三角関数の難問積分を脳死解法で解いてみた

デプシロンイルタ＠高校積分 さんが Twitterで出題していた積分の問題 を解きます。

私「分母が奇数乗だから見当がつかないな・・・」
私「うーん、これは$ \tan{\frac{x}{2}} = t $！(脳死)」
私「で、そのあとどうするんだ」

導出

というわけで$ \sin{x}, \cos{x}, dx $を$ t $で表すところから始めます。

$$ \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}} = 1 $$

なので、

$$ \frac{1}{\tan^2{\frac{x}{2}}} + 1 = \frac{1}{\sin^2{\frac{x}{2}}} $$

$$ 1 + \tan^2{\frac{x}{2}} = \frac{1}{\cos^2{\frac{x}{2}}} $$

よって

$$ \sin^2{\frac{x}{2}} = \frac{1}{1 + \frac{1}{t^2}} = \frac{t^2}{1 + t^2} $$

$$ \cos^2{\frac{x}{2}} = \frac{1}{1 + t^2} $$

であるから

$$ \sin{x} = 2\sin{\frac{x}{2}}\cos{\frac{x}{2}} = 2\sqrt{\frac{t^2}{(1 + t^2)^2}} = \frac{2t}{1 + t^2} \left(\because 0 \leq x \leq \frac{\pi}{4}で\sin{x} \geq 0, t \geq 0 \right)$$

$$ \cos{x} = \cos^2{\frac{x}{2}} - \sin^2{\frac{x}{2}} = \frac{1 - t^2}{1 + t^2} $$

$$ \frac{1}{2 \cos^2{\frac{x}{2}}} dx = dt \hspace{3mm} \therefore dx = \frac{2}{1 + t^2} dt $$

と求まりました。

計算

$ \tan{\frac{x}{2}} = t $とし、$ a = \tan{\frac{\pi}{8}} $とする。

\begin{align*} & \int_0^{\frac{\pi}{4}} \frac{\sin^2{x}}{\sin^3{x} + \cos^3{x}} dx \\ =& \int_0^a \frac{\left(\frac{2t}{1 + t^2}\right)^2}{\left(\frac{2t}{1 + t^2}\right)^3 + \left(\frac{1 - t^2}{1 + t^2}\right)^3} \cdot \frac{2}{1 + t^2} dt \\ =& \int_0^a \frac{(2t)^2 (1 + t^2)}{(2t)^3 + (1-t^2)^3} \cdot \frac{2}{1 + t^2} dt \\ =& \int_0^a \frac{2(2t)^2}{(2t)^3 + (1-t^2)^3} dt \\ =& \int_0^a \frac{8t^2}{-t^6+3t^4+8t^3-3t^2+1} dt \end{align*}

部分分数分解

最後に出てきた分数を部分分数分解するために、分母を因数分解します。そのために、(分母)=0の解を複素数の範囲で求めます。

\begin{align*} -t^6+3t^4+8t^3-3t^2+1 &= 0 \\ t^3\left(-t^3 + 3t + 8 - \frac{3}{t} + \frac{1}{t^3} \right) &= 0 \\ -t^3 + 3t + 8 - \frac{3}{t} + \frac{1}{t^3} &= 0 (\because t = 0は解ではない)\\ -t^3 + 3t - \frac{3}{t} + \frac{1}{t^3} &= -8 \\ t^3 - 3t + \frac{3}{t} - \frac{1}{t^3} &= 8 \\ \left(t - \frac{1}{t} \right)^3 &= 8 \\ t - \frac{1}{t} &= 2, 2e^{\frac{\tau}{3}}, 2e^{\frac{2\tau}{3}} (\tau = 2\pi) \\ t - \frac{1}{t} &= 2, -1 + \sqrt{3}i, -1 - \sqrt{3}i \end{align*}

