難問の積分を解いてみた
問題 \( I=\displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{x^2+1+\sqrt{x^4+x^2+1}} \> dx \)を求めよ。
まず，
\[ \begin{array}{r@{\>}l}
x^4+x^2+1 & =x^4+2x^2+1-x^2 \\
& = (x^2+1)^2-x^2 \\
& = \{ (x^2 + 1)+x \}\{ (x^2 + 1)-x \} \\
& = (x^2+x+1)(x^2-x+1)
\end{array} \]
と変形できるので，
\[ \displaystyle I=\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{x^2+1+\sqrt{(x^2+x+1)(x^2-x+1)}} \> dx \]
二重根号を外すことを考えて，
\[ \begin{array}{rl}
I & \displaystyle =\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{\sqrt{2x^2+2+2\sqrt{(x^2+x+1)(x^2-x+1)}}}{\sqrt{2}} \> dx \\
& \displaystyle = \dfrac{1}{\sqrt{2}} \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{(x^2+x+1)+(x^2-x+1)+2\sqrt{(x^2+x+1)(x^2-x+1)}} \> dx \\
& \displaystyle = \dfrac{1}{\sqrt{2}} \int_{-\frac{1}{2}}^{\frac{1}{2}} (\sqrt{x^2+x+1}+\sqrt{x^2-x+1}) \> dx \\
& \displaystyle = \dfrac{1}{\sqrt{2}} \left( \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{x^2+x+1} \> dx + \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{x^2-x+1}) \> dx \right) \\
\end{array}\]
ここで，ルートの中の平方完成を考えて，
\[ \begin{array}{rl}
I & \displaystyle = \dfrac{1}{\sqrt{2}} \left\{ \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{ \left( x + \dfrac{1}{2} \right)^2+\dfrac{3}{4}} \> dx+\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{ \left( x-\dfrac{1}{2} \right)^2+\dfrac{3}{4}} \> dx \right\} \\
\end{array}\]
ここで，\(x+\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}\tan{\theta_1}\)，\(x-\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}\tan{\theta_2}\)とする。
また，\(\tan t = \dfrac{2}{\sqrt{3}} \left(0 \leqq t \leqq \dfrac{\pi}{2}\right)\)を満たす \( t \) を \(\alpha\) とすると，
\[\begin{array}{rl}
\dfrac{dx}{d\theta_1} & = \dfrac{\sqrt{3}}{2}\dfrac{1}{\cos^2\theta_1} \\
\dfrac{dx}{d\theta_2} & = \dfrac{\sqrt{3}}{2}\dfrac{1}{\cos^2\theta_2}
\end{array}\]
\[\begin{array}{c|ccc}
x & -\dfrac{1}{2} & \rightarrow & \dfrac{1}{2} \\ \hline
\theta_1 & 0 & \rightarrow & \alpha
\end{array} \hspace{5mm}
\begin{array}{c|ccc}
x & -\dfrac{1}{2} & \rightarrow & \dfrac{1}{2} \\ \hline
\theta_2 & -\alpha & \rightarrow & 0
\end{array}\]
となるので，
\[ \begin{array}{rl}
I & \displaystyle = \dfrac{1}{\sqrt{2}} \left( \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{ \dfrac{3}{4}\tan^2\theta_1+\dfrac{3}{4}} \> dx+\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{ \dfrac{3}{4}\tan^2\theta_2+\dfrac{3}{4}} \> dx \right) \\
& \displaystyle = \dfrac{1}{\sqrt{2}} \left( \int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{\sqrt{3}}{2}\sqrt{1+\tan^2\theta_1} \> dx+\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{\sqrt{3}}{2}\sqrt{1+\tan^2\theta_2} \> dx \right) \\
& \displaystyle = \dfrac{\sqrt{3}}{2\sqrt{2}} \left( \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\dfrac{1}{\cos^2\theta_1}} \> dx+\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\dfrac{1}{\cos^2\theta_2}} \> dx \right) \\
