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

はい、承知いたしました。与えられた積分問題を解いていきます。

1. 問題の内容

以下の5つの不定積分を計算します。

(1) $\int \frac{x}{\sqrt{5-x}} dx$

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

(3) $\int (\sin x + \cos x)^2 dx$

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

(5) $\int (\sin x + \sin 2x)^2 dx$

2. 解き方の手順

(1) $\int \frac{x}{\sqrt{5-x}} dx$

置換積分を行います。$t = 5 - x$ とおくと、$x = 5 - t$、$dx = -dt$ となります。

$\int \frac{x}{\sqrt{5-x}} dx = \int \frac{5-t}{\sqrt{t}} (-dt) = - \int (5t^{-1/2} - t^{1/2}) dt = - (5 \cdot 2t^{1/2} - \frac{2}{3}t^{3/2}) + C = -10\sqrt{t} + \frac{2}{3}t\sqrt{t} + C = -10\sqrt{5-x} + \frac{2}{3}(5-x)\sqrt{5-x} + C = -10\sqrt{5-x} + \frac{10}{3}\sqrt{5-x} - \frac{2}{3}x\sqrt{5-x} + C = -\frac{20}{3}\sqrt{5-x} - \frac{2}{3}x\sqrt{5-x} + C = -\frac{2}{3}(10+x)\sqrt{5-x} + C$

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

$\cos x = t$とおくと、$\sin x dx = -dt$となる。

$\int \frac{\sin x \cos^2 x}{1 + \cos x} dx = \int \frac{t^2}{1+t} (-dt) = - \int \frac{t^2}{t+1} dt = - \int (t-1+\frac{1}{t+1}) dt = -(\frac{1}{2}t^2 - t + \ln|t+1|) + C = -\frac{1}{2}\cos^2 x + \cos x - \ln|\cos x + 1| + C$

(3) $\int (\sin x + \cos x)^2 dx$

$(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x = 1 + \sin 2x$

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

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

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

(5) $\int (\sin x + \sin 2x)^2 dx$

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

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

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

$\int 4\sin^2 x \cos^2 xdx = \int (1-\cos 2x) + (1+\cos2x)/2 *$

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

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

3. 最終的な答え

(1) $\int \frac{x}{\sqrt{5-x}} dx = -\frac{2}{3}(x+10)\sqrt{5-x} + C$

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

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

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

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