$ t - \frac{1}{t} = 2 $のとき$ t^2 - 2t - 1 = 0 $より$ t = 1 \pm \sqrt{2} $

$ t - \frac{1}{t} = -1 \pm_1 \sqrt{3}i $のとき(ここから同じ添え字は複号同順、異なる添え字は複号任意)

\begin{align*} t - \frac{1}{t} &= -1 \pm_1 \sqrt{3}i \\ t^2 + (1 \mp_1 \sqrt{3}i) t - 1 &= 0 \\ \left(t + \frac{1 \mp_1 \sqrt{3}i}{2}\right)^2 - \frac{-1 \mp_1 \sqrt{3}i}{2} - 1 &= 0 \\ \left(t + \frac{1 \mp_1 \sqrt{3}i}{2}\right)^2 &= \frac{-1 \mp_1 \sqrt{3}i}{2} + 1 \\ \left(t + \frac{1 \mp_1 \sqrt{3}i}{2}\right)^2 &= \frac{1 \mp_1 \sqrt{3}i}{2} \\ t + \frac{1 \mp_1 \sqrt{3}i}{2} &= \pm_2 \frac{\sqrt{3} \mp_1 i}{2} \\ t &= - \frac{1 \mp_1 \sqrt{3}i}{2} \pm_2 \frac{\sqrt{3} \mp_1 i}{2} \\ t &= - \frac{1 \pm_2 \sqrt{3}}{2} \pm_1 \left( \frac{\sqrt{3} \pm_2 1}{2} \right)i \end{align*}

(ここまで同じ添え字は複号同順、異なる添え字は複号任意)

よって、

\begin{align*} & -t^6+3t^4+8t^3-3t^2+1 \\ =& -(t - (1+\sqrt{2}))(t - (1-\sqrt{2}))\\ \cdot&\left(t - \left(- \frac{1 + \sqrt{3}}{2} + \left( \frac{\sqrt{3} + 1}{2} \right)i\right)\right)\left(t - \left(- \frac{1 + \sqrt{3}}{2} - \left( \frac{\sqrt{3} + 1}{2} \right)i\right)\right)\\ \cdot&\left(t - \left(- \frac{1 - \sqrt{3}}{2} + \left( \frac{\sqrt{3} - 1}{2} \right)i\right)\right)\left(t - \left(- \frac{1 - \sqrt{3}}{2} - \left( \frac{\sqrt{3} - 1}{2} \right)i\right)\right)\\ =& -(t - (1+\sqrt{2}))(t - (1-\sqrt{2}))(t^2 + (1+\sqrt{3})t + (2+\sqrt{3}))(t^2 + (1-\sqrt{3})t + (2-\sqrt{3})) \end{align*}

と実数の範囲で因数分解できます。そこで、

\begin{align*} & \frac{8t^2}{-t^6+3t^4+8t^3-3t^2+1} \\ =& \frac{-8t^2}{-(-t^6+3t^4+8t^3-3t^2+1)} \\ =& \frac{A}{t - (1+\sqrt{2})} + \frac{B}{t - (1-\sqrt{2})} + \frac{Ct+D}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} + \frac{Et+F}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} \end{align*}

となる実数$ A $～$ F $を求めます。

$$ (t - (1-\sqrt{2}))(t^2 + (1+\sqrt{3})t + (2+\sqrt{3}))(t^2 + (1-\sqrt{3})t + (2-\sqrt{3})) \\ = t^5 + (1+\sqrt{2}) t^4 + 2\sqrt{2} t^3 + (-4+2\sqrt{2}) t^2 + (3-2\sqrt{2}) t + (-1+\sqrt{2}) $$

$$ (t - (1+\sqrt{2}))(t^2 + (1+\sqrt{3})t + (2+\sqrt{3}))(t^2 + (1-\sqrt{3})t + (2-\sqrt{3})) \\ = t^5 + (1-\sqrt{2}) t^4 - 2\sqrt{2} t^3 + (-4-2\sqrt{2}) t^2 + (3+2\sqrt{2}) t + (-1-\sqrt{2}) $$

