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

1. 問題の内容

次の不定積分を求めます。

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

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

2. 解き方の手順

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

被積分関数を部分分数分解します。

$\frac{1 + \sin x}{\sin x (1 + \cos x)} = \frac{A}{\sin x} + \frac{B}{1 + \cos x}$ とおくと、

$1 + \sin x = A(1 + \cos x) + B \sin x$

$x = 0$ のとき、 $1 = 2A$ より $A = \frac{1}{2}$

$x = \frac{\pi}{2}$ のとき、 $2 = A + B$ より $B = 2 - A = \frac{3}{2}$

したがって、

$\frac{1 + \sin x}{\sin x (1 + \cos x)} = \frac{1}{2\sin x} + \frac{3}{2(1 + \cos x)}$

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

$= \frac{1}{2} \int \csc x dx + \frac{3}{2} \int \frac{1}{1 + \cos x} dx$

$\int \csc x dx = -\ln|\csc x + \cot x| + C_1$

$\int \frac{1}{1 + \cos x} dx = \int \frac{1 - \cos x}{1 - \cos^2 x} dx = \int \frac{1 - \cos x}{\sin^2 x} dx = \int \csc^2 x dx - \int \frac{\cos x}{\sin^2 x} dx = -\cot x + \csc x + C_2$

したがって、

$\int \frac{1 + \sin x}{\sin x (1 + \cos x)} dx = -\frac{1}{2}\ln|\csc x + \cot x| + \frac{3}{2}(-\cot x + \csc x) + C$

$= -\frac{1}{2}\ln|\csc x + \cot x| - \frac{3}{2}\cot x + \frac{3}{2}\csc x + C$

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

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

$\sin^2 x + \cos^2 x = 1$ より、 $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - t^2}$

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

この積分は複雑なので、別の方法を試みます。

$t = \tan(\frac{x}{2})$ とおくと、

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

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

$= \int \frac{4t}{(1+t^2+2t)(3+3t^2+1-t^2)} dt = \int \frac{4t}{(1+t)^2 (4+2t^2)} dt = \int \frac{2t}{(1+t)^2 (2+t^2)} dt$

部分分数分解すると、$\frac{2t}{(1+t)^2 (2+t^2)} = \frac{A}{1+t} + \frac{B}{(1+t)^2} + \frac{Ct+D}{2+t^2}$

$2t = A(1+t)(2+t^2) + B(2+t^2) + (Ct+D)(1+t)^2$

$2t = A(2+t^2+2t+t^3) + B(2+t^2) + (Ct+D)(1+2t+t^2)$

$2t = At^3 + (2A+C)t^2 + (2A+2C+D)t + (2A+2B+D)$

$A=0, 2A+C=0, 2A+2C+D=2, 2A+2B+D=0$

$A=0, C=0, D=2, B=-1$

$\frac{2t}{(1+t)^2 (2+t^2)} = \frac{-1}{(1+t)^2} + \frac{2}{2+t^2}$

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

$\int \frac{2}{2+t^2} dt = \int \frac{2}{2(1+(\frac{t}{\sqrt{2}})^2)} dt = \int \frac{1}{1+(\frac{t}{\sqrt{2}})^2} dt = \sqrt{2} \arctan(\frac{t}{\sqrt{2}}) + C_2$

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

3. 最終的な答え

(1) $-\frac{1}{2}\ln|\csc x + \cot x| - \frac{3}{2}\cot x + \frac{3}{2}\csc x + C$

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