数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 三角関数(正弦と余弦)、累乗、漸化式

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4の解答を求めてみる。

1. $\begin{array}{l} \frac{d}{dx} (\frac{\sin^{m + 1} x \cos^{n \cdot 1} x}{m + n} + \frac{n - 1}{m + n} \int \sin^{m} x \cos^{n - 2} x dx) \\ = \frac{(m + 1) \sin^{m} x \cos^{n} x + (\sin^{m + 1} x) (n - 1) (\cos^{n - 2} x) (- \sin x)}{m + n} \\ + \frac{(n - 1) \sin^{m} x \cos^{n - 2} x}{m + n} \\ = \frac{(m + 1) \sin^{m} x \cos^{n} x - (n - 1) (\sin^{n + 2} x \cos^{n - 2} x) + (n - 1) \sin^{m} x \cos^{n - 2} x}{m + n} \\ = \frac{\sin^{m} x \cos^{n - 2} x ((m + 1) \cos^{2} x - (n - 1) \sin^{2} x + (n - 1))}{m + n} \\ = \frac{\sin^{m} x \cos^{n - 2} x (m \cos^{2} x - n \sin^{2} x + n)}{m + n} \\ = \frac{\sin^{m} x \cos^{n - 2} x (m \cos^{2} x - n (1 - \cos^{2} x) + n)}{m + n} \\ = \sin^{m} x \cos^{n} x \end{array}$
  
  よって、
  
  $I (m, n) = \frac{\sin^{m + 1} x \cos^{n - 1} x}{m + n} + \frac{n - 1}{m + n} I (m, n - 2)$
  
  が成り立つ。
  
  $\begin{array}{l} \frac{d}{dx} (- \frac{\sin^{m - 1} x \cos^{n + 1} x}{m + n} + \frac{m - 1}{m + n} I (m - 2, n)) \\ = \frac{- (m - 1) \sin^{m - 2} x \cos^{n + 2} x + (\sin^{m} x) (n + 1) \cos^{n} x}{m + n} \\ + \frac{(m - 1) (\sin^{m - 2} x \cos^{n} x)}{m + n} \\ = \frac{\sin^{m - 2} x \cos^{n} x (- (m - 1) \cos^{2} x + (n + 1) \sin^{2} x + (m - 1))}{m + n} \\ = \frac{\sin^{m - 2} x \cos^{n} x (- m (1 - \sin^{2} x) + n \sin^{2} x + m)}{m + n} \\ = \sin^{m} x \cos^{n} x \\ I (m, n) = - \frac{\sin^{m - 1} x \cos^{n + 1} x}{n 1 + n} + \frac{m - 1}{m + n} I (m - 2, n) \end{array}$
2. $\begin{array}{l} \frac{d}{dx} (- \frac{\sin^{m + 1} x \cos^{n + 1} x}{n + 1} + \frac{m + n + 2}{n + 1} I (m, n + 2)) \\ = \frac{- (m + 1) \sin^{m} x \cos^{n + 2} x + (n + 1) \sin^{m + 2} x \cos^{n} x}{n + 1} \\ + \frac{(m + n + 2) \sin^{m} x \cos^{n + 2} x}{n + 1} \\ = \frac{\sin^{m} x \cos^{n} x (- (m + 1) \cos^{2} x + (n + 1) \sin^{2} x + (m + n + 2) \cos^{2} x)}{n + 1} \\ = \frac{\sin^{m} x \cos^{n} x (n + 1) (\cos^{2} x + \sin^{2} x)}{n + 1} \\ = \sin^{n} x \cos^{n} x \end{array}$
  
