次の不定積分を求めます。 (1) $\int \cos^4 x dx$ (2) $\int \sin 2x \cdot \cos 2x \cdot \cos 4x dx$ (3) $\int \cos^3 x dx$ (4) $\int \frac{dx}{\sin x}$ (5) $\int \cos x \cos 3x dx$ (6) $\int \frac{dx}{1 + \sin x}$ (7) $\int x \cos^2 x dx$

1. 問題の内容

次の不定積分を求めます。

(1) $\int \cos^4 x dx$

(2) $\int \sin 2x \cdot \cos 2x \cdot \cos 4x dx$

(3) $\int \cos^3 x dx$

(4) $\int \frac{dx}{\sin x}$

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

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

(7) $\int x \cos^2 x dx$

2. 解き方の手順

(1) $\int \cos^4 x dx$

$\cos^2 x = \frac{1 + \cos 2x}{2}$ を用いる。

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

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

$\cos^4 x = \frac{1}{4}(1 + 2\cos 2x + \frac{1 + \cos 4x}{2}) = \frac{1}{4}(1 + 2\cos 2x + \frac{1}{2} + \frac{1}{2}\cos 4x) = \frac{1}{4}(\frac{3}{2} + 2\cos 2x + \frac{1}{2}\cos 4x) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$

$\int \cos^4 x dx = \int (\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x) dx = \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$

(2) $\int \sin 2x \cdot \cos 2x \cdot \cos 4x dx$

$\sin 2x \cos 2x = \frac{1}{2}\sin 4x$

$\int \frac{1}{2}\sin 4x \cos 4x dx = \int \frac{1}{4}\sin 8x dx = -\frac{1}{32}\cos 8x + C$

(3) $\int \cos^3 x dx$

$\cos^3 x = \cos^2 x \cos x = (1 - \sin^2 x)\cos x$

$\int (1 - \sin^2 x)\cos x dx$

$u = \sin x$, $du = \cos x dx$

$\int (1 - u^2)du = u - \frac{u^3}{3} + C = \sin x - \frac{\sin^3 x}{3} + C$

(4) $\int \frac{dx}{\sin x}$

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

$u = \cos x$, $du = -\sin x dx$

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

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

$\int \frac{dx}{\sin x} = \frac{1}{2}\ln|\tan^2 \frac{x}{2}| + C = \ln|\tan \frac{x}{2}| + C$

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

$\cos x \cos 3x = \frac{1}{2}(\cos(3x+x) + \cos(3x-x)) = \frac{1}{2}(\cos 4x + \cos 2x)$

$\int \frac{1}{2}(\cos 4x + \cos 2x) dx = \frac{1}{8}\sin 4x + \frac{1}{4}\sin 2x + C$

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

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

(7) $\int x \cos^2 x dx$

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

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

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

$\int x \cos 2x dx = x \frac{\sin 2x}{2} - \int \frac{\sin 2x}{2} dx = \frac{x\sin 2x}{2} + \frac{\cos 2x}{4}$

$\int x \cos^2 x dx = \frac{1}{4}x^2 + \frac{1}{2}(\frac{x\sin 2x}{2} + \frac{\cos 2x}{4}) + C = \frac{x^2}{4} + \frac{x\sin 2x}{4} + \frac{\cos 2x}{8} + C$

3. 最終的な答え

(1) $\frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$

(2) $-\frac{1}{32}\cos 8x + C$

(3) $\sin x - \frac{\sin^3 x}{3} + C$

(4) $\ln|\tan \frac{x}{2}| + C$

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

(6) $\tan x - \sec x + C$

(7) $\frac{x^2}{4} + \frac{x\sin 2x}{4} + \frac{\cos 2x}{8} + C$