$\int_{0}^{\frac{\pi}{3}} x \sin(x - \frac{\pi}{3}) dx$ を計算します。

1. 問題の内容

$\int_{0}^{\frac{\pi}{3}} x \sin(x - \frac{\pi}{3}) dx$ を計算します。

2. 解き方の手順

まず、三角関数の加法定理を用いて $\sin(x - \frac{\pi}{3})$ を展開します。

$\sin(x - \frac{\pi}{3}) = \sin(x) \cos(\frac{\pi}{3}) - \cos(x) \sin(\frac{\pi}{3}) = \frac{1}{2} \sin(x) - \frac{\sqrt{3}}{2} \cos(x)$

したがって、積分は次のようになります。

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

部分積分を用いてそれぞれの積分を計算します。

$\int x \sin(x) dx$:

$u = x$, $dv = \sin(x) dx$

$du = dx$, $v = -\cos(x)$

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

$\int x \cos(x) dx$:

$u = x$, $dv = \cos(x) dx$

$du = dx$, $v = \sin(x)$

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

したがって、

$\frac{1}{2} \int_{0}^{\frac{\pi}{3}} x \sin(x) dx = \frac{1}{2} [-x \cos(x) + \sin(x)]_{0}^{\frac{\pi}{3}} = \frac{1}{2} [-\frac{\pi}{3} \cos(\frac{\pi}{3}) + \sin(\frac{\pi}{3}) - (0)] = \frac{1}{2} [-\frac{\pi}{3} (\frac{1}{2}) + \frac{\sqrt{3}}{2}] = -\frac{\pi}{12} + \frac{\sqrt{3}}{4}$

$\frac{\sqrt{3}}{2} \int_{0}^{\frac{\pi}{3}} x \cos(x) dx = \frac{\sqrt{3}}{2} [x \sin(x) + \cos(x)]_{0}^{\frac{\pi}{3}} = \frac{\sqrt{3}}{2} [\frac{\pi}{3} \sin(\frac{\pi}{3}) + \cos(\frac{\pi}{3}) - (0 + 1)] = \frac{\sqrt{3}}{2} [\frac{\pi}{3} (\frac{\sqrt{3}}{2}) + \frac{1}{2} - 1] = \frac{\sqrt{3}}{2} [\frac{\pi \sqrt{3}}{6} - \frac{1}{2}] = \frac{\pi}{4} - \frac{\sqrt{3}}{4}$

$\int_{0}^{\frac{\pi}{3}} x \sin(x - \frac{\pi}{3}) dx = (-\frac{\pi}{12} + \frac{\sqrt{3}}{4}) - (\frac{\pi}{4} - \frac{\sqrt{3}}{4}) = -\frac{\pi}{12} - \frac{3\pi}{12} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = -\frac{4\pi}{12} + \frac{2\sqrt{3}}{4} = -\frac{\pi}{3} + \frac{\sqrt{3}}{2}$

3. 最終的な答え

$-\frac{\pi}{3} + \frac{\sqrt{3}}{2}$