sin(mx)sin(nx), cos(mx)cos(nx), sin(mx)cos(nx) の0から2πまでの積分

準備

準備の準備

$$\begin{array}{rcl}\cos (\alpha +\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)\\ \cos (\alpha -\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)\\ \sin (\alpha +\beta )& =& \sin \left(\alpha \right)\cos \left(\beta \right)+\cos \left(\alpha \right)\sin \left(\beta \right)\\ \sin (\alpha -\beta )& =& \sin \left(\alpha \right)\cos \left(\beta \right)-\cos \left(\alpha \right)\sin \left(\beta \right)\end{array}$$

準備1

$$\begin{array}{r}\\ \cos (\alpha -\beta )-\cos (\alpha +\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)-\{\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)\}\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& \cancel{\cos \left(\alpha \right)\cos \left(\beta \right)}+\sin \left(\alpha \right)\sin \left(\beta \right)\cancel{-\cos \left(\alpha \right)\cos \left(\beta \right)}+\sin \left(\alpha \right)\sin \left(\beta \right)\\ & =& 2\sin \left(\alpha \right)\sin \left(\beta \right)\\ \sin \left(\alpha \right)\sin \left(\beta \right)& =& \frac{\cos (\alpha -\beta )-\cos (\alpha +\beta )}{2}\end{array}$$

準備2

$$\begin{array}{r}\\ \cos (\alpha -\beta )+\cos (\alpha +\beta )& =& \cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)+\{\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)\}\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& \cos \left(\alpha \right)\cos \left(\beta \right)+\cancel{\sin \left(\alpha \right)\sin \left(\beta \right)}+\cos \left(\alpha \right)\cos \left(\beta \right)-\cancel{\sin \left(\alpha \right)\sin \left(\beta \right)}\\ & =& 2\cos \left(\alpha \right)\cos \left(\beta \right)\\ \cos \left(\alpha \right)\cos \left(\beta \right)& =& \frac{\cos (\alpha -\beta )+\cos (\alpha +\beta )}{2}\end{array}$$

準備3

$$\begin{array}{rcl}\cos \left(2x\right)& =& \cos (x+x)\\ & =& \cos \left(x\right)\cos \left(x\right)-\sin \left(x\right)\sin \left(x\right)\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& {\cos }^2\left(x\right)-{\sin }^2(x)\\ & =& (1-{\sin }^2(x))-{\sin }^2(x)\\ & =& 1-2{\sin }^2(x)\\ \cos \left(2x\right)-1& =& -2{\sin }^2(x)\\ 1-\cos \left(2x\right)& =& 2{\sin }^2(x)\\ {\sin }^2(x)& =& \frac{1-\cos \left(2x\right)}{2}=\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\end{array}$$

準備4

$$\begin{array}{rcl}\cos \left(2x\right)& =& \cos (x+x)\\ & =& \cos \left(x\right)\cos \left(x\right)-\sin \left(x\right)\sin \left(x\right)\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& {\cos }^2\left(x\right)-{\sin }^2(x)\\ & =& {\cos }^2\left(x\right)-(1-{\cos }^2\left(x\right))\\ & =& 2{\cos }^2(x)-1\\ \cos \left(2x\right)+1& =& 2{\cos }^2(x)\\ {\cos }^2(x)& =& \frac{1+\cos \left(2x\right)}{2}=\frac{1}{2}+\frac{\cos \left(2x\right)}{2}\end{array}$$

準備5

$$\begin{array}{r}\\ \sin (\alpha -\beta )+\sin (\alpha +\beta )& =& \sin \left(\alpha \right)\cos \left(\beta \right)-\cos \left(\alpha \right)\sin \left(\beta \right)+\{\sin \left(\alpha \right)\cos \left(\beta \right)+\cos \left(\alpha \right)\sin \left(\beta \right)\}\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& \sin \left(\alpha \right)\cos \left(\beta \right)\cancel{-\cos \left(\alpha \right)\sin \left(\beta \right)}+\sin \left(\alpha \right)\cos \left(\beta \right)\cancel{+\cos \left(\alpha \right)\sin \left(\beta \right)}\\ & =& 2\sin \left(\alpha \right)\cos \left(\beta \right)\\ \sin \left(\alpha \right)\cos \left(\beta \right)& =& \frac{\sin (\alpha -\beta )+\sin (\alpha +\beta )}{2}\end{array}$$

