数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 置換積分法、三角関数(正弦と余弦と正接)、有理関数の積分、部分分数分解、対数関数

学習環境

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)、8.1(不定積分の計算)、問題7の解答を求めてみる。

1. $\begin{array}{l} \sin x \\ = \sin (\frac{x}{2} + \frac{x}{2}) \\ = 2 \sin \frac{x}{2} \cos \frac{x}{2} \\ = \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2}} \\ = \frac{2 \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}{\frac{\sin^{2} \frac{x}{2}}{\cos^{2} \frac{x}{2}} + 1} \\ = \frac{2 t}{1 + t^{2}} \\ \cos x \\ = \cos (\frac{x}{2} + \frac{x}{2}) \\ = \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2} \\ = \frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2}} \\ = \frac{1 - t^{2}}{1 + t^{2}} \\ \frac{dt}{dx} = \frac{1}{\cos^{2} \frac{x}{2}} \cdot \frac{1}{2} \\ = \frac{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} \\ = \frac{1 + t^{2}}{2} \\ \frac{dy}{dt} = \frac{2}{1 + t^{2}} \end{array}$
  
  よって、
  
  $\begin{array}{l} \int \frac{dx}{1 + \cos x} \\ = \int \frac{1}{1 + \frac{1 - t^{2}}{1 + t^{2}}} \cdot \frac{2}{1 + t^{2}} dt \\ = \int \frac{2}{1 + t^{2} + 1 - t^{2}} dt \\ = \int dt \\ = t \\ = \tan \frac{x}{2} \end{array}$
2. 置換積分と部分分数分解。
  
  $\begin{array}{l} \int \frac{dx}{\cos x} \\ = \int \frac{1 + t^{2}}{1 - t^{2}} \cdot \frac{2}{1 + t^{2}} dt \\ = \int \frac{2}{1 - t^{2}} dx \\ = 2 \int \frac{1}{(1 + t) (1 - t)} dt \\ \frac{A}{1 + t} + \frac{B}{1 - t} \\ = \frac{(B - A) t + (A + B)}{1 - t^{2}} \\ - A + B = 0 \\ A + B = 1 \\ B = \frac{1}{2} \\ A = \frac{1}{2} \\ \int \frac{dx}{\cos x} \\ = \int \frac{dt}{1 + t} + \int \frac{dt}{1 - t} \\ = \log |1 + t| - \log |1 - t| \\ = \log |\frac{1 + t}{1 - t}| \\ = \log |\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}| \end{array}$
3. $\begin{array}{l} \int \frac{\sin x}{1 + \sin x} dx \\ = \int \frac{2 t}{1 + t^{2}} \cdot \frac{1}{1 + \frac{2 t}{1 + t^{2}}} \cdot \frac{2}{1 + t^{2}} dt \\ = \int \frac{4 t}{(1 + t^{2}) (t^{2} + 2 t + 1)} dt \\ = 4 \int \frac{t}{(1 + t^{2}) {(1 + t)}^{2}} dt \\ \frac{A t + B}{t^{2} + 1} + \frac{C}{t + 1} + \frac{D}{{(t + 1)}^{2}} \\ (A t + B) {(t + 1)}^{2} + C (t^{2} + 1) (t + 1) + D (t^{2} + 1) \\ = (A t + B) (t^{2} + 2 t + 1) + C (t^{3} + t^{2} + t + 1) + D (t^{2} + 1) \\ = (A + C) t^{3} + (2 A + B + C + D) t^{2} + (A + 2 B + C) t + (B + C + D) \\ {\begin{cases} A + C = 0 \\ 2 A + B + C + D = 0 \\ A + 2 B + C = 1 \\ B + C + D = 0 \end{cases} \\ C = - A \\ 2 B = 1 \\ B = \frac{1}{2} \\ \frac{1}{2} - A + D = 0 \\ D = A - \frac{1}{2} \\ A + \frac{1}{2} + A - \frac{1}{2} = 0 \\ A = 0 \\ C = 0 \\ D = - \frac{1}{2} \\ \int \frac{t}{(1 + t^{2}) {(1 + t)}^{2}} dt \\ = \frac{1}{2} (\int \frac{dt}{1 + t^{2}} - \int \frac{dt}{{(1 + t)}^{2}}) \\ = \frac{1}{2} (\arctan t + \frac{1}{1 + t}) \\ = \frac{1}{2} (\arctan (\tan \frac{x}{2}) + \frac{1}{1 + \tan \frac{x}{2}}) \\ = \frac{1}{2} (\frac{x}{2} + \frac{1}{1 + \tan \frac{x}{2}}) \\ \int \frac{\sin x}{1 + \sin x} dx \\ = 2 (\frac{x}{2} + \frac{1}{1 + \tan \frac{x}{2}}) \\ = x + \frac{2}{1 + \tan \frac{x}{2}} \end{array}$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import pprint, symbols, Derivative, sin, cos, tan, atan, log, plot print('7.') x = symbols('x') fs = [1 / (1 + cos(x)), 1 / cos(x), sin(x) / (1 + sin(x))] gs = [tan(x / 2), log(abs((1 + tan(x / 2)) / (1 - tan(x / 2)))), x + 2 / (1 + tan(x / 2))] s = {x: 1} class MyTestCase(TestCase): def test1(self): self.assertEqual(float(fs[0].subs(s)), float(Derivative(gs[0], x, 1).doit().subs(s))) def test2(self): self.assertEqual(float(fs[1].subs(s)), float(Derivative(log((1 + tan(x / 2)) / (1 - tan(x / 2))), x, 1).doit().subs(s))) def test3(self): self.assertEqual(float(fs[2].subs(s)), float(Derivative(gs[2], x, 1).doit().subs(s))) p = plot(*[f for f in fs + gs], (x, -5, 5), ylim=(-5, 5), legend=True, show=False) colors = ['red', 'green', 'blue', 'brown', 'orange', 'purple', 'pink', 'gray', 'skyblue', 'yellow'] for o, color in zip(p, colors): o.line_color = color p.show() p.save('sample7.png') if __name__ == '__main__': main()

% ./sample7.py -v 7. test1 (__main__.MyTestCase) ... ok test2 (__main__.MyTestCase) ... ok test3 (__main__.MyTestCase) ... ok ---------------------------------------------------------------------- Ran 3 tests in 0.044s OK %

Kamimura's blog

2019年12月30日月曜日

数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 置換積分法、三角関数(正弦と余弦と正接)、有理関数の積分、部分分数分解、対数関数

0 コメント:

コメントを投稿