数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 置換積分法、三角関数(正弦、余弦、正接)、平方、平方根、部分積分法、指数関数

解析入門(上)
楽天ブックス Yahoo!

学習環境

• Surface
• Windows 10 Pro (OS)
• Nebo(Windows アプリ)
• iPad
• MyScript Nebo - MyScript(iPad アプリ(iOS))
• 参考書籍
  • Pythonからはじめる数学入門 (楽天ブックス Yahoo!)
  • JavaScriptリファレンス 第6版 (楽天ブックス Yahoo!)
  • インタラクティブ・データビジュアライゼーション (楽天ブックス Yahoo!)

解析入門(上) (松坂和夫 数学入門シリーズ 4) (松坂 和夫(著)、岩波書店)の第8章(積分の計算)、8.2(定積分の計算)、問題1の解答を求めてみる。

1. 
   
   1. 
      
      $x=a\left(\sin t\right)$

      とおくと、

      $\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dt}}=a\left(\cos t\right)\\ {\int }_0^a\sqrt{a^2-x^2}\mathrm{dx}\\ =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{a^2-a^2{\sin }^{2}t}a\left(\cos t\right)\mathrm{dt}\\ =a^2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1-{\sin }^{2}t}\cos t\mathrm{dt}\\ =a^2\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}t\mathrm{dt}\\ =a^2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos 2t+1}{2}\mathrm{dt}\\ =\frac{a^2}{2}{\left[\frac{1}{2}\sin 2t+t\right]}_0^{\frac{\pi }{2}}\\ =\frac{a^2}{2}·\frac{\pi }{2}\\ =\frac{\pi a^2}{4}\end{array}$
   2. 
      
      $\begin{array}{l}{\int }_0^a\frac{\mathrm{dx}}{\sqrt{a^2-x^2}}\\ =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{a\left(\cos t\right)}{a\left(\cos t\right)}\mathrm{dt}\\ =\frac{\pi }{2}\end{array}$
   3. 
      
      $\begin{array}{l}{\int }_0^1\frac{\mathrm{dx}}{1+x+x^2}\\ ={\int }_0^1\frac{\mathrm{dx}}{{\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}}\\ x+\frac{1}{2}=\frac{\sqrt{3}}{2}t\\ \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\sqrt{3}}{2}\\ {\int }_0^1\frac{\mathrm{dy}}{1+x+x^2}\\ =\frac{\sqrt{3}}{2}\underset{\frac{1}{\sqrt{3}}}{\overset{\sqrt{3}}{\int }}\frac{\mathrm{dt}}{\frac{3}{4}{t}^2+\frac{3}{4}}\\ =\frac{2\sqrt{3}}{3}\left(\arctan \sqrt{3}-\arctan \frac{1}{\sqrt{3}}\right)\\ =\frac{2\sqrt{3}}{3}\left(\frac{\pi }{3}-\frac{\pi }{6}\right)\\ =\frac{2\sqrt{3}}{3}·\frac{\pi }{6}\\ =\frac{\sqrt{3}}{9}\pi \\ =\frac{\pi }{3\sqrt{3}}\end{array}$
   4. 
      
      $\begin{array}{l}{\int }_{-1}^1\frac{\mathrm{dy}}{{\left(1+x^2\right)}^2}\\ =2{\int }_0^1\frac{\mathrm{dx}}{{\left(1+x^2\right)}^2}\\ \int \frac{\mathrm{dx}}{{\left(1+x^2\right)}^2}\\ =\frac{1}{2\left(2-1\right)}\left(\frac{x}{x^2+1}+\left(2·2-3\right)\int \frac{\mathrm{dx}}{1+x^2}\right)\\ =\frac{1}{2}\left(\frac{x}{x^2+1}+\arctan x\right)\\ {\int }_{-1}^1\frac{\mathrm{dx}}{{\left(1+x^2\right)}^2}\\ ={\left[\frac{x}{x^2+1}+\arctan x\right]}_0^1\\ =\frac{1}{2}+\arctan 1-\arctan 0\\ =\frac{1}{2}+\frac{\pi }{4}\end{array}$
   5. 
      
