次の定積分の値を求める問題です。ここでは、(1) $\int_{0}^{3} \frac{x+1}{x^2+3} dx$, (2) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin x \cos x} dx$, (3) $\int_{0}^{1} \frac{1}{e^x + e^{-x}} dx$, (4) $\int_{1}^{e} x \ln x dx$, (5) $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx$, $J = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} dx$ について解きます。

1. 問題の内容

次の定積分の値を求める問題です。ここでは、(1) $\int_{0}^{3} \frac{x+1}{x^2+3} dx$, (2) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin x \cos x} dx$, (3) $\int_{0}^{1} \frac{1}{e^x + e^{-x}} dx$, (4) $\int_{1}^{e} x \ln x dx$, (5) $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx$, $J = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} dx$ について解きます。

2. 解き方の手順

(1) $\int_{0}^{3} \frac{x+1}{x^2+3} dx$

$x^2+3 = t$ と置換すると、$2x dx = dt$, $x dx = \frac{1}{2} dt$.

$x=0$ のとき $t=3$, $x=3$ のとき $t=12$.

$\int_{0}^{3} \frac{x+1}{x^2+3} dx = \int_{0}^{3} \frac{x}{x^2+3} dx + \int_{0}^{3} \frac{1}{x^2+3} dx = \frac{1}{2}\int_{3}^{12} \frac{1}{t} dt + \int_{0}^{3} \frac{1}{x^2+(\sqrt{3})^2} dx$.

$\frac{1}{2} \left[ \ln t \right]_{3}^{12} + \frac{1}{\sqrt{3}} \left[ \arctan \frac{x}{\sqrt{3}} \right]_{0}^{3} = \frac{1}{2}(\ln 12 - \ln 3) + \frac{1}{\sqrt{3}}(\arctan \sqrt{3} - \arctan 0) = \frac{1}{2} \ln \frac{12}{3} + \frac{1}{\sqrt{3}} (\frac{\pi}{3} - 0) = \frac{1}{2} \ln 4 + \frac{\pi}{3\sqrt{3}} = \ln 2 + \frac{\pi}{3\sqrt{3}}$.

(2) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin x \cos x} dx$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin x \cos x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{2}{2 \sin x \cos x} dx = 2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin 2x} dx = 2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc 2x dx$.

$t = 2x$ とすると $dt = 2 dx$. $x = \frac{\pi}{6}$ のとき $t = \frac{\pi}{3}$, $x = \frac{\pi}{3}$ のとき $t = \frac{2\pi}{3}$.

$\int \csc x dx = -\ln |\csc x + \cot x|$ なので、

$2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc 2x dx = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \csc t dt = \left[ -\ln |\csc t + \cot t| \right]_{\frac{\pi}{3}}^{\frac{2\pi}{3}} = -\ln |\csc \frac{2\pi}{3} + \cot \frac{2\pi}{3}| + \ln |\csc \frac{\pi}{3} + \cot \frac{\pi}{3}| = -\ln |\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}| + \ln |\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}| = -\ln \frac{1}{\sqrt{3}} + \ln \frac{3}{\sqrt{3}} = \ln \sqrt{3} + \ln \sqrt{3} = 2 \ln \sqrt{3} = \ln 3$.

(3) $\int_{0}^{1} \frac{1}{e^x + e^{-x}} dx = \int_{0}^{1} \frac{e^x}{e^{2x} + 1} dx$

$u = e^x$ とすると、$du = e^x dx$. $x = 0$ のとき $u = 1$, $x = 1$ のとき $u = e$.

$\int_{1}^{e} \frac{1}{u^2+1} du = \left[ \arctan u \right]_{1}^{e} = \arctan e - \arctan 1 = \arctan e - \frac{\pi}{4}$.

(4) $\int_{1}^{e} x \ln x dx$

部分積分を用いて、

$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x dx = \frac{x^2}{2} \ln x - \frac{x^2}{4}$.

$\int_{1}^{e} x \ln x dx = \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_{1}^{e} = (\frac{e^2}{2} \ln e - \frac{e^2}{4}) - (\frac{1^2}{2} \ln 1 - \frac{1^2}{4}) = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2+1}{4}$.

(5) $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx$, $J = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} dx$

$I + J = \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x} dx = \int_{0}^{\frac{\pi}{2}} 1 dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$.

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx$ において、$x = \frac{\pi}{2} - t$ と置換すると、$dx = -dt$. $x=0$ のとき $t = \frac{\pi}{2}$, $x = \frac{\pi}{2}$ のとき $t = 0$.

$I = \int_{\frac{\pi}{2}}^{0} \frac{\sin(\frac{\pi}{2}-t)}{\sin(\frac{\pi}{2}-t) + \cos(\frac{\pi}{2}-t)} (-dt) = \int_{0}^{\frac{\pi}{2}} \frac{\cos t}{\cos t + \sin t} dt = J$.

したがって、$I = J$ である。

$I + J = 2I = \frac{\pi}{2}$ より、$I = \frac{\pi}{4}$。 よって、$J = \frac{\pi}{4}$。

3. 最終的な答え

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

(2) $\ln 3$

(3) $\arctan e - \frac{\pi}{4}$

(4) $\frac{e^2+1}{4}$

(5) $I = \frac{\pi}{4}$, $J = \frac{\pi}{4}$