)座標は\(g\)を用いて求める．
$$\begin{eqnarray}
y&=&\frac{x}{t}-1
\\&=&\frac{\frac{t}{\sqrt{t^2+1}}}{t}-1
\\&=&\frac{1}{\sqrt{t^2+1}}-1
\end{eqnarray}$$
面積\(A_1\)(=四角形\(ABCO\))を求める．
$$\begin{eqnarray}
A_1&=&\overline{OA}\cdot\overline{AB}\;\cdots\;長方形の面積
\\&=&t\cdot\frac{1}{2}\frac{1}{1+t^2}
\\&=&\frac{t}{2\left(1+t^2\right)}
\end{eqnarray}$$
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\(x\lt0\)の領域の着色は間違い))を求める．
$$\begin{eqnarray}
2y&=&\frac{1}{1+x^2}
\\1+x^2&=&\frac{1}{2y}
\\x^2&=&\frac{1}{2y}-1
\\x^2&=&\frac{1}{2y}-1
\\&=&\frac{1-2y}{2y}
\\x&=&\sqrt{\frac{1-2y}{2y}}
\end{eqnarray}$$
$$\begin{eqnarray}
A_2&=&\int_{\frac{1}{2\left(1+t^2\right)}}^{\frac{1}{2}} x\mathrm{d}y
\\&=&\int_{\frac{1}{2\left(1+t^2\right)}}^{\frac{1}{2}} \sqrt{\frac{1-2y}{2y}} \mathrm{d}y
\\&=&\int_{\sqrt{\frac{1}{\left(1+t^2\right)}}-1}^{0}
\sqrt{\frac{1-\left(u+1\right)^2}{\cancel{2y}}}\cancel{\sqrt{2y}}\mathrm{d}u
\\&&\;\cdots\;u=\sqrt{2y}-1,\;\frac{\mathrm{d}u}{\mathrm{d}y}
=\frac{1}{2}\frac{1}{\sqrt{2y}}\cdot2=\frac{1}{\sqrt{2y}},\;\mathrm{d}y=\sqrt{2y}\mathrm{d}u
\\&&\;\cdots\;\frac{1}{2}\rightarrow\sqrt{2\cdot \frac{1}{2}}-1=0
\\&&\;\cdots\;\frac{1}{2\left(1+t^2\right)}\rightarrow\sqrt{2\cdot \frac{1}{2\left(1+t^2\right)}}-1
=\frac{1}{\sqrt{1+t^2}}-1
\\&=&\int_{\frac{1}{\sqrt{1+t^2}}-1}^{0}
\sqrt{1-\left(u+1\right)^2}\mathrm{d}u
\end{eqnarray}$$
面積\(B_1\)(=三角形\(PQR\))を求める．
$$\begin{eqnarray}
B_1&=&\frac{1}{2}\overline{QR}\cdot\overline{PR}\;\cdots\;三角形の面積
\\&=&\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+1}}\cdot\left\{1+\left(\frac{1}{\sqrt{t^2+1}}-1\right)\right\}
\\&=&\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+1}}\cdot\frac{1}{\sqrt{t^2+1}}
\\&=&\frac{t}{2\left(1+t^2\right)}
\end{eqnarray}$$
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．
$$\begin{eqnarray}
x^2+\left(y+1\right)^2&=&1
\\x^2&=&1-\left(y+1\right)^2
\\x&=&\sqrt{1-\left(y+1\right)^2}
\end{eqnarray}$$
$$\begin{eqnarray}
B_2&=&\int_{\frac{1}{\sqrt{t^2+1}}-1}^{0} x \mathrm{d}y
\\&=&\int_{\frac{1}{\sqrt{t^2+1}}-1}^{0} \sqrt{1-\left(y+1\right)^2} \mathrm{d}y
\end{eqnarray}$$
扇PQRの面積を求める．
$$\begin{eqnarray}
扇PQR&=&半径1の面積\cdot比率=\left(1\cdot1\cdot\pi\right)\cdot\frac{\theta}{2\pi}
&=&\frac{\theta}{2}
\end{eqnarray}$$
\(\theta\)と\(t\)の関係式を求める．
$$\begin{eqnarray}
\tan{\left(\theta\right)}&=&\frac{t-0}{0-(-1)}=\frac{t}{1}=t
\\\tan^{-1}{\left(t\right)}&=&\theta
\end{eqnarray}$$
扇PQRの面積を介して．
$$\begin{eqnarray}
扇PQR&=&\frac{\theta}{2}
\\&=&\frac{\tan^{-1}\left(t\right)}{2}
\\&=&B_1+B_2
\\&=&A_1+A_2\;\cdots\;B_1=A_1,\;B_2=A_2
\\&=&\int_0^t \frac{1}{2}\frac{1}{1+x^2}\mathrm{d}x
=凾数fの区間[0,t]における定積分
\end{eqnarray}$$
$$\begin{eqnarray}
\int_0^t \frac{1}{1+x^2}\mathrm{d}x&=&\tan^{-1}\left(t\right)
\end{eqnarray}$$
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