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

1. 問題の内容

与えられた4つの積分を計算します。

(1) $\int (\csc x + \tan x) \cos x \, dx$

(2) $\int \frac{\sin^2 4x}{\cos^2 4x} \, dx$

(3) $\int \cos 5x \cos 3x \, dx$

(4) $\int \cos^2 \frac{x}{4} \, dx$

2. 解き方の手順

(1) $\int (\csc x + \tan x) \cos x \, dx$

まず、$\csc x = \frac{1}{\sin x}$と$\tan x = \frac{\sin x}{\cos x}$を使って式を簡略化します。

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

$\int \cot x \, dx = \ln |\sin x|$

$\int \sin x \, dx = -\cos x$

したがって、$\int (\cot x + \sin x) \, dx = \ln |\sin x| - \cos x + C$

(2) $\int \frac{\sin^2 4x}{\cos^2 4x} \, dx$

$\frac{\sin^2 4x}{\cos^2 4x} = \tan^2 4x$

$\tan^2 4x = \sec^2 4x - 1$

$\int \tan^2 4x \, dx = \int (\sec^2 4x - 1) \, dx = \int \sec^2 4x \, dx - \int 1 \, dx$

$\int \sec^2 4x \, dx = \frac{1}{4} \tan 4x$

$\int 1 \, dx = x$

したがって、$\int (\sec^2 4x - 1) \, dx = \frac{1}{4} \tan 4x - x + C$

(3) $\int \cos 5x \cos 3x \, dx$

積和の公式: $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$

$\cos 5x \cos 3x = \frac{1}{2} [\cos(5x+3x) + \cos(5x-3x)] = \frac{1}{2} [\cos 8x + \cos 2x]$

$\int \cos 5x \cos 3x \, dx = \frac{1}{2} \int (\cos 8x + \cos 2x) \, dx = \frac{1}{2} (\int \cos 8x \, dx + \int \cos 2x \, dx)$

$\int \cos 8x \, dx = \frac{1}{8} \sin 8x$

$\int \cos 2x \, dx = \frac{1}{2} \sin 2x$

したがって、$\frac{1}{2} (\frac{1}{8} \sin 8x + \frac{1}{2} \sin 2x) + C = \frac{1}{16} \sin 8x + \frac{1}{4} \sin 2x + C$

(4) $\int \cos^2 \frac{x}{4} \, dx$

$\cos^2 A = \frac{1 + \cos 2A}{2}$

$\cos^2 \frac{x}{4} = \frac{1 + \cos \frac{x}{2}}{2}$

$\int \cos^2 \frac{x}{4} \, dx = \int \frac{1 + \cos \frac{x}{2}}{2} \, dx = \frac{1}{2} \int (1 + \cos \frac{x}{2}) \, dx = \frac{1}{2} (\int 1 \, dx + \int \cos \frac{x}{2} \, dx)$

$\int 1 \, dx = x$

$\int \cos \frac{x}{2} \, dx = 2 \sin \frac{x}{2}$

したがって、$\frac{1}{2} (x + 2 \sin \frac{x}{2}) + C = \frac{x}{2} + \sin \frac{x}{2} + C$

3. 最終的な答え

(1) $\ln |\sin x| - \cos x + C$

(2) $\frac{1}{4} \tan 4x - x + C$

(3) $\frac{1}{16} \sin 8x + \frac{1}{4} \sin 2x + C$

(4) $\frac{x}{2} + \sin \frac{x}{2} + C$