& \displaystyle = \dfrac{\sqrt{3}}{2\sqrt{2}} \left( \int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{1}{\cos\theta_1} \> dx+\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{1}{\cos\theta_2} \> dx \right) \\
& \displaystyle = \dfrac{\sqrt{3}}{2\sqrt{2}} \left( \int_{0}^{\alpha} \dfrac{1}{\cos\theta_1} \dfrac{\sqrt{3}}{2}\dfrac{d\theta_1}{\cos^2\theta_1}+\int_{-\alpha}^{0} \dfrac{1}{\cos\theta_2} \dfrac{\sqrt{3}}{2}\dfrac{d\theta_2}{\cos^2\theta_2} \right) \\
& \displaystyle = \dfrac{3}{4\sqrt{2}} \left( \int_{0}^{\alpha} \dfrac{d\theta_1}{\cos^3\theta_1}+\int_{-\alpha}^{0} \dfrac{d\theta_2}{\cos^3\theta_2} \right) \\
& \displaystyle = \dfrac{3}{4\sqrt{2}} \int_{-\alpha}^{\alpha} \dfrac{d\theta}{\cos^3\theta} \\
& \displaystyle = \dfrac{3}{2\sqrt{2}} \int_{0}^{\alpha} \dfrac{d\theta}{\cos^3\theta} \\
\end{array} \]
ここで，\(\dfrac{1}{\cos^3\theta}\)の不定積分を考える。
\[\begin{array}{rl}
\displaystyle \int \dfrac{d\theta}{\cos^3\theta}
& \displaystyle =\int \dfrac{1}{\cos^2\theta} \dfrac{1}{\cos\theta} d\theta \\
& \displaystyle =\int (\tan\theta)’\dfrac{1}{\cos\theta} d\theta \\
& \displaystyle = \dfrac{\tan\theta}{\cos\theta}-\int \tan\theta \left( \dfrac{1}{\cos\theta} \right)’ d\theta \\
& \displaystyle = \dfrac{\tan\theta}{\cos\theta}-\int \tan\theta \dfrac{\tan\theta}{\cos\theta} d\theta \\
& \displaystyle = \dfrac{\tan\theta}{\cos\theta}-\int \tan^2\theta \dfrac{1}{\cos\theta} d\theta \\
& \displaystyle = \dfrac{\tan\theta}{\cos\theta}-\int \left( \dfrac{1}{\cos^2\theta}-1 \right) \dfrac{1}{\cos\theta} d\theta \\
& \displaystyle = \dfrac{\tan\theta}{\cos\theta}-\int \dfrac{1}{\cos^3\theta} \> d\theta + \int \dfrac{1}{\cos\theta} \> d\theta \\
\end{array}\]
\[\begin{array}{rl}
\displaystyle 2\int \dfrac{1}{\cos^3\theta} \> d\theta
& \displaystyle = \dfrac{\tan\theta}{\cos\theta}+\int \dfrac{1}{\cos\theta} \> d\theta \\
\displaystyle \int \dfrac{1}{\cos^3\theta} \> d\theta
& \displaystyle = \dfrac{1}{2}\dfrac{\tan\theta}{\cos\theta}+\dfrac{1}{2}\int \dfrac{1}{\cos\theta} \> d\theta
\end{array}\]
さらに，\(\dfrac{1}{\cos\theta}\)の不定積分を考えると，
\[\begin{array}{rl}
\displaystyle \int \dfrac{d\theta}{\cos\theta}
& \displaystyle = \int \dfrac{\cos\theta}{\cos^2\theta} \> d\theta \\
& \displaystyle = \int \dfrac{\cos\theta}{1-\sin^2\theta} \> d\theta \\
& \displaystyle = \int \dfrac{\cos\theta}{(1+\sin\theta)(1-\sin\theta)} \> d\theta \\
& \displaystyle = \int \dfrac{1}{2} \left( \dfrac{\cos\theta}{1+\sin\theta} + \dfrac{\cos\theta}{1-\sin\theta}\right) \> d\theta \\
& \displaystyle = \dfrac{1}{2} \int \left( \dfrac{(1+\sin\theta)’}{1+\sin\theta}-\dfrac{(1-\sin\theta)’}{1-\sin\theta} \right) \> d\theta \\
