数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 漸化式による計算、三角関数(正弦と余弦、正接)、累乗、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5の解答を求めてみる。

1. 以下積分定数は省略。
  
  $\begin{array}{l} \int \tan^{3} x dx \\ = \int \frac{\sin^{3} x}{\cos^{3} x} dx \\ = \int \sin^{3} x \cos^{- 3} x dx \\ = I (3, - 3) \\ = - \frac{\sin^{4} x \cos^{- 2} x}{- 2} + \frac{2}{- 2} I (3, - 1) \\ = \frac{\sin^{4} x}{2 \cos^{2} x} - (- \frac{\sin^{2} x}{2} + \frac{2}{2} I (1, - 1)) \\ = \frac{\sin^{4} x}{2 \cos^{2} x} + \frac{\sin^{2} x}{2} - \int \sin x \cos^{- 1} x dx \\ = \frac{\sin^{4} x}{2 \cos^{2} x} + \frac{\sin^{2} x}{2} + \log |\cos x| \\ = \frac{\sin^{2} x (\sin^{2} x + \cos^{2} x)}{2 \cos^{2} x} + \log |\cos x| \\ = \frac{\tan^{2} x}{2} + \log |\cos x| \end{array}$
2. $\begin{array}{l} \int \sin^{2} x \cos^{2} x dx \\ = \frac{\sin^{3} x \cos x}{4} + \frac{1}{4} I (2, 0) \\ = \frac{\sin^{3} x \cos x}{4} + \frac{1}{4} \int \sin^{2} x \cos^{0} x dx \\ = \frac{\sin^{3} x \cos x}{4} + \frac{1}{4} (- \frac{\sin x \cos x}{2} + \frac{1}{2} \int \sin^{0} x \cos^{0} x d) \\ = \frac{\sin^{3} x \cos x}{4} - \frac{\sin x \cos x}{8} + \frac{1}{8} x \\ = \frac{1}{8} (\sin x \cos x (2 \sin^{2} x - 1) + x) \\ = \frac{1}{8} (\sin x \cos x (\cos 0 - \cos 2 x - 1) + x) \\ = \frac{1}{8} (x - \sin x \cos x \cos 2 x) \end{array}$
3. $\begin{array}{l} \int \frac{dx}{\sin^{3} x \cos^{3} x} \\ = I (- 3, - 3) \\ = - \frac{\sin^{- 2} x \cos^{- 2} x}{- 2} + \frac{- 6 + 2}{- 2} I (- 3, - 1) \\ = \frac{\sin^{- 2} x \cos^{- 2} x}{2} + 2 (\frac{\sin^{- 2} x \cos^{0} x}{- 2} + \frac{- 2}{- 2} I (- 1, - 1)) \\ = \frac{\sin^{- 2} x \cos^{- 2} x}{2} - \sin^{- 2} x + 2 \int \frac{1}{\sin x \cos x} dx \\ = \frac{\sin^{- 2} x \cos^{- 2} x}{2} - \sin^{- 2} x + 2 \log |\tan x| \\ = \frac{1}{2 \sin^{2} x \cos^{2} x} - \frac{1}{\sin^{2} x} + 2 \log |\tan x| \end{array}$

#!/usr/bin/env python3 from sympy import pprint, symbols, Integral, sin, cos, tan, log, plot print('5.') x = symbols('x') m, n = symbols('m, n', integer=True) fs = [tan(x) ** 3, sin(x) ** 2 * cos(x) ** 2, 1 / (sin(x) ** 3 * cos(x) ** 3)] gs = [tan(x) ** 2 / 2 + log(abs(cos(x))), (x - sin(x) * cos(x) * cos(2 * x)) / 8, 1 / (2 * sin(x) ** 2 * cos(x) ** 2) - 1 / sin(x) ** 2 + 2 * log(abs(tan(x)))] for i, f in enumerate(fs, 1): print(f'({i})') I = Integral(f, x) for o in [I, I.doit()]: pprint(o.simplify()) print() p = plot(*fs, *gs, (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 + gs, colors): pprint(o) p.show() p.save('sample5.png')

% ./sample5.py 5. (1) ⌠ ⎮ 3 ⎮ tan (x) dx ⌡ 1 log(cos(x)) + ───────── 2 2⋅cos (x) (2) ⌠ ⎮ ⎛1 cos(4⋅x)⎞ ⎮ ⎜─ - ────────⎟ dx ⎮ ⎝8 8 ⎠ ⌡ x sin(4⋅x) ─ - ──────── 8 32 (3) ⌠ ⎮ 1 ⎮ ─────────────── dx ⎮ 3 3 ⎮ sin (x)⋅cos (x) ⌡ ⎛ 2 ⎞ 1 1 2⋅log(sin(x)) - log⎝-cos (x)⎠ + ─────── - ───────────────── 2 2 2 cos (x) 2⋅sin (x)⋅cos (x) ⎛ 3 ⎞ ⎝tan (x), red⎠ ⎛ 2 2 ⎞ ⎝sin (x)⋅cos (x), green⎠ ⎛ 1 ⎞ ⎜───────────────, blue⎟ ⎜ 3 3 ⎟ ⎝sin (x)⋅cos (x) ⎠ ⎛ 2 ⎞ ⎜ tan (x) ⎟ ⎜log(│cos(x)│) + ───────, brown⎟ ⎝ 2 ⎠ ⎛x sin(x)⋅cos(x)⋅cos(2⋅x) ⎞ ⎜─ - ──────────────────────, orange⎟ ⎝8 8 ⎠ ⎛ 1 1 ⎞ ⎜2⋅log(│tan(x)│) - ─────── + ─────────────────, purple⎟ ⎜ 2 2 2 ⎟ ⎝ sin (x) 2⋅sin (x)⋅cos (x) ⎠ %

Kamimura's blog

2019年12月27日金曜日

数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 漸化式による計算、三角関数(正弦と余弦、正接)、累乗、対数関数

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