与えられた4つの不定積分を計算します。 (1) $\int (\sin x - 5\cos x) dx$ (2) $\int (\tan^2 x - 1) dx$ (3) $\int \frac{\tan^2 x + 2}{\sin^2 x} dx$ (4) $\int \frac{1}{\tan^2 x} dx$

1. 問題の内容

与えられた4つの不定積分を計算します。

(1) $\int (\sin x - 5\cos x) dx$

(2) $\int (\tan^2 x - 1) dx$

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

(4) $\int \frac{1}{\tan^2 x} dx$

2. 解き方の手順

(1) $\int (\sin x - 5\cos x) dx$

$\sin x$ の積分は $-\cos x$ で、$\cos x$ の積分は $\sin x$ です。

$\int (\sin x - 5\cos x) dx = \int \sin x dx - 5\int \cos x dx = -\cos x - 5\sin x + C$

(2) $\int (\tan^2 x - 1) dx$

三角関数の公式 $\tan^2 x + 1 = \frac{1}{\cos^2 x}$ を利用すると、$\tan^2 x - 1 = \frac{1}{\cos^2 x} - 2$ となります。

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

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

$\tan x = \frac{\sin x}{\cos x}$ なので、$\frac{\tan^2 x + 2}{\sin^2 x} = \frac{\frac{\sin^2 x}{\cos^2 x} + 2}{\sin^2 x} = \frac{\sin^2 x + 2\cos^2 x}{\sin^2 x \cos^2 x} = \frac{\sin^2 x + \cos^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1 + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x} + \frac{1}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} + \frac{1}{\sin^2 x} = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} + \frac{1}{\sin^2 x} = \frac{1}{\cos^2 x} + \frac{2}{\sin^2 x}$

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

(4) $\int \frac{1}{\tan^2 x} dx$

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

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

3. 最終的な答え

(1) $-\cos x - 5\sin x + C$

(2) $\tan x - 2x + C$

(3) $\tan x - 2\cot x + C$

(4) $-\cot x - x + C$