$$ (t - (1+\sqrt{2}))(t - (1-\sqrt{2}))(t^2 + (1-\sqrt{3})t + (2-\sqrt{3})) \\ = t^4 + (-1-\sqrt{3}) t^3 + (-1+\sqrt{3}) t^2 + (-5+3\sqrt{3}) t + (-2+\sqrt{3}) $$

$$ (t - (1+\sqrt{2}))(t - (1-\sqrt{2}))(t^2 + (1+\sqrt{3})t + (2+\sqrt{3})) \\ = t^4 + (-1+\sqrt{3}) t^3 + (-1-\sqrt{3}) t^2 + (-5-3\sqrt{3}) t + (-2-\sqrt{3}) $$

なので、

$$ \frac{A}{t - (1+\sqrt{2})} + \frac{B}{t - (1-\sqrt{2})} + \frac{Ct+D}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} + \frac{Et+F}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} $$

の分子は

\begin{align*} & (A+B+C+E) t^5 \\ + & ((1+\sqrt{2})A + (1-\sqrt{2})B + (-1-\sqrt{3})C + D + (-1+\sqrt{3})E + F) t^4 \\ + & (2\sqrt{2}A - 2\sqrt{2}B + (-1+\sqrt{3})C + (-1-\sqrt{3})D + (-1-\sqrt{3})E + (-1+\sqrt{3})F) t^3 \\ + & ((-4+2\sqrt{2})A + (-4-2\sqrt{2})B + (-5+3\sqrt{3})C + (-1+\sqrt{3})D + (-5-3\sqrt{3})E + (-1-\sqrt{3})F) t^2 \\ + & ((3-2\sqrt{2})A + (3+2\sqrt{2})B + (-2+\sqrt{3})C + (-5+3\sqrt{3})D + (-2-\sqrt{3})E + (-5-3\sqrt{3})F ) t^1 \\ + & ((-1+\sqrt{2})A + (-1-\sqrt{2})B + (-2+\sqrt{3})D + (-2-\sqrt{3})F t^0 \end{align*}

となります。

あとは、これが$ t^2 $の係数だけ$ -8 $になって、残りは$ 0 $になればよいですね。これはこのエグい連立方程式を何らかの方法で解くことで求まります。ここでは結果だけ示します。

$$ A = -\frac{\sqrt{2}}{6}, B=\frac{\sqrt{2}}{6}, C=-\frac{\sqrt{3}}{3}, D=\frac{-\sqrt{3}-2}{3}, E=\frac{\sqrt{3}}{3}, F=\frac{\sqrt{3}-2}{3} $$

よって、

\begin{align*} & \frac{8t^2}{-t^6+3t^4+8t^3-3t^2+1} \\ =& -\frac{\sqrt{2}}{6} \cdot \frac{1}{t - (1+\sqrt{2})} \\ +& \frac{\sqrt{2}}{6} \cdot \frac{1}{t - (1-\sqrt{2})} \\ +& \left(-\frac{\sqrt{3}}{3} t + \frac{-\sqrt{3}-2}{3} \right)\frac{1}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} \\ +& \left(\frac{\sqrt{3}}{3} t + \frac{\sqrt{3}-2}{3} \right)\frac{1}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} \\ \end{align*}

と表せます。

tan(π/8)を求める

積分のために$ \tan{\frac{\pi}{8}} $を求めます。と言ってもこれは簡単で、

\begin{align*} \frac{2x}{1-x^2} &= 1 \\ 2x &= 1-x^2 \\ x^2 + 2x - 1 &= 0 \\ x &= -1 \pm \sqrt{2} \\ \end{align*}

