与えられた3つの定積分の値を求めます。 (1) $\int_{0}^{\frac{\pi}{4}} \frac{x}{\cos^2 x} dx$ (2) $\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x} dx$ (3) $\int_{0}^{2} \frac{2x+1}{\sqrt{x^2+4}} dx$

1. 問題の内容

与えられた3つの定積分の値を求めます。

(1) $\int_{0}^{\frac{\pi}{4}} \frac{x}{\cos^2 x} dx$

(2) $\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x} dx$

(3) $\int_{0}^{2} \frac{2x+1}{\sqrt{x^2+4}} dx$

2. 解き方の手順

(1)

部分積分を用いて計算します。

$u = x, dv = \frac{1}{\cos^2 x} dx$とすると、$du = dx, v = \tan x$となります。

よって、

$\int_{0}^{\frac{\pi}{4}} \frac{x}{\cos^2 x} dx = [x\tan x]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} \tan x dx$

$= \frac{\pi}{4} \tan \frac{\pi}{4} - 0\tan 0 - \int_{0}^{\frac{\pi}{4}} \tan x dx = \frac{\pi}{4} - \int_{0}^{\frac{\pi}{4}} \tan x dx$

$\int \tan x dx = \int \frac{\sin x}{\cos x} dx$

$t = \cos x$ とおくと $dt = -\sin x dx$より、

$\int \tan x dx = \int \frac{-dt}{t} = -\log |t| + C = -\log |\cos x| + C$

したがって、

$\int_{0}^{\frac{\pi}{4}} \tan x dx = [-\log |\cos x|]_{0}^{\frac{\pi}{4}} = -\log(\cos \frac{\pi}{4}) + \log(\cos 0) = -\log(\frac{1}{\sqrt{2}}) + \log(1) = -\log(2^{-\frac{1}{2}}) + 0 = \frac{1}{2}\log 2$

よって、

$\int_{0}^{\frac{\pi}{4}} \frac{x}{\cos^2 x} dx = \frac{\pi}{4} - \frac{1}{2}\log 2$

(2)

$\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{\cos^2 x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{1-\sin^2 x} dx$

$t = \sin x$ とおくと $dt = \cos x dx$

$x=0$のとき$t=0$、$x=\frac{\pi}{4}$のとき$t=\frac{1}{\sqrt{2}}$

$\int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{1-t^2} dt = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{(1-t)(1+t)} dt = \frac{1}{2} \int_{0}^{\frac{1}{\sqrt{2}}} (\frac{1}{1-t} + \frac{1}{1+t}) dt$

$= \frac{1}{2} [-\log|1-t| + \log|1+t|]_{0}^{\frac{1}{\sqrt{2}}} = \frac{1}{2}[\log|\frac{1+t}{1-t}|]_{0}^{\frac{1}{\sqrt{2}}} = \frac{1}{2} \log(\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}) - \frac{1}{2}\log(1) = \frac{1}{2} \log(\frac{\sqrt{2}+1}{\sqrt{2}-1}) = \frac{1}{2} \log(\frac{(\sqrt{2}+1)^2}{2-1}) = \frac{1}{2} \log((\sqrt{2}+1)^2) = \log(\sqrt{2}+1)$

(3)

$\int_{0}^{2} \frac{2x+1}{\sqrt{x^2+4}} dx = \int_{0}^{2} \frac{2x}{\sqrt{x^2+4}} dx + \int_{0}^{2} \frac{1}{\sqrt{x^2+4}} dx$

$\int_{0}^{2} \frac{2x}{\sqrt{x^2+4}} dx$ について、 $t = x^2+4$ とおくと、$dt = 2x dx$

$x=0$のとき$t=4$、$x=2$のとき$t=8$

$\int_{4}^{8} \frac{1}{\sqrt{t}} dt = [2\sqrt{t}]_{4}^{8} = 2\sqrt{8} - 2\sqrt{4} = 4\sqrt{2} - 4$

$\int_{0}^{2} \frac{1}{\sqrt{x^2+4}} dx = \int_{0}^{2} \frac{1}{\sqrt{x^2+2^2}} dx$

$x = 2\tan \theta$とおくと $dx = \frac{2}{\cos^2 \theta} d\theta$

$x=0$のとき$\theta=0$、$x=2$のとき$\tan \theta = 1$より$\theta=\frac{\pi}{4}$

$\int_{0}^{\frac{\pi}{4}} \frac{1}{\sqrt{4\tan^2 \theta + 4}} \frac{2}{\cos^2 \theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{1}{2\sqrt{\tan^2 \theta + 1}} \frac{2}{\cos^2 \theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{1}{\sqrt{\frac{1}{\cos^2 \theta}}} \frac{1}{\cos^2 \theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{1}{\frac{1}{\cos \theta}} \frac{1}{\cos^2 \theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{\cos \theta}{\cos^2 \theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{\cos \theta}{1 - \sin^2 \theta} d\theta$

(2)より、$\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x} dx = \log (\sqrt{2} + 1)$

よって、$\int_{0}^{2} \frac{2x+1}{\sqrt{x^2+4}} dx = 4\sqrt{2} - 4 + \log (\sqrt{2}+1)$

3. 最終的な答え

(1) $\frac{\pi}{4} - \frac{1}{2}\log 2$

(2) $\log(\sqrt{2}+1)$

(3) $4\sqrt{2} - 4 + \log (\sqrt{2}+1)$