& \displaystyle = \dfrac{1}{2} \left( \log|1+\sin\theta|-\log|1-\sin\theta| \right)+C \\
& \displaystyle = \dfrac{1}{2} \log\dfrac{1+\sin\theta}{1-\sin\theta}+C
\end{array}\]
なので，
\[\begin{array}{rl}
\displaystyle \int \dfrac{1}{\cos^3\theta} \> d\theta
& \displaystyle = \dfrac{1}{2}\dfrac{\tan\theta}{\cos\theta}+\dfrac{1}{4}\log\dfrac{1+\sin\theta}{1-\sin\theta}+C
\end{array}\]
\(\tan\alpha=\dfrac{2}{\sqrt{3}}\)より\(\sin\alpha=\dfrac{2}{\sqrt{7}}\)，\(\cos\alpha=\dfrac{\sqrt{3}}{\sqrt{7}}\)なので，
\[\begin{array}{rl}
I & \displaystyle = \dfrac{3}{2\sqrt{2}} \int_{0}^{\alpha} \dfrac{d\theta}{\cos^3\theta} \\
& =\dfrac{3}{2\sqrt{2}} \left[ \dfrac{1}{2}\dfrac{\tan\theta}{\cos\theta}+\dfrac{1}{4}\log\dfrac{1+\sin\theta}{1-\sin\theta} \right]_0^\alpha \\
& =\dfrac{3}{2\sqrt{2}} \left\{ \left( \dfrac{1}{2}\dfrac{2}{\sqrt{3}}\dfrac{\sqrt{7}}{\sqrt{3}}+\dfrac{1}{4}\log\dfrac{1+\frac{2}{\sqrt{7}}}{1-\frac{2}{\sqrt{7}}} \right)-0 \right\} \\
& =\dfrac{3}{2\sqrt{2}} \left( \dfrac{\sqrt{7}}{3}+\dfrac{1}{4}\log\dfrac{\sqrt{7}+2}{\sqrt{7}-2} \right) \\
& =\dfrac{\sqrt{14}}{4}+\dfrac{3\sqrt{2}}{16}\log\dfrac{\sqrt{7}+2}{\sqrt{7}-2}
\end{array}\]
問題は次の動画から。自分でも解いてみたので解答を残すことにしました。
\( \displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\left( x+\dfrac{1}{2} \right)^2+\dfrac{3}{4}} dx\)からの変数変換で，\(y=x+\dfrac{1}{2}\)とおいて，\(y+\sqrt{y^2+\dfrac{3}{4}} = t\)とするのは初めて見た気がするので，覚えておこうと思います。
\[\begin{array}{rl}
y+\sqrt{y^2+\dfrac{3}{4}} & =t \\
\sqrt{y^2+\dfrac{3}{4}} & =t-y \\
y^2+\dfrac{3}{4} & =t^2-2ty+y^2 \\
2ty & =t^2-\dfrac{3}{4} \\
8ty & =4t^2-3 \\
y & =\dfrac{4t^2-3}{8t} \\
\dfrac{dy}{dt} & = \dfrac{8t \cdot 8t-(4t^2-3) \cdot 8}{64t^2} \\
&= \dfrac{8t^2-4t^2+3}{8t^2} \\
&= \dfrac{4t^2+3}{8t^2} \\
\end{array}\]
となるから，
\[\begin{array}{rl}
\displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\left( x+\dfrac{1}{2} \right)^2+\dfrac{3}{4}} dx
& \displaystyle = \int_0^1 \sqrt{y^2+\dfrac{3}{4}} \> dy \\
& \displaystyle = \int_{0}^{1} (t-y) dy \\
& \displaystyle = \int_{\frac{\sqrt{3}}{2}}^{1+\frac{\sqrt{7}}{2}} \left( t-\dfrac{4t^2-3}{8t} \right) \dfrac{4t^2+3}{8t^2} dt \\
& \displaystyle = \int_{\frac{\sqrt{3}}{2}}^{1+\frac{\sqrt{7}}{2}} \dfrac{4t^2+3}{8t} \dfrac{4t^2+3}{8t^2} dt \\
& \displaystyle = \int_{\frac{\sqrt{3}}{2}}^{1+\frac{\sqrt{7}}{2}} \dfrac{(4t^2+3)^2}{64t^3} dt \\
& \displaystyle = \int_{\frac{\sqrt{3}}{2}}^{1+\frac{\sqrt{7}}{2}} \dfrac{16t^4+24t^2+9}{64t^3} dt \\
& \displaystyle = \dfrac{1}{64} \int_{\frac{\sqrt{3}}{2}}^{1+\frac{\sqrt{7}}{2}} (16t+24t^{-1}+9t^{-3}) dt \\
& \displaystyle = \dfrac{1}{64} \left[ 8t^2+24\log|t|-\dfrac{9}{2}t^{-2} \right]_{\frac{\sqrt{3}}{2}}^{1+\frac{\sqrt{7}}{2}} \\
& = \cdots \\
& = \dfrac{\sqrt{7}}{4}+\dfrac{3}{8}\log\dfrac{2+\sqrt{7}}{\sqrt{3}}
\end{array}\]
式は複雑ですが，三角関数を経由しないのがいいですね。
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