  （証明終）
3. $\begin{array}{l} \frac{d}{dx} (\frac{\sin^{m + 1} x \cos^{n + 1} x}{m + 1} + \frac{m + n + 2}{m + 1} I (m + 2, n)) \\ = \frac{(m + 1) \sin^{m} x \cos^{n + 2} x - (n + 1) \sin^{m + 2} x \cos^{n} x}{m + 1} \\ + \frac{(m + n + 2) \sin^{n + 2} x \cos^{n} x}{m + 1} \\ = \frac{\sin^{m} x \cos^{n} x ((m + 1) \cos^{2} x - (n + 1) \sin^{2} x + (m + n + 2) \sin^{2} x)}{m + 1} \\ = \frac{\sin^{m} x \cos^{n} x ((m + 1) \cos^{2} x + (m + 1) \sin^{2} x)}{m + 1} \\ = \sin^{m} x \cos^{n} x \end{array}$
  
  （証明終）

#!/usr/bin/env python3 from sympy import pprint, symbols, Integral, sin, cos, plot print('4.') x = symbols('x') m, n = symbols('m, n', integer=True) f = sin(x) ** m * cos(x) ** n fs = [f.subs({m: m0, n: n0}) for m0 in range(-1, 2) for n0 in range(-1, 3)] for g in fs: I = Integral(g, x) for o in [I, I.doit()]: pprint(o) print() p = plot(*fs, (x, -5, 5), ylim=(-5, 5), legend=False, show=False) colors = ['red', 'green', 'blue', 'brown', 'orange', 'purple', 'pink', 'gray', 'skyblue', 'yellow'] for o, color in zip(p, colors): o.line_color = color for o in zip(fs, colors): pprint(o) p.show() p.save('sample4.png')

% ./sample4.py 4. ⌠ ⎮ 1 ⎮ ───────────── dx ⎮ sin(x)⋅cos(x) ⌡ ⎛ 2 ⎞ log⎝sin (x) - 1⎠ - ──────────────── + log(sin(x)) 2 ⌠ ⎮ 1 ⎮ ────── dx ⎮ sin(x) ⌡ log(cos(x) - 1) log(cos(x) + 1) ─────────────── - ─────────────── 2 2 ⌠ ⎮ cos(x) ⎮ ────── dx ⎮ sin(x) ⌡ log(sin(x)) ⌠ ⎮ 2 ⎮ cos (x) ⎮ ─────── dx ⎮ sin(x) ⌡ log(cos(x) - 1) log(cos(x) + 1) ─────────────── - ─────────────── + cos(x) 2 2 ⌠ ⎮ 1 ⎮ ────── dx ⎮ cos(x) ⌡ log(sin(x) - 1) log(sin(x) + 1) - ─────────────── + ─────────────── 2 2 ⌠ ⎮ 1 dx ⌡ x ⌠ ⎮ cos(x) dx ⌡ sin(x) ⌠ ⎮ 2 ⎮ cos (x) dx ⌡ x sin(x)⋅cos(x) ─ + ───────────── 2 2 ⌠ ⎮ sin(x) ⎮ ────── dx ⎮ cos(x) ⌡ -log(cos(x)) ⌠ ⎮ sin(x) dx ⌡ -cos(x) ⌠ ⎮ sin(x)⋅cos(x) dx ⌡ 2 sin (x) ─────── 2 ⌠ ⎮ 2 ⎮ sin(x)⋅cos (x) dx ⌡ 3 -cos (x) ───────── 3 ⎛ 1 ⎞ ⎜─────────────, red⎟ ⎝sin(x)⋅cos(x) ⎠ ⎛ 1 ⎞ ⎜──────, green⎟ ⎝sin(x) ⎠ ⎛cos(x) ⎞ ⎜──────, blue⎟ ⎝sin(x) ⎠ ⎛ 2 ⎞ ⎜cos (x) ⎟ ⎜───────, brown⎟ ⎝ sin(x) ⎠ ⎛ 1 ⎞ ⎜──────, orange⎟ ⎝cos(x) ⎠ (1, purple) (cos(x), pink) ⎛ 2 ⎞ ⎝cos (x), gray⎠ ⎛sin(x) ⎞ ⎜──────, skyblue⎟ ⎝cos(x) ⎠ (sin(x), yellow) %

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2019年12月26日木曜日

数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 三角関数(正弦と余弦)、累乗、漸化式

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