次の定積分の値を求めます。 (1) $\int_0^1 e^{2x} dx$ (2) $\int_0^1 \frac{1}{\sqrt{2-x^2}} dx$ (3) $\int_0^{\pi} x\sin x dx$ (4) $\int_0^2 \frac{2x}{x^2+4} dx$ (5) $\int_0^{\frac{\pi}{2}} \cos^4 x dx$

1. 問題の内容

次の定積分の値を求めます。

(1) $\int_0^1 e^{2x} dx$

(2) $\int_0^1 \frac{1}{\sqrt{2-x^2}} dx$

(3) $\int_0^{\pi} x\sin x dx$

(4) $\int_0^2 \frac{2x}{x^2+4} dx$

(5) $\int_0^{\frac{\pi}{2}} \cos^4 x dx$

2. 解き方の手順

(1)

$\int e^{2x} dx = \frac{1}{2} e^{2x} + C$

したがって、

$\int_0^1 e^{2x} dx = \left[ \frac{1}{2} e^{2x} \right]_0^1 = \frac{1}{2} e^2 - \frac{1}{2} e^0 = \frac{1}{2}e^2 - \frac{1}{2} = \frac{e^2-1}{2}$

(2)

$x = \sqrt{2} \sin \theta$ とおくと、$dx = \sqrt{2} \cos \theta d\theta$

$x=0$のとき$\theta=0$, $x=1$のとき$\sin \theta = \frac{1}{\sqrt{2}}$ より $\theta = \frac{\pi}{4}$

$\sqrt{2-x^2} = \sqrt{2-2\sin^2 \theta} = \sqrt{2} \sqrt{1-\sin^2 \theta} = \sqrt{2} \cos \theta$

$\int_0^1 \frac{1}{\sqrt{2-x^2}} dx = \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{2} \cos \theta} \sqrt{2} \cos \theta d\theta = \int_0^{\frac{\pi}{4}} 1 d\theta = \left[ \theta \right]_0^{\frac{\pi}{4}} = \frac{\pi}{4}$

(3)

部分積分を行う。$\int x \sin x dx = x(-\cos x) - \int 1 \cdot (-\cos x) dx = -x\cos x + \sin x + C$

したがって、

$\int_0^{\pi} x \sin x dx = \left[ -x\cos x + \sin x \right]_0^{\pi} = (-\pi \cos \pi + \sin \pi) - (0\cos 0 + \sin 0) = (-\pi(-1) + 0) - (0+0) = \pi$

(4)

$t = x^2+4$ とおくと、$dt = 2x dx$

$x=0$のとき$t=4$, $x=2$のとき$t=8$

$\int_0^2 \frac{2x}{x^2+4} dx = \int_4^8 \frac{1}{t} dt = \left[ \ln |t| \right]_4^8 = \ln 8 - \ln 4 = \ln \frac{8}{4} = \ln 2$

(5)

$\cos^2 x = \frac{1+\cos 2x}{2}$

$\cos^4 x = (\cos^2 x)^2 = (\frac{1+\cos 2x}{2})^2 = \frac{1}{4}(1+2\cos 2x + \cos^2 2x) = \frac{1}{4}(1+2\cos 2x + \frac{1+\cos 4x}{2}) = \frac{1}{4}(1+2\cos 2x + \frac{1}{2}+\frac{1}{2}\cos 4x) = \frac{1}{4}(\frac{3}{2}+2\cos 2x + \frac{1}{2}\cos 4x) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$

$\int \cos^4 x dx = \int (\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x) dx = \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$

$\int_0^{\frac{\pi}{2}} \cos^4 x dx = \left[ \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x \right]_0^{\frac{\pi}{2}} = (\frac{3}{8}\cdot\frac{\pi}{2} + \frac{1}{4}\sin \pi + \frac{1}{32}\sin 2\pi) - (0+0+0) = \frac{3\pi}{16} + 0 + 0 = \frac{3\pi}{16}$

3. 最終的な答え

(1) $\frac{e^2-1}{2}$

(2) $\frac{\pi}{4}$

(3) $\pi$

(4) $\ln 2$

(5) $\frac{3\pi}{16}$