準備6

$$\begin{array}{rcl}\sin \left(2x\right)& =& \sin (x+x)\\ & =& \sin \left(x\right)\cos \left(x\right)+\cos \left(x\right)\sin \left(x\right)\cdots \mathrm{準}\mathrm{備}\mathrm{の}\mathrm{準}\mathrm{備}\\ & =& 2\sin \left(x\right)\cos \left(x\right)\\ \sin \left(x\right)\cos \left(x\right)& =& \frac{\sin \left(2x\right)}{2}\end{array}$$

$\sin \left(mx\right)\sin \left(nx\right)$

$m,n\in \mathbb{N},m\ne n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m\ne n\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos (mx-nx)-\cos (mx+nx)]\mathrm{d}x\cdots \mathrm{準}\mathrm{備}1\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos \left((m-n)x\right)-\cos \left((m+n)x\right)]\mathrm{d}x\\ & =& \frac{1}{2}[{\int }_0^{2\pi }\cos \left((m-n)x\right)\mathrm{d}x-{\int }_0^{2\pi }\cos \left((m+n)x\right)\mathrm{d}x]\\ & =& \frac{1}{2}[{[\frac{1}{m-n}\sin \left((m-n)x\right)]}_0^{2\pi }-{[\frac{1}{m+n}\sin \left((m+n)x\right)]}_0^{2\pi }]\\ & =& \frac{1}{2}[\frac{1}{m-n}[\sin \left((m-n)2\pi \right)-\sin \left((m-n)0\right)]-\frac{1}{m+n}[\sin \left((m+n)2\pi \right)-\sin \left((m+n)0\right)]]\\ & =& \frac{1}{2}[\frac{1}{m-n}[0-0]-\frac{1}{m+n}[0-0]]\\ & =& \frac{1}{2}[\frac{0}{m-n}-\frac{0}{m+n}]\\ & =& \frac{1}{2}0\\ & =& 0\end{array}$$

$m,n\in \mathbb{N},m=n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m=n\\ & =& {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(mx\right)\mathrm{d}x\\ & =& {\int }_0^{2\pi }{\sin }^2\left(mx\right)\mathrm{d}x\\ & =& \frac{1}{m}{\int }_0^{2m\pi }{\sin }^2\left(u\right)\mathrm{d}u\cdots u=mx,\frac{\mathrm{d}u}{\mathrm{d}x}=m,\mathrm{d}x=\frac{1}{m}\mathrm{d}u\\ & =& \frac{1}{m}{\int }_0^{2m\pi }[\frac{1}{2}-\frac{1}{2}\cos \left(2u\right)]\mathrm{d}u\cdots \mathrm{準}\mathrm{備}3\\ & =& \frac{1}{2m}[{\int }_0^{2m\pi }\mathrm{d}u-{\int }_0^{2m\pi }\cos \left(2u\right)\mathrm{d}u]\\ & =& \frac{1}{2m}[{\left[u\right]}_0^{2m\pi }-{[\frac{1}{2}\sin \left(2u\right)]}_0^{2m\pi }]\\ & =& \frac{1}{2m}[[2m\pi -0]-[\frac{1}{2}\sin (2\cdot 2m\pi )-\frac{1}{2}\sin (2\cdot 0)]]\\ & =& \frac{1}{2m}[2m\pi -[0-0]]\\ & =& \frac{1}{\cancel{2m}}\cancel{2m}\pi \\ & =& \pi \end{array}$$

