## 問題の解答

## 問題の解答

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1. 問題の内容

以下の4つの定積分の値を計算します。

(1) $\int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx$, (m, n は整数)

(2) $\int_{0}^{T} \sin(\frac{2m\pi}{T}t) \sin(\frac{2n\pi}{T}t) dt$, (m, n は整数)

(3) $\int_{0}^{\frac{1}{2}} \arcsin(x) dx$

(4) $\int_{0}^{1} \frac{\arctan(x)}{1+x^2} dx$

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2. 解き方の手順

**(1) $\int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx$**

三角関数の積を和に変換する公式 $\cos(A)\cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$ を用います。

$\int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx = \frac{1}{2}\int_{-\pi}^{\pi} [\cos((m+n)x) + \cos((m-n)x)] dx$

$m \neq n$ のとき、

$\int_{-\pi}^{\pi} \cos((m+n)x) dx = \frac{1}{m+n} [\sin((m+n)x)]_{-\pi}^{\pi} = 0$

$\int_{-\pi}^{\pi} \cos((m-n)x) dx = \frac{1}{m-n} [\sin((m-n)x)]_{-\pi}^{\pi} = 0$

したがって、 $m \neq n$ のとき、 $\int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx = 0$

$m = n$ のとき、

$\int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx = \frac{1}{2}\int_{-\pi}^{\pi} [\cos(2mx) + 1] dx$

$= \frac{1}{2} [\frac{1}{2m}\sin(2mx) + x]_{-\pi}^{\pi} = \frac{1}{2} [(\frac{1}{2m}\sin(2m\pi) + \pi) - (\frac{1}{2m}\sin(-2m\pi) - \pi)] = \frac{1}{2} [0 + \pi - 0 + \pi] = \pi$

**(2) $\int_{0}^{T} \sin(\frac{2m\pi}{T}t) \sin(\frac{2n\pi}{T}t) dt$**

三角関数の積を和に変換する公式 $\sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$ を用います。

$\int_{0}^{T} \sin(\frac{2m\pi}{T}t) \sin(\frac{2n\pi}{T}t) dt = \frac{1}{2} \int_{0}^{T} [\cos(\frac{2(m-n)\pi}{T}t) - \cos(\frac{2(m+n)\pi}{T}t)] dt$

$m \neq n$ のとき、

$\int_{0}^{T} \cos(\frac{2(m-n)\pi}{T}t) dt = \frac{T}{2(m-n)\pi}[\sin(\frac{2(m-n)\pi}{T}t)]_{0}^{T} = \frac{T}{2(m-n)\pi}[\sin(2(m-n)\pi) - \sin(0)] = 0$

$\int_{0}^{T} \cos(\frac{2(m+n)\pi}{T}t) dt = \frac{T}{2(m+n)\pi}[\sin(\frac{2(m+n)\pi}{T}t)]_{0}^{T} = \frac{T}{2(m+n)\pi}[\sin(2(m+n)\pi) - \sin(0)] = 0$

したがって、 $m \neq n$ のとき、 $\int_{0}^{T} \sin(\frac{2m\pi}{T}t) \sin(\frac{2n\pi}{T}t) dt = 0$

$m = n$ のとき、

$\int_{0}^{T} \sin(\frac{2m\pi}{T}t) \sin(\frac{2n\pi}{T}t) dt = \frac{1}{2} \int_{0}^{T} [1 - \cos(\frac{4m\pi}{T}t)] dt$

$= \frac{1}{2} [t - \frac{T}{4m\pi}\sin(\frac{4m\pi}{T}t)]_{0}^{T} = \frac{1}{2} [(T - \frac{T}{4m\pi}\sin(4m\pi)) - (0 - \frac{T}{4m\pi}\sin(0))] = \frac{1}{2} [T - 0 - 0 + 0] = \frac{T}{2}$

**(3) $\int_{0}^{\frac{1}{2}} \arcsin(x) dx$**

部分積分を行います。 $u = \arcsin(x)$, $dv = dx$ とすると、$du = \frac{1}{\sqrt{1-x^2}}dx$, $v = x$

$\int_{0}^{\frac{1}{2}} \arcsin(x) dx = [x\arcsin(x)]_{0}^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}} dx$

$= [\frac{1}{2}\arcsin(\frac{1}{2}) - 0\arcsin(0)] - \int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}} dx = \frac{1}{2} \cdot \frac{\pi}{6} - \int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}} dx = \frac{\pi}{12} - \int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}} dx$

ここで、$w = 1-x^2$ と置換すると、$dw = -2x dx$, $x dx = -\frac{1}{2} dw$

$x=0$ のとき $w=1$, $x=\frac{1}{2}$ のとき $w=1-\frac{1}{4} = \frac{3}{4}$

$\int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}} dx = \int_{1}^{\frac{3}{4}} \frac{-\frac{1}{2}}{\sqrt{w}} dw = -\frac{1}{2} \int_{1}^{\frac{3}{4}} w^{-\frac{1}{2}} dw = -\frac{1}{2} [2w^{\frac{1}{2}}]_{1}^{\frac{3}{4}} = -[w^{\frac{1}{2}}]_{1}^{\frac{3}{4}} = -[\sqrt{\frac{3}{4}} - \sqrt{1}] = -( \frac{\sqrt{3}}{2} - 1) = 1 - \frac{\sqrt{3}}{2}$

したがって、 $\int_{0}^{\frac{1}{2}} \arcsin(x) dx = \frac{\pi}{12} - (1 - \frac{\sqrt{3}}{2}) = \frac{\pi}{12} - 1 + \frac{\sqrt{3}}{2} = \frac{\pi + 6\sqrt{3} - 12}{12}$

**(4) $\int_{0}^{1} \frac{\arctan(x)}{1+x^2} dx$**

$u = \arctan(x)$ と置換すると、$du = \frac{1}{1+x^2} dx$

$x=0$ のとき $u=\arctan(0)=0$, $x=1$ のとき $u=\arctan(1)=\frac{\pi}{4}$

$\int_{0}^{1} \frac{\arctan(x)}{1+x^2} dx = \int_{0}^{\frac{\pi}{4}} u du = [\frac{1}{2}u^2]_{0}^{\frac{\pi}{4}} = \frac{1}{2} (\frac{\pi}{4})^2 - 0 = \frac{1}{2} \frac{\pi^2}{16} = \frac{\pi^2}{32}$

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3. 最終的な答え

(1) $m \neq n$ のとき $0$, $m = n$ のとき $\pi$

(2) $m \neq n$ のとき $0$, $m = n$ のとき $\frac{T}{2}$

(3) $\frac{\pi + 6\sqrt{3} - 12}{12}$

(4) $\frac{\pi^2}{32}$