定積分 $\int_{0}^{\frac{3\pi}{4}} \{(1+x)\sin x + (1-x)\cos x\} dx$ を計算する。

1. 問題の内容

定積分 $\int_{0}^{\frac{3\pi}{4}} \{(1+x)\sin x + (1-x)\cos x\} dx$ を計算する。

2. 解き方の手順

まず、積分の中身を展開します。

$(1+x)\sin x + (1-x)\cos x = \sin x + x\sin x + \cos x - x\cos x$

積分を分解します。

$\int_{0}^{\frac{3\pi}{4}} \{(1+x)\sin x + (1-x)\cos x\} dx = \int_{0}^{\frac{3\pi}{4}} (\sin x + \cos x + x\sin x - x\cos x) dx$

$= \int_{0}^{\frac{3\pi}{4}} \sin x dx + \int_{0}^{\frac{3\pi}{4}} \cos x dx + \int_{0}^{\frac{3\pi}{4}} x\sin x dx - \int_{0}^{\frac{3\pi}{4}} x\cos x dx$

部分積分を使って、$\int x\sin x dx$ と $\int x\cos x dx$ を計算します。

$\int x\sin x dx = -x\cos x + \int \cos x dx = -x\cos x + \sin x + C_1$

$\int x\cos x dx = x\sin x - \int \sin x dx = x\sin x + \cos x + C_2$

したがって、

$\int_{0}^{\frac{3\pi}{4}} x\sin x dx = [-x\cos x + \sin x]_{0}^{\frac{3\pi}{4}} = -\frac{3\pi}{4}\cos(\frac{3\pi}{4}) + \sin(\frac{3\pi}{4}) - (0) = -\frac{3\pi}{4}(-\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2} = \frac{3\pi\sqrt{2}}{8} + \frac{\sqrt{2}}{2}$

$\int_{0}^{\frac{3\pi}{4}} x\cos x dx = [x\sin x + \cos x]_{0}^{\frac{3\pi}{4}} = \frac{3\pi}{4}\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4}) - (0 + 1) = \frac{3\pi}{4}(\frac{\sqrt{2}}{2}) - \frac{\sqrt{2}}{2} - 1 = \frac{3\pi\sqrt{2}}{8} - \frac{\sqrt{2}}{2} - 1$

$\int_{0}^{\frac{3\pi}{4}} \sin x dx = [-\cos x]_{0}^{\frac{3\pi}{4}} = -\cos(\frac{3\pi}{4}) + \cos(0) = -(-\frac{\sqrt{2}}{2}) + 1 = \frac{\sqrt{2}}{2} + 1$

$\int_{0}^{\frac{3\pi}{4}} \cos x dx = [\sin x]_{0}^{\frac{3\pi}{4}} = \sin(\frac{3\pi}{4}) - \sin(0) = \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2}$

したがって、

$\int_{0}^{\frac{3\pi}{4}} (\sin x + \cos x + x\sin x - x\cos x) dx = (\frac{\sqrt{2}}{2} + 1) + (\frac{\sqrt{2}}{2}) + (\frac{3\pi\sqrt{2}}{8} + \frac{\sqrt{2}}{2}) - (\frac{3\pi\sqrt{2}}{8} - \frac{\sqrt{2}}{2} - 1) = \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + \frac{3\pi\sqrt{2}}{8} + \frac{\sqrt{2}}{2} - \frac{3\pi\sqrt{2}}{8} + \frac{\sqrt{2}}{2} + 1 = 2 + 2\sqrt{2}$

3. 最終的な答え

$2\sqrt{2} + 2$