$\cos \left(mx\right)\cos \left(nx\right)$

$m,n\in \mathbb{N},m\ne n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m\ne n\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos (mx-nx)+\cos (mx+nx)]\mathrm{d}x\cdots \mathrm{準}\mathrm{備}2\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\cos \left((m-n)x\right)+\cos \left((m+n)x\right)]\mathrm{d}x\\ & =& \frac{1}{2}[{\int }_0^{2\pi }\cos \left((m-n)x\right)\mathrm{d}x+{\int }_0^{2\pi }\cos \left((m+n)x\right)\mathrm{d}x]\\ & =& \frac{1}{2}[{[\frac{1}{m-n}\sin \left((m-n)x\right)]}_0^{2\pi }+{[\frac{1}{m+n}\sin \left((m+n)x\right)]}_0^{2\pi }]\\ & =& \frac{1}{2}[\frac{1}{m-n}[\sin \left((m-n)2\pi \right)-\sin \left((m-n)0\right)]+\frac{1}{m+n}[\sin \left((m+n)2\pi \right)-\sin \left((m+n)0\right)]]\\ & =& \frac{1}{2}[\frac{1}{m-n}[0-0]+\frac{1}{m+n}[0-0]]\\ & =& \frac{1}{2}[\frac{0}{m-n}+\frac{0}{m+n}]\\ & =& \frac{1}{2}0\\ & =& 0\end{array}$$

$m,n\in \mathbb{N},m=n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m=n\\ & =& {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(mx\right)\mathrm{d}x\\ & =& {\int }_0^{2\pi }{\cos }^2\left(mx\right)\mathrm{d}x\\ & =& \frac{1}{m}{\int }_0^{2m\pi }{\cos }^2\left(u\right)\mathrm{d}u\cdots u=mx,\frac{\mathrm{d}u}{\mathrm{d}x}=m,\mathrm{d}x=\frac{1}{m}\mathrm{d}u\\ & =& \frac{1}{m}{\int }_0^{2m\pi }[\frac{1}{2}+\frac{1}{2}\cos \left(2u\right)]\mathrm{d}u\cdots \mathrm{準}\mathrm{備}4\\ & =& \frac{1}{2m}[{\int }_0^{2m\pi }\mathrm{d}u+{\int }_0^{2m\pi }\cos \left(2u\right)\mathrm{d}u]\\ & =& \frac{1}{2m}[{\left[u\right]}_0^{2m\pi }+{[\frac{1}{2}\sin \left(2u\right)]}_0^{2m\pi }]\\ & =& \frac{1}{2m}[[2m\pi -0]+[\frac{1}{2}\sin (2\cdot 2m\pi )-\frac{1}{2}\sin (2\cdot 0)]]\\ & =& \frac{1}{2m}[2m\pi +[0-0]]\\ & =& \frac{1}{\cancel{2m}}\cancel{2m}\pi \\ & =& \pi \end{array}$$

$\sin \left(mx\right)\cos \left(nx\right)$

$m,n\in \mathbb{N},m\ne n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m\ne n\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\sin (mx-nx)+\sin (mx+nx)]\mathrm{d}x\cdots \mathrm{準}\mathrm{備}5\\ & =& {\int }_0^{2\pi }\frac{1}{2}[\sin ((m-n)x)+\sin ((m+n)x)]\mathrm{d}x\\ & =& \frac{1}{2}[{\int }_0^{2\pi }\sin ((m-n)x)\mathrm{d}x+{\int }_0^{2\pi }\sin ((m+n)x)\mathrm{d}x]\\ & =& \frac{1}{2}[{[\frac{-1}{m-n}\cos ((m-n)x)]}_0^{2\pi }+{[\frac{-1}{m+n}\cos ((m+n)x)]}_0^{2\pi }]\\ & =& \frac{1}{2}[\frac{-1}{m-n}[\cos ((m-n)2\pi )-\cos ((m-n)0)]+\frac{-1}{m+n}[\cos ((m-n)2\pi )-\cos ((m-n)0)]]\\ & =& \frac{1}{2}[\frac{-1}{m-n}[1-1]+\frac{-1}{m+n}[1-1]]\\ & =& \frac{1}{2}[\frac{-1}{m-n}0+\frac{-1}{m+n}0]\\ & =& \frac{1}{2}[0+0]\\ & =& \frac{1}{2}0\\ & =& 0\end{array}$$

