1. $\int \frac{1}{\cos x} dx$

はい、承知いたしました。3つの積分問題を解きます。

1. 問題の内容

1. $\int \frac{1}{\cos x} dx$

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

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

2. 解き方の手順

1. $\int \frac{1}{\cos x} dx$ の積分

$\frac{1}{\cos x} = \frac{\cos x}{\cos^2 x} = \frac{\cos x}{1 - \sin^2 x}$

$u = \sin x$ と置換すると、$du = \cos x dx$ なので、

$\int \frac{1}{\cos x} dx = \int \frac{\cos x}{1 - \sin^2 x} dx = \int \frac{1}{1 - u^2} du$

$\frac{1}{1 - u^2} = \frac{1}{(1 - u)(1 + u)} = \frac{A}{1 - u} + \frac{B}{1 + u}$

$1 = A(1 + u) + B(1 - u)$

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

$u = -1$ のとき、$1 = 2B$ より $B = \frac{1}{2}$

$\int \frac{1}{1 - u^2} du = \frac{1}{2} \int (\frac{1}{1 - u} + \frac{1}{1 + u}) du = \frac{1}{2} (-\ln|1 - u| + \ln|1 + u|) + C = \frac{1}{2} \ln|\frac{1 + u}{1 - u}| + C$

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

2. $\int \frac{1}{2 + 3 \cos x} dx$ の積分

$t = \tan \frac{x}{2}$ と置換する。

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

$\int \frac{1}{2 + 3 \cos x} dx = \int \frac{1}{2 + 3 \frac{1 - t^2}{1 + t^2}} \frac{2}{1 + t^2} dt = \int \frac{1}{\frac{2(1 + t^2) + 3(1 - t^2)}{1 + t^2}} \frac{2}{1 + t^2} dt = \int \frac{2}{2 + 2t^2 + 3 - 3t^2} dt = \int \frac{2}{5 - t^2} dt$

$\frac{2}{5 - t^2} = \frac{2}{(\sqrt{5} - t)(\sqrt{5} + t)} = \frac{A}{\sqrt{5} - t} + \frac{B}{\sqrt{5} + t}$

$2 = A(\sqrt{5} + t) + B(\sqrt{5} - t)$

$t = \sqrt{5}$ のとき、$2 = 2\sqrt{5}A$ より $A = \frac{1}{\sqrt{5}}$

$t = -\sqrt{5}$ のとき、$2 = 2\sqrt{5}B$ より $B = \frac{1}{\sqrt{5}}$

$\int \frac{2}{5 - t^2} dt = \frac{1}{\sqrt{5}} \int (\frac{1}{\sqrt{5} - t} + \frac{1}{\sqrt{5} + t}) dt = \frac{1}{\sqrt{5}} (\ln|\sqrt{5} + t| - \ln|\sqrt{5} - t|) + C = \frac{1}{\sqrt{5}} \ln|\frac{\sqrt{5} + t}{\sqrt{5} - t}| + C$

$\frac{1}{\sqrt{5}} \ln|\frac{\sqrt{5} + \tan \frac{x}{2}}{\sqrt{5} - \tan \frac{x}{2}}| + C$

3. $\int \frac{\sin x}{2(1 + \sin x)} dx$ の積分

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

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

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

3. 最終的な答え

1. $\int \frac{1}{\cos x} dx = \ln|\sec x + \tan x| + C$

2. $\int \frac{1}{2 + 3 \cos x} dx = \frac{1}{\sqrt{5}} \ln|\frac{\sqrt{5} + \tan \frac{x}{2}}{\sqrt{5} - \tan \frac{x}{2}}| + C$

3. $\int \frac{\sin x}{2(1 + \sin x)} dx = \frac{1}{2} (\sec x - \tan x + x) + C$