      $\begin{array}{l}\tan \frac{\theta }{2}=t\\ \frac{1}{2{\cos }^2\frac{\theta }{2}}=\frac{\mathrm{dt}}{d\theta }\\ {\int }_0^{\pi }\frac{d\theta }{1+a\left(\cos \theta \right)}\\ ={\int }_0^{\infty }\frac{1}{1+a\left(\cos \theta \right)}·2{\cos }^2\frac{\theta }{2}\mathrm{dt}\\ =\underset{b\to \infty }{\lim }{\int }_0^b\frac{2{\cos }^2\frac{\theta }{2}}{1+a\left({\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}\right)}\mathrm{dt}\\ =\underset{b\to \infty }{\lim }{\int }_0^b\frac{2{\cos }^2\frac{\theta }{2}}{{\sin }^2\frac{\theta }{2}+{\cos }^2\frac{\theta }{2}+a\left({\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}\right)}\mathrm{dt}\\ =\underset{b\to \infty }{\lim }{\int }_0^b\frac{2}{t^2+1+a\left(1-t^2\right)}\mathrm{dt}\\ =2\underset{b\to \infty }{\lim }{\int }_0^b\frac{1}{\left(1-a\right)t^2+1+a}\mathrm{dt}\\ =\frac{2}{1-a}\underset{b\to \infty }{\lim }{\int }_0^b\frac{1}{t^2+\frac{1+a}{1-a}}\mathrm{dt}\\ =\frac{2}{1-a}·\sqrt{\frac{1-a}{1+a}}\underset{b\to \infty }{\lim }{\left[\arctan \sqrt{\frac{1-a}{1+a}}t\right]}_0^b\\ =\frac{2}{\sqrt{1-a^2}}\underset{b\to \infty }{\lim }\arctan \sqrt{\frac{1-a}{1+a}}b\\ =\frac{2}{\sqrt{1-a^2}}·\frac{\pi }{2}\\ =\frac{\pi }{\sqrt{1-a^2}}\end{array}$
   6. 
      
      $\begin{array}{l}\int \frac{x\arcsin x}{\sqrt{1-x^2}}\mathrm{dx}\\ =x{\arcsin }^{2}x-\int \left(\arcsin x+\frac{x}{\sqrt{1-x^2}}\right)\arcsin x\mathrm{dx}\\ =x{\arcsin }^{2}x-\int {\arcsin }^{2}x\mathrm{dx}-\int \frac{x\arcsin x}{\sqrt{1-x^2}}\mathrm{dx}\\ \int \frac{x\arcsin x}{\sqrt{1-x^2}}\\ =\frac{1}{2}\left(x{\arcsin }^{2}x-\int {\arcsin }^{2}x\mathrm{dx}\right)\\ t=\arcsin x\\ \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{1}{\sqrt{1-x^2}}\\ \sin t=x\\ \int {\arcsin }^{2}x\mathrm{dx}\\ =\int t\sqrt[2]{1-x^2}\mathrm{dt}\\ =\int t\sqrt[2]{1-{\sin }^{2}t}\mathrm{dt}\\ =\int t^2\cos t\mathrm{dt}\\ =t^2\sin t-2\int t\sin t\mathrm{dt}\\ =t^2\sin t-2\left(-t\cos t+\int \cos t\mathrm{dt}\right)\\ {\int }_{-1}^1\frac{x\arcsin x}{\sqrt{1-x^2}}\mathrm{dx}\\ =\frac{1}{2}{\left[x{\arcsin }^{2}x\right]}_{-1}^1-\frac{1}{2}{\left[t^2\sin t\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-{\left[t\cos t\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+{\left[\sin t\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\\ =\frac{1}{2}\left({\arcsin }^{2}1+{\arcsin }^2\left(-1\right)\right)-\frac{1}{2}·\frac{{\pi }^2}{4}·2+2\\ =\frac{1}{2}\left(\frac{{\pi }^2}{4}+\frac{{\pi }^2}{4}\right)-\frac{{\pi }^2}{4}+2\\ =2\end{array}$
   7. 
      
      $\begin{array}{l}\int e^{-ax}\sin bx\mathrm{dx}\\ =-\frac{1}{a}{e}^{-ax}\sin bx+\frac{b}{a}\int e^{-ax}\cos bx\mathrm{dx}\\ =-\frac{1}{a}{e}^{-ax}\sin bx+\frac{b}{a}\left(-\frac{1}{a}{e}^{-ax}\cos bx-\frac{b}{a}\int e^{-ax}\sin bx\mathrm{dx}\right)\\ =-\frac{1}{a}{e}^{-ax}\sin bx-\frac{b}{a^2}{e}^{-ax}\cos bx-\frac{b^2}{a^2}\int e^{-ax}\sin bx\mathrm{dx}\\ \left(1+\frac{b^2}{a^2}\right)\int e^{-ax}\sin bx\mathrm{dx}=-\frac{1}{a}{e}^{-ax}\sin bx-\frac{b}{a^2}{e}^{-ax}\cos bx\\ \int e^{-ax}\sin bx\mathrm{dx}=-\frac{a^2}{a^2+b^2}·\frac{1}{a}\left(e^{-ax}\sin bx+\frac{b}{a}{e}^{-ax}\cos bx\right)\\ {\int }_0^{+\infty }{e}^{-ax}\sin bxdx\\ =\underset{A\to \infty }{\lim }{\int }_0^A{e}^{-ax}\sin bx\mathrm{dx}\\ =-\frac{a}{a^2+b^2}\underset{A\to \infty }{\lim }{\left[e^{-ax}\left(\sin bx+\frac{b}{a}\cos bx\right)\right]}_0^A\\ =-\frac{a}{a^2+b^2}\underset{A\to \infty }{\lim }\left(e^{-aA}\left(\sin bA+\frac{b}{a}\cos bA\right)-\left(\sin 0+\frac{b}{a}\cos 0\right)\right)\\ =-\frac{a}{a^2+b^2}·\left(-\frac{b}{a}\right)\\ =\frac{b}{a^2+b^2}\end{array}$