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

はい、承知いたしました。問題の定積分を解いていきます。

1. 問題の内容

以下の4つの定積分の値を求めます。

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

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

(3) $\int_0^{\pi/4} \cos^3 x dx$

(4) $\int_0^{\sqrt{3}} \frac{1 + x}{1 + x^2} dx$

2. 解き方の手順

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

部分積分を2回用います。

$u = x^2$, $dv = e^{2x} dx$ とすると、$du = 2x dx$, $v = \frac{1}{2} e^{2x}$

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

次に、$\int x e^{2x} dx$ を部分積分します。

$u = x$, $dv = e^{2x} dx$ とすると、$du = dx$, $v = \frac{1}{2} e^{2x}$

$\int x e^{2x} dx = \frac{1}{2} x e^{2x} - \int \frac{1}{2} e^{2x} dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}$

したがって、

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

$\int_0^2 x^2 e^{2x} dx = \left[ \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} \right]_0^2 = \frac{1}{2} (4) e^4 - \frac{1}{2} (2) e^4 + \frac{1}{4} e^4 - \frac{1}{4} = 2 e^4 - e^4 + \frac{1}{4} e^4 - \frac{1}{4} = e^4 + \frac{1}{4} e^4 - \frac{1}{4} = \frac{5}{4} e^4 - \frac{1}{4}$

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

$x = a \sin \theta$ と置換すると、$dx = a \cos \theta d\theta$

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

積分範囲は、$x = 0 \Rightarrow \theta = 0$, $x = a/2 \Rightarrow \sin \theta = 1/2 \Rightarrow \theta = \pi/6$

$\int_0^{a/2} \frac{dx}{\sqrt{a^2 - x^2}} = \int_0^{\pi/6} \frac{a \cos \theta}{a \cos \theta} d\theta = \int_0^{\pi/6} d\theta = [\theta]_0^{\pi/6} = \frac{\pi}{6}$

(3) $\int_0^{\pi/4} \cos^3 x dx$

$\cos^3 x = \cos x (1 - \sin^2 x)$

$\int_0^{\pi/4} \cos^3 x dx = \int_0^{\pi/4} \cos x (1 - \sin^2 x) dx$

$u = \sin x$ と置換すると、$du = \cos x dx$

積分範囲は、$x = 0 \Rightarrow u = 0$, $x = \pi/4 \Rightarrow u = \frac{1}{\sqrt{2}}$

$\int_0^{\pi/4} \cos x (1 - \sin^2 x) dx = \int_0^{\frac{1}{\sqrt{2}}} (1 - u^2) du = \left[ u - \frac{1}{3} u^3 \right]_0^{\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}} - \frac{1}{3} \frac{1}{2\sqrt{2}} = \frac{1}{\sqrt{2}} - \frac{1}{6\sqrt{2}} = \frac{6 - 1}{6\sqrt{2}} = \frac{5}{6\sqrt{2}} = \frac{5\sqrt{2}}{12}$

(4) $\int_0^{\sqrt{3}} \frac{1 + x}{1 + x^2} dx$

$\int_0^{\sqrt{3}} \frac{1 + x}{1 + x^2} dx = \int_0^{\sqrt{3}} \frac{1}{1 + x^2} dx + \int_0^{\sqrt{3}} \frac{x}{1 + x^2} dx$

$\int_0^{\sqrt{3}} \frac{1}{1 + x^2} dx = [\arctan x]_0^{\sqrt{3}} = \arctan \sqrt{3} - \arctan 0 = \frac{\pi}{3}$

$\int_0^{\sqrt{3}} \frac{x}{1 + x^2} dx = \frac{1}{2} \int_0^{\sqrt{3}} \frac{2x}{1 + x^2} dx = \frac{1}{2} [\ln (1 + x^2)]_0^{\sqrt{3}} = \frac{1}{2} (\ln 4 - \ln 1) = \frac{1}{2} \ln 4 = \ln 2$

$\int_0^{\sqrt{3}} \frac{1 + x}{1 + x^2} dx = \frac{\pi}{3} + \ln 2$

3. 最終的な答え

(1) $\frac{5}{4} e^4 - \frac{1}{4}$

(2) $\frac{\pi}{6}$

(3) $\frac{5\sqrt{2}}{12}$

(4) $\frac{\pi}{3} + \ln 2$