与えられた4つの積分を計算します。 (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)

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

(2)

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

$\tan^2 x = \sec^2 x - 1$であるから、

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

(3)

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

(4)

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

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) $-\cot x - x + C$