数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 部分積分法、置換積分、三角関数(逆正弦関数、逆正接関数、正弦、余弦)、対数関数、累乗、累乗根

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)、8.2(定積分の計算)、問題5の解答を求めてみる。

1. $\begin{array}{l} x = \frac{a + b}{2} + t \\ \frac{b - a}{2} = h \end{array}$
  
  とおくと、
  
  $\begin{array}{l} \int_{a}^{b} \frac{dx}{\sqrt{(x - a) (x - b)}} \\ \int_{\frac{3 a + b}{2}}^{\frac{a + 3 b}{2}} \frac{dt}{\sqrt{((\frac{a + b}{2} + t) - a) (b - (\frac{a + b}{2} + t))}} \\ = \int_{\frac{3 a + b}{2}}^{\frac{a + 3 b}{2}} \frac{dt}{\sqrt{(\frac{b - a}{2} + t) (\frac{b - a}{2} - t)}} \\ = \int_{\frac{3 a + b}{2}}^{\frac{a + 3 b}{2}} \frac{dt}{\sqrt{(h + t) (h - t)}} \\ = \int_{\frac{3 a + b}{2}}^{\frac{a + 3 b}{2}} \frac{dt}{\sqrt{h^{2} - t^{2}}} dt \\ = {[\arcsin \frac{t}{h}]}_{\frac{3 a + b}{2}}^{\frac{a + 3 b}{2}} \\ = \arcsin \frac{a + 3 b}{2 h} - \arcsin \frac{3 a + b}{2 h} \\ = \arcsin \frac{a + 3 b}{2} \cdot \frac{2}{b - a} - \arcsin \frac{3 a + b}{2} \cdot \frac{2}{b - a} \\ = \arcsin \frac{a + b}{b - a} - \arcsin \frac{3 a + b}{b - a} \end{array}$
2. $\begin{array}{l} α < 1 \\ \int \frac{\log x}{x^{α}} dx \\ = \frac{x^{1 - α}}{1 - α} \log x - \int \frac{x^{1 - α}}{1 - α} \cdot \frac{1}{x} dx \\ = \frac{1}{1 - α} (x^{1 - α} \log x - \int x^{- α} dx) \\ = \frac{1}{1 - α} (x^{1 - α} \log x - \frac{x^{1 - α}}{1 - α}) \\ \int_{0}^{1} \frac{\log x}{x^{x}} dx \\ = \frac{1}{1 - α} \lim_{b \to + 0} {[x^{1 - α} \log x - \frac{x^{1 - α}}{1 - α}]}_{b}^{1} \\ = \frac{1}{1 - α} \lim_{b \to + 0} (- \frac{1}{1 - α} - b^{1 - α} \log b + \frac{b^{1 - α}}{1 - α}) \\ = \frac{1}{1 - α} (- \frac{1}{1 - α}) \\ = - \frac{1}{{(1 - α)}^{2}} \end{array}$
3. $\begin{array}{l} \sqrt{\frac{1 + x}{1 - x}} = t \\ \frac{dt}{dx} = \frac{\frac{\sqrt{1 - x}}{2 \sqrt{1 + x}} + \frac{\sqrt{1 + x}}{2 \sqrt{1 - x}}}{1 - x} \\ = \frac{1 - x + 1 + x}{2 (1 - x) \sqrt{1 - x^{2}}} \\ = \frac{1}{(1 - x) \sqrt{1 - x^{2}}} \\ \frac{1 + x}{1 - x} = t^{2} \\ 1 + x = t^{2} (1 - x) \\ x = \frac{t^{2} - 1}{t^{2} + 1} \\ \int \frac{dx}{(1 - 2 a x + a^{2}) \sqrt{1 - x^{2}}} \\ = \int \frac{1 - x}{1 - 2 a x + a^{2}} dt \\ = \int \frac{1 - \frac{t^{2} - 1}{t^{2} + 1}}{1 - 2 a \cdot \frac{t^{2} - 1}{t^{2} + 1} + a^{2}} dt \\ = \int \frac{t^{2} + 1 - (t^{2} - 1)}{t^{2} + 1 - 2 a (t^{2} - 1) + a^{2} (t^{2} + 1)} dt \\ = 2 \int \frac{dt}{(a^{2} - 2 a + 1) t^{2} + a^{2} + 2 a + 1} \\ = 2 \int \frac{dt}{{(a - 1)}^{2} t^{2} + {(a + 1)}^{2}} \\ = \frac{2}{{(a - 1)}^{2}} \int \frac{dt}{t^{2} + {(\frac{a + 1}{a - 1})}^{2}} \\ = \frac{2}{{(a - 1)}^{2}} \cdot \frac{a - 1}{a + 1} \arctan \frac{a - 1}{a + 1} t \\ = \frac{2}{a^{2} - 1} \arctan \frac{a - 1}{a + 1} t \\ \int_{- 1}^{1} \frac{dx}{(1 - 2 a x + a^{2}) \sqrt{1 - λ^{2}}} \\ = \frac{2}{a^{2} - 1} \lim_{b \to \infty} {[\arctan \frac{a - 1}{a + 1} t]}_{0}^{b} \\ = \frac{2}{a^{2} - 1} \lim_{b \to \infty} \arctan \frac{a - 1}{a + 1} b \\ = |\frac{2}{a^{2} - 1} \frac{π}{2}| \\ = \frac{π}{|a^{2} - 1|} \\ |a| \neq 1 \end{array}$
4. $\begin{array}{l} \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ \\ = \int_{0}^{\frac{π}{2}} (1 - \frac{\cos θ}{\sin θ + \cos θ}) d θ \\ = \frac{π}{2} - \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ + \cos θ} d θ \\ = \frac{π}{2} - \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ \\ \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ = \frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4} \end{array}$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import symbols, Integral, sqrt, pi, log, sin, cos, plot print('5.') x = symbols('x') class MyTestCase(TestCase): def test1(self): a = 1 b = 2 self.assertEqual(Integral(1 / sqrt((x - a) * (b - x)), (x, a, b)).doit(), pi) def test2(self): alpha = -1 self.assertEqual(Integral(log(x) / x ** alpha, (x, 0, 1)).doit(), -1 / (1 - alpha) ** 2) def test3(self): a = 2 self.assertEqual(float(Integral(1 / ((1 - 2 * a * x + a ** 2) * sqrt(1 - x ** 2)), (x, -1, 1)).doit()), float(pi / abs(1 - a ** 2))) def test4(self): self.assertEqual(Integral(sin(x) / (sin(x) + cos(x)), (x, 0, pi / 2)).doit(), pi / 4) p = plot(1 / sqrt((x - 1) * (2 - x)), log(x) / x ** -2, 1 / ((1 - 2 * 3 * x + 3 ** 2) * sqrt(1 - x ** 2)), sin(x) / (sin(x) + cos(x)), cos(x) / (sin(x) + cos(x)), (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('sample5.png') if __name__ == '__main__': main()

% ./sample5.py -v 5. test1 (__main__.MyTestCase) ... ok test2 (__main__.MyTestCase) ... ok test3 (__main__.MyTestCase) ... ok test4 (__main__.MyTestCase) ... ok ---------------------------------------------------------------------- Ran 4 tests in 18.068s OK %

Kamimura's blog

2020年1月13日月曜日

数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 部分積分法、置換積分、三角関数(逆正弦関数、逆正接関数、正弦、余弦)、対数関数、累乗、累乗根

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