以下の3つの定積分の値を求める問題です。 (1) $\int_{-2}^{2} \frac{x^2 - 4}{x^2 + 4} dx$ (2) $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x \cos x}{\sin^3 x} dx$ (3) $\int_{0}^{\frac{\pi}{2}} \cos^2 x \sin^6 x dx$

1. 問題の内容

以下の3つの定積分の値を求める問題です。

(1) $\int_{-2}^{2} \frac{x^2 - 4}{x^2 + 4} dx$

(2) $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x \cos x}{\sin^3 x} dx$

(3) $\int_{0}^{\frac{\pi}{2}} \cos^2 x \sin^6 x dx$

2. 解き方の手順

(1)

まず、被積分関数を変形します。

$\frac{x^2 - 4}{x^2 + 4} = \frac{x^2 + 4 - 8}{x^2 + 4} = 1 - \frac{8}{x^2 + 4}$

よって、積分は次のようになります。

$\int_{-2}^{2} \frac{x^2 - 4}{x^2 + 4} dx = \int_{-2}^{2} (1 - \frac{8}{x^2 + 4}) dx$

$= \int_{-2}^{2} 1 dx - 8 \int_{-2}^{2} \frac{1}{x^2 + 4} dx$

$= [x]_{-2}^{2} - 8 \int_{-2}^{2} \frac{1}{x^2 + 2^2} dx$

ここで、$x = 2\tan\theta$ と置換すると、$dx = 2\sec^2\theta d\theta$ となり、$x^2 + 4 = 4\tan^2\theta + 4 = 4\sec^2\theta$ となります。

$x=-2$ のとき $\tan\theta = -1$ なので $\theta = -\pi/4$

$x=2$ のとき $\tan\theta = 1$ なので $\theta = \pi/4$

したがって、

$\int_{-2}^{2} \frac{1}{x^2 + 4} dx = \int_{-\pi/4}^{\pi/4} \frac{1}{4\sec^2\theta} 2\sec^2\theta d\theta = \frac{1}{2} \int_{-\pi/4}^{\pi/4} d\theta = \frac{1}{2} [\theta]_{-\pi/4}^{\pi/4} = \frac{1}{2}(\frac{\pi}{4} - (-\frac{\pi}{4})) = \frac{\pi}{4}$

したがって、

$\int_{-2}^{2} \frac{x^2 - 4}{x^2 + 4} dx = [x]_{-2}^{2} - 8 \int_{-2}^{2} \frac{1}{x^2 + 4} dx = (2 - (-2)) - 8(\frac{\pi}{4}) = 4 - 2\pi$

(2)

$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x \cos x}{\sin^3 x} dx$

$u = x$, $dv = \frac{\cos x}{\sin^3 x} dx$ と置換積分を行う。

$du = dx$

$v = \int \frac{\cos x}{\sin^3 x} dx$

$t = \sin x$ とおくと, $dt = \cos x dx$.

$\int \frac{\cos x}{\sin^3 x} dx = \int \frac{1}{t^3} dt = \int t^{-3} dt = \frac{t^{-2}}{-2} = -\frac{1}{2\sin^2 x}$

よって、$v = -\frac{1}{2\sin^2 x}$

$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x \cos x}{\sin^3 x} dx = [x (-\frac{1}{2\sin^2 x})]_{\frac{\pi}{4}}^{\frac{\pi}{3}} - \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (-\frac{1}{2\sin^2 x}) dx$

$= [-\frac{x}{2\sin^2 x}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} + \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sin^2 x} dx$

$= [-\frac{x}{2\sin^2 x}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} + \frac{1}{2} [-\cot x]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$

$= [-\frac{x}{2\sin^2 x} - \frac{1}{2} \cot x]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$

$= (-\frac{\pi/3}{2(3/4)} - \frac{1}{2}\frac{1}{\sqrt{3}}) - (-\frac{\pi/4}{2(1/2)} - \frac{1}{2}(1))$

$= (-\frac{2\pi}{9} - \frac{\sqrt{3}}{6}) - (-\frac{\pi}{4} - \frac{1}{2})$

$= -\frac{2\pi}{9} - \frac{\sqrt{3}}{6} + \frac{\pi}{4} + \frac{1}{2}$

$= \frac{\pi}{36} - \frac{\sqrt{3}}{6} + \frac{1}{2}$

$= \frac{\pi - 6\frac{\sqrt{3}}{6} + 18}{36} = \frac{\pi - \sqrt{3} + 18}{36}$

(3)

$\int_{0}^{\frac{\pi}{2}} \cos^2 x \sin^6 x dx = \int_{0}^{\frac{\pi}{2}} (\cos x)^2 (\sin x)^6 dx$

Wallisの公式を使う。

$\int_0^{\frac{\pi}{2}} \sin^m x \cos^n x dx = \frac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})} = \frac{\Gamma(\frac{6+1}{2})\Gamma(\frac{2+1}{2})}{2\Gamma(\frac{6+2+2}{2})} = \frac{\Gamma(\frac{7}{2})\Gamma(\frac{3}{2})}{2\Gamma(5)}$

$\Gamma(5) = 4! = 24$

$\Gamma(\frac{3}{2}) = \frac{1}{2}\Gamma(\frac{1}{2}) = \frac{\sqrt{\pi}}{2}$

$\Gamma(\frac{7}{2}) = \frac{5}{2} \Gamma(\frac{5}{2}) = \frac{5}{2} \frac{3}{2} \Gamma(\frac{3}{2}) = \frac{15}{4} \frac{\sqrt{\pi}}{2} = \frac{15\sqrt{\pi}}{8}$

$\int_{0}^{\frac{\pi}{2}} \cos^2 x \sin^6 x dx = \frac{\frac{15\sqrt{\pi}}{8}\frac{\sqrt{\pi}}{2}}{2\cdot 24} = \frac{\frac{15\pi}{16}}{48} = \frac{15\pi}{16 \cdot 48} = \frac{15\pi}{768} = \frac{5\pi}{256}$

3. 最終的な答え

(1) $4 - 2\pi$

(2) $\frac{\pi - \sqrt{3} + 18}{36}$

(3) $\frac{5\pi}{256}$