$m,n\in \mathbb{N},m=n$

$$\begin{array}{rcl}& & {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(nx\right)\mathrm{d}x\cdots m,n\in \mathbb{N},m=n\\ & =& {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(mx\right)\mathrm{d}x\\ & =& {\int }_0^{2\pi }\frac{1}{2}\sin \left(2mx\right)\mathrm{d}x\cdots \mathrm{準}\mathrm{備}6\\ & =& \frac{1}{2}{\int }_0^{4m\pi }\frac{1}{2m}\sin \left(u\right)\mathrm{d}u\cdots u=2mx,\frac{\mathrm{d}u}{\mathrm{d}x}=2m,\mathrm{d}x=\frac{1}{2m}\mathrm{d}u\\ & =& \frac{1}{4m}{[-\cos \left(u\right)]}_0^{4m\pi }\\ & =& \frac{1}{4m}[-\cos \left(4m\pi \right)-(-\cos \left(0\right))]\\ & =& \frac{1}{4m}[-1-(-1)]\\ & =& \frac{1}{4m}[-1+1]\\ & =& \frac{1}{4m}0\\ & =& 0\end{array}$$

まとめ

$$\begin{array}{r}m,n\in \mathbb{N}\\ {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x& =& \{\begin{array}{l}\pi \cdots m=n\\ 0\cdots m\ne n\end{array}\\ {\int }_0^{2\pi }\cos \left(mx\right)\cos \left(nx\right)\mathrm{d}x& =& \{\begin{array}{l}\pi \cdots m=n\\ 0\cdots m\ne n\end{array}\\ {\int }_0^{2\pi }\sin \left(mx\right)\cos \left(nx\right)\mathrm{d}x& =& 0\end{array}$$

単位円の半分に収まる(?)正方形の面積は有理数(一辺の長さは無理数(sin(arctan(2))=2/√5))

点$C$の位置

$$\begin{array}{rcl}& & C(\overline{OF},\overline{CF})\\ & =& C(\cos (\theta ),\sin (\theta ))\end{array}$$

角$\theta$との関係

$$\begin{array}{rcl}\frac{\overline{CF}}{\overline{OF}}& =& 2\cdots \mathrm{正}\mathrm{方}\mathrm{形}\mathrm{と}\mathrm{な}\mathrm{る}\mathrm{た}\mathrm{め}\\ & =& \frac{\sin (\theta )}{\cos (\theta )}\\ & =& \tan (\theta )\\ {\tan }^{-1}(2)& =& \theta \end{array}$$

線分$OF$の長さ

$$\begin{array}{rcl}\overline{OF}& =& \cos ({\tan }^{-1}\left(2\right))\\ & =& \frac{1}{\sqrt{\tan {({\tan }^{-1}(2))}^2+1}}\\ & & \cdots \cos (\theta )=\frac{1}{\sqrt{{\tan }^2\left(\theta \right)+1}}\\ & =& \frac{1}{\sqrt{2^2+1}}\\ & =& \frac{1}{\sqrt{5}}\end{array}$$

線分$CF$の長さ

$$\begin{array}{r}\overline{CF}\\ & =& \sin (\theta )\\ & =& 2\cos (\theta )\\ & =& 2\frac{1}{\sqrt{5}}\\ & =& \frac{2}{\sqrt{5}}\end{array}$$

面積$A_1$

$$\begin{array}{rcl}{A}_1& =& {\overline{CF}}^2\\ & =& {\left(\frac{2}{\sqrt{5}}\right)}^2\\ & =& \frac{2^2}{{\left(\sqrt{5}\right)}^2}\\ & =& \frac{4}{5}(=0.8)\end{array}$$

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

---

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．

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