(5) $\int_{0}^{\frac{3\pi}{2}} \cos^{7}x \, dx$ (6) $\int_{\pi}^{2\pi} \sin^{8}x \, dx$ (7) $\int_{0}^{\frac{\pi}{2}} 2\cos 2x \cos x \, dx$ (8) $\int_{0}^{1} x^{2} \sin^{-1}x \, dx$

はい、承知しました。画像にある4つの積分問題を解きます。

1. 問題の内容

(5) $\int_{0}^{\frac{3\pi}{2}} \cos^{7}x \, dx$

(6) $\int_{\pi}^{2\pi} \sin^{8}x \, dx$

(7) $\int_{0}^{\frac{\pi}{2}} 2\cos 2x \cos x \, dx$

(8) $\int_{0}^{1} x^{2} \sin^{-1}x \, dx$

2. 解き方の手順

(5) $\int_{0}^{\frac{3\pi}{2}} \cos^{7}x \, dx$

$\cos^{7}x$は奇関数なので、$\cos^{7}x = \cos^{6}x \cdot \cos x = (\cos^{2}x)^{3} \cdot \cos x = (1-\sin^{2}x)^{3} \cos x$と変形する。

$u = \sin x$と置くと、$du = \cos x dx$となる。積分範囲は$x=0$のとき$u = \sin 0 = 0$、$x = \frac{3\pi}{2}$のとき$u = \sin \frac{3\pi}{2} = -1$となる。

$\int_{0}^{\frac{3\pi}{2}} \cos^{7}x \, dx = \int_{0}^{-1} (1-u^{2})^{3} \, du = \int_{0}^{-1} (1 - 3u^{2} + 3u^{4} - u^{6}) \, du$

$= [u - u^{3} + \frac{3}{5}u^{5} - \frac{1}{7}u^{7}]_{0}^{-1} = (-1 - (-1) + \frac{3}{5}(-1) - \frac{1}{7}(-1)) - (0) = -1 + 1 - \frac{3}{5} + \frac{1}{7} = -\frac{21}{35} + \frac{5}{35} = -\frac{16}{35}$.

(6) $\int_{\pi}^{2\pi} \sin^{8}x \, dx$

$\sin(x + \pi) = -\sin x$より、$\sin^{8}(x + \pi) = \sin^{8}x$である。

$I = \int_{\pi}^{2\pi} \sin^{8}x \, dx = \int_{0}^{\pi} \sin^{8}(x + \pi) \, dx = \int_{0}^{\pi} \sin^{8}x \, dx$.

$\sin(\pi - x) = \sin x$より、$\sin^{8}(\pi - x) = \sin^{8}x$。

$I = \int_{0}^{\pi} \sin^{8}x \, dx = \int_{0}^{\frac{\pi}{2}} \sin^{8}x \, dx + \int_{\frac{\pi}{2}}^{\pi} \sin^{8}x \, dx = \int_{0}^{\frac{\pi}{2}} \sin^{8}x \, dx + \int_{0}^{\frac{\pi}{2}} \sin^{8}(\pi-x) \, dx = 2\int_{0}^{\frac{\pi}{2}} \sin^{8}x \, dx$.

$\int_{0}^{\frac{\pi}{2}} \sin^{n}x \, dx = \frac{n-1}{n} \frac{n-3}{n-2} ... \frac{1}{2} \frac{\pi}{2}$ (if n is even)

$\int_{0}^{\frac{\pi}{2}} \sin^{8}x \, dx = \frac{7}{8} \frac{5}{6} \frac{3}{4} \frac{1}{2} \frac{\pi}{2} = \frac{35\pi}{256}$.

Thus, $I = 2\int_{0}^{\frac{\pi}{2}} \sin^{8}x \, dx = 2(\frac{35\pi}{256}) = \frac{35\pi}{128}$.

(7) $\int_{0}^{\frac{\pi}{2}} 2\cos 2x \cos x \, dx$

$2\cos 2x \cos x = \cos(2x+x) + \cos(2x-x) = \cos 3x + \cos x$.

$\int_{0}^{\frac{\pi}{2}} 2\cos 2x \cos x \, dx = \int_{0}^{\frac{\pi}{2}} (\cos 3x + \cos x) \, dx = [\frac{1}{3}\sin 3x + \sin x]_{0}^{\frac{\pi}{2}} = (\frac{1}{3}\sin \frac{3\pi}{2} + \sin \frac{\pi}{2}) - (0) = \frac{1}{3}(-1) + 1 = 1 - \frac{1}{3} = \frac{2}{3}$.

(8) $\int_{0}^{1} x^{2} \sin^{-1}x \, dx$

部分積分を用いる。$u = \sin^{-1}x$, $dv = x^{2} dx$とすると、$du = \frac{1}{\sqrt{1-x^{2}}} dx$, $v = \frac{x^{3}}{3}$。

$\int_{0}^{1} x^{2} \sin^{-1}x \, dx = [\frac{x^{3}}{3} \sin^{-1}x]_{0}^{1} - \int_{0}^{1} \frac{x^{3}}{3} \frac{1}{\sqrt{1-x^{2}}} \, dx = \frac{1}{3} \sin^{-1}(1) - 0 - \frac{1}{3} \int_{0}^{1} \frac{x^{3}}{\sqrt{1-x^{2}}} \, dx = \frac{1}{3} \frac{\pi}{2} - \frac{1}{3} \int_{0}^{1} \frac{x^{3}}{\sqrt{1-x^{2}}} \, dx$.

$t = 1-x^{2}$と置くと、$x^{2} = 1 - t$, $dt = -2x \, dx$。$x = 0$のとき$t = 1$, $x = 1$のとき$t = 0$。

$\int_{0}^{1} \frac{x^{3}}{\sqrt{1-x^{2}}} \, dx = \int_{0}^{1} \frac{x^{2}x}{\sqrt{1-x^{2}}} \, dx = \int_{1}^{0} \frac{1-t}{\sqrt{t}} (-\frac{1}{2}) \, dt = \frac{1}{2} \int_{0}^{1} \frac{1-t}{\sqrt{t}} \, dt = \frac{1}{2} \int_{0}^{1} (t^{-\frac{1}{2}} - t^{\frac{1}{2}}) \, dt = \frac{1}{2} [2t^{\frac{1}{2}} - \frac{2}{3}t^{\frac{3}{2}}]_{0}^{1} = \frac{1}{2} (2 - \frac{2}{3}) = \frac{1}{2} (\frac{4}{3}) = \frac{2}{3}$.

$\int_{0}^{1} x^{2} \sin^{-1}x \, dx = \frac{\pi}{6} - \frac{1}{3} (\frac{2}{3}) = \frac{\pi}{6} - \frac{2}{9}$.

3. 最終的な答え

(5) $-\frac{16}{35}$

(6) $\frac{35\pi}{128}$

(7) $\frac{2}{3}$

(8) $\frac{\pi}{6} - \frac{2}{9}$