与えられた6つの不定積分を計算する問題です。 (1) $\int \frac{dx}{1+\cos x}$ (2) $\int \frac{dx}{2+\cos x}$ (3) $\int \frac{1+\cos x}{(1+\sin x)^2} dx$ (4) $\int \frac{\cos x}{4+5\sin x} dx$ (5) $\int \frac{1}{2+\tan x} dx$ (6) $\int \frac{dx}{1+\sin x - \cos x}$

1. 問題の内容

与えられた6つの不定積分を計算する問題です。

(1) $\int \frac{dx}{1+\cos x}$

(2) $\int \frac{dx}{2+\cos x}$

(3) $\int \frac{1+\cos x}{(1+\sin x)^2} dx$

(4) $\int \frac{\cos x}{4+5\sin x} dx$

(5) $\int \frac{1}{2+\tan x} dx$

(6) $\int \frac{dx}{1+\sin x - \cos x}$

2. 解き方の手順

(1) $\int \frac{dx}{1+\cos x}$

半角の公式 $\cos x = 2\cos^2\frac{x}{2}-1 = 1-2\sin^2\frac{x}{2}$ を用いて、

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

$\int \frac{dx}{1+\cos x} = \int \frac{dx}{2\cos^2\frac{x}{2}} = \frac{1}{2} \int \sec^2\frac{x}{2} dx = \frac{1}{2} \cdot 2 \tan\frac{x}{2} + C = \tan\frac{x}{2} + C$

(2) $\int \frac{dx}{2+\cos x}$

$t = \tan\frac{x}{2}$ とおくと、$\cos x = \frac{1-t^2}{1+t^2}$ かつ $dx = \frac{2dt}{1+t^2}$

$\int \frac{dx}{2+\cos x} = \int \frac{1}{2+\frac{1-t^2}{1+t^2}} \frac{2dt}{1+t^2} = \int \frac{2dt}{2(1+t^2)+1-t^2} = \int \frac{2dt}{t^2+3} = \frac{2}{\sqrt{3}}\arctan\frac{t}{\sqrt{3}} + C = \frac{2}{\sqrt{3}}\arctan\frac{\tan\frac{x}{2}}{\sqrt{3}} + C$

(3) $\int \frac{1+\cos x}{(1+\sin x)^2} dx$

$\int \frac{1+\cos x}{(1+\sin x)^2} dx = \int \frac{1}{(1+\sin x)^2} dx + \int \frac{\cos x}{(1+\sin x)^2} dx$

$t = 1+\sin x$ とおくと、$dt = \cos x dx$

$\int \frac{\cos x}{(1+\sin x)^2} dx = \int \frac{dt}{t^2} = -\frac{1}{t} = -\frac{1}{1+\sin x}$

また、$\frac{1+\cos x}{(1+\sin x)^2} = \frac{1}{(1+\sin x)^2} + \frac{\cos x}{(1+\sin x)^2}$

$\int \frac{1+\cos x}{(1+\sin x)^2} dx = \int \frac{1+\cos x}{(1+\sin x)^2} dx = \int \frac{1}{(1+\sin x)^2} dx + \int \frac{\cos x}{(1+\sin x)^2} dx$

ここで、$\int \frac{\cos x}{(1+\sin x)^2} dx = -\frac{1}{1+\sin x}$

$\int \frac{1}{1+\sin x}dx = \int \frac{1-\sin x}{1-\sin^2 x} dx = \int \frac{1-\sin x}{\cos^2 x} dx = \int \sec^2 x - \sec x \tan x dx = \tan x - \sec x + C$

$\int \frac{dx}{(1+\sin x)^2} = \int \frac{1}{(1+\sin x)^2} dx = \frac{\sin x - 1}{\cos^2 x (1+\sin x)} + C$

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

(4) $\int \frac{\cos x}{4+5\sin x} dx$

$t = 4+5\sin x$ とおくと、$dt = 5\cos x dx$, $\cos x dx = \frac{dt}{5}$

$\int \frac{\cos x}{4+5\sin x} dx = \int \frac{1}{t} \frac{dt}{5} = \frac{1}{5} \ln |t| + C = \frac{1}{5} \ln |4+5\sin x| + C$

(5) $\int \frac{1}{2+\tan x} dx = \int \frac{\cos x}{2\cos x + \sin x} dx$

$I = \int \frac{\cos x}{2\cos x + \sin x} dx$ , $J = \int \frac{\sin x}{2\cos x + \sin x} dx$ とする。

$2I+J = \int \frac{2\cos x + \sin x}{2\cos x + \sin x} dx = \int dx = x+C_1$

$-I+2J = \int \frac{-(\cos x - 2\sin x)}{2\cos x + \sin x} dx$

$t = 2\cos x + \sin x$, $dt = -2\sin x + \cos x dx = -(\cos x - 2\sin x) dx$

$-I+2J = \int \frac{dt}{t} = \ln |t| + C_2 = \ln |2\cos x + \sin x| + C_2$

$4I+2J = 2x + 2C_1$, $-I+2J = \ln |2\cos x + \sin x| + C_2$

$5I = 2x - \ln |2\cos x + \sin x| + C_3$

$I = \frac{2}{5}x - \frac{1}{5}\ln |2\cos x + \sin x| + C$

(6) $\int \frac{dx}{1+\sin x - \cos x}$

$t = \tan\frac{x}{2}$ とおくと、$\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$, $dx = \frac{2dt}{1+t^2}$

$\int \frac{dx}{1+\sin x - \cos x} = \int \frac{1}{1+\frac{2t}{1+t^2} - \frac{1-t^2}{1+t^2}} \frac{2dt}{1+t^2} = \int \frac{2dt}{1+t^2 + 2t - (1-t^2)} = \int \frac{2dt}{2t^2+2t} = \int \frac{dt}{t(t+1)} = \int (\frac{1}{t} - \frac{1}{t+1}) dt = \ln |t| - \ln |t+1| + C = \ln |\frac{t}{t+1}| + C = \ln |\frac{\tan\frac{x}{2}}{\tan\frac{x}{2}+1}| + C$

3. 最終的な答え

(1) $\tan\frac{x}{2} + C$

(2) $\frac{2}{\sqrt{3}}\arctan\frac{\tan\frac{x}{2}}{\sqrt{3}} + C$

(3) $\frac{1}{1+\sin x} + C$

(4) $\frac{1}{5} \ln |4+5\sin x| + C$

(5) $\frac{2}{5}x - \frac{1}{5}\ln |2\cos x + \sin x| + C$

(6) $\ln |\frac{\tan\frac{x}{2}}{\tan\frac{x}{2}+1}| + C$