次の不定積分を求めます。 (1) $\int (1 + \sin x)^3 \cos x dx$ (2) $\int \frac{(1 - \cos x) \sin x}{1 + \cos x} dx$ (3) $\int \frac{\sin 2x}{1 + \sin^2 x} dx$ (4) $\int \frac{\tan x}{1 + \cos x} dx$ (5) $\int \frac{\tan x}{1 + \tan x} dx$ (6) $\int (1 + \tan x)^3 dx$

1. 問題の内容

次の不定積分を求めます。

(1) $\int (1 + \sin x)^3 \cos x dx$

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

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

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

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

(6) $\int (1 + \tan x)^3 dx$

2. 解き方の手順

(1) $t = 1 + \sin x$ とおくと、$dt = \cos x dx$ なので、

$\int (1 + \sin x)^3 \cos x dx = \int t^3 dt = \frac{1}{4} t^4 + C = \frac{1}{4} (1 + \sin x)^4 + C$

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

ここで、$\sin x = \frac{\sin x (1 + \cos x)}{1 + \cos x} = \frac{\sin x + \sin x \cos x}{1 + \cos x}$ より、

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

$t = 1 + \cos x$ とおくと、$dt = - \sin x dx$ なので、

$2 \int \frac{\sin x}{1 + \cos x} dx = -2 \int \frac{dt}{t} = -2 \log |t| + C_1 = -2 \log |1 + \cos x| + C_1 = -2 \log (1 + \cos x) + C_1$ (∵ $1 + \cos x > 0$)

また、$\int \sin x dx = - \cos x + C_2$ なので、

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

(3) $t = 1 + \sin^2 x$ とおくと、$dt = 2 \sin x \cos x dx = \sin 2x dx$ なので、

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

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

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

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

ここで、$\frac{1}{t(1 + t)} = \frac{A}{t} + \frac{B}{1 + t}$ と部分分数分解すると、$1 = A(1 + t) + Bt = (A + B) t + A$ より、$A = 1$, $B = -1$ なので、

$\frac{1}{t(1 + t)} = \frac{1}{t} - \frac{1}{1 + t}$

したがって、$- \int \frac{dt}{t (1 + t)} = - \int (\frac{1}{t} - \frac{1}{1 + t}) dt = - \log |t| + \log |1 + t| + C = - \log |\cos x| + \log |1 + \cos x| + C = \log \left| \frac{1 + \cos x}{\cos x} \right| + C = \log \left| \frac{1}{\cos x} + 1 \right| + C = \log |\sec x + 1| + C$

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

ここで、$\int \frac{\sin x}{\sin x + \cos x} dx = A \int \frac{\sin x + \cos x}{\sin x + \cos x} dx + B \int \frac{\cos x - \sin x}{\sin x + \cos x} dx$

$\sin x = A (\sin x + \cos x) + B (\cos x - \sin x) = (A - B) \sin x + (A + B) \cos x$ より、$A - B = 1$, $A + B = 0$ なので、$A = \frac{1}{2}$, $B = - \frac{1}{2}$

したがって、$\int \frac{\sin x}{\sin x + \cos x} dx = \frac{1}{2} \int 1 dx - \frac{1}{2} \int \frac{\cos x - \sin x}{\sin x + \cos x} dx$

$t = \sin x + \cos x$ とおくと、$dt = (\cos x - \sin x) dx$ なので、

$\frac{1}{2} \int 1 dx - \frac{1}{2} \int \frac{\cos x - \sin x}{\sin x + \cos x} dx = \frac{1}{2} x - \frac{1}{2} \int \frac{dt}{t} = \frac{1}{2} x - \frac{1}{2} \log |t| + C = \frac{1}{2} x - \frac{1}{2} \log |\sin x + \cos x| + C$

(6) $\int (1 + \tan x)^3 dx = \int (1 + 3 \tan x + 3 \tan^2 x + \tan^3 x) dx = \int (1 + 3 \tan x + 3 \tan^2 x + \tan x \tan^2 x) dx$

$= \int (1 + 3 \tan x + 3 \tan^2 x + \tan x (\sec^2 x - 1)) dx = \int (1 + 2 \tan x + 3 \tan^2 x + \tan x \sec^2 x) dx$

$= \int (1 + 2 \tan x + 3 (\sec^2 x - 1) + \tan x \sec^2 x) dx = \int (-2 + 2 \tan x + 3 \sec^2 x + \tan x \sec^2 x) dx$

$= -2x + 2 \int \tan x dx + 3 \tan x + \frac{1}{2} \tan^2 x + C$

$= -2x + 2 \int \frac{\sin x}{\cos x} dx + 3 \tan x + \frac{1}{2} \tan^2 x + C$

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

$2 \int \frac{\sin x}{\cos x} dx = -2 \int \frac{dt}{t} = -2 \log |t| + C_1 = -2 \log |\cos x| + C_1$

したがって、$\int (1 + \tan x)^3 dx = -2x - 2 \log |\cos x| + 3 \tan x + \frac{1}{2} \tan^2 x + C$

3. 最終的な答え

(1) $\frac{1}{4} (1 + \sin x)^4 + C$

(2) $\cos x - 2 \log (1 + \cos x) + C$

(3) $\log (1 + \sin^2 x) + C$

(4) $\log |\sec x + 1| + C$

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

(6) $-2x - 2 \log |\cos x| + 3 \tan x + \frac{1}{2} \tan^2 x + C$