と$ \tan{\frac{\pi}{8}} > 0 $から、$ \tan{\frac{\pi}{8}} = \sqrt{2} - 1 $がわかります。

積分の続き

あとはウイニングランです。

第1項

\begin{align*} & \int_0^a -\frac{\sqrt{2}}{6} \cdot \frac{1}{t - (1+\sqrt{2})} dt \\ =& -\frac{\sqrt{2}}{6} \int_0^a \frac{1}{t - (1+\sqrt{2})} dt \\ =& -\frac{\sqrt{2}}{6} [\log{|t - (1+\sqrt{2})|}]_0^a \\ =& -\frac{\sqrt{2}}{6} \log\frac{2}{1+\sqrt{2}} \end{align*}

第2項

\begin{align*} & \int_0^a \frac{\sqrt{2}}{6} \cdot \frac{1}{t - (1-\sqrt{2})} dt \\ =& \frac{\sqrt{2}}{6} \int_0^a \frac{1}{t - (1-\sqrt{2})} dt \\ =& \frac{\sqrt{2}}{6} [\log{|t - (1-\sqrt{2})|}]_0^a \\ =& \frac{\sqrt{2}}{6} \log\frac{2(\sqrt{2}-1)}{\sqrt{2}-1} \\ =& \frac{\sqrt{2}}{6} \log{2} \end{align*}

第3項

$$ \int \frac{1}{(x+p)^2+q^2} dx = \frac{1}{q} \arctan\left(\frac{x+p}{q}\right)+(積分定数), \frac{2}{1+\sqrt{3}} = \sqrt{3}-1 $$
に注意して積分を行います。

\begin{align*} & \left(-\frac{\sqrt{3}}{3} t + \frac{-\sqrt{3}-2}{3} \right)\frac{1}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} \\ =& \frac{-\frac{\sqrt{3}}{6}(2t + (1+\sqrt{3})) + \frac{-1-\sqrt{3}}{6}}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} \\ =& \frac{-\frac{\sqrt{3}}{6}(2t + (1+\sqrt{3}))}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} + \frac{\frac{-1-\sqrt{3}}{6}}{\left(t + \frac{1+\sqrt{3}}{2} \right)^2 + \frac{2+\sqrt{3}}{2}} \\ =& \frac{-\frac{\sqrt{3}}{6}(2t + (1+\sqrt{3}))}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} + \frac{\frac{-1-\sqrt{3}}{6}}{\left(t + \frac{1+\sqrt{3}}{2} \right)^2 + \left(\frac{1+\sqrt{3}}{2}\right)^2} \\ \end{align*}

であるから

\begin{align*} & \int_0^a \left(-\frac{\sqrt{3}}{3} t + \frac{-\sqrt{3}-2}{3} \right)\frac{1}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} dt \\ =& \int_0^a \left( \frac{-\frac{\sqrt{3}}{6}(2t + (1+\sqrt{3}))}{t^2 + (1+\sqrt{3})t + (2+\sqrt{3})} + \frac{\frac{-1-\sqrt{3}}{6}}{\left(t + \frac{1+\sqrt{3}}{2} \right)^2 + \left(\frac{1+\sqrt{3}}{2}\right)^2} \right) dt \\ =& -\frac{\sqrt{3}}{6} \left[ \log({t^2 + (1+\sqrt{3})t + (2+\sqrt{3})}) \right]_0^a + \frac{-1-\sqrt{3}}{6} \cdot \frac{2}{1+\sqrt{3}} \left[ \arctan\left(t(\sqrt{3}-1) + 1 \right) \right]_0^a \\ =& -\frac{\sqrt{3}}{6}(\log{(4-\sqrt{2}+\sqrt{6})} - \log{(2+\sqrt{3})}) - \frac{1}{3}\left( \arctan\left(a(\sqrt{3}-1) + 1 \right) - \frac{\pi}{4} \right) \end{align*}

第4項

$$ \int \frac{1}{(x+p)^2+q^2} dx = \frac{1}{q} \arctan\left(\frac{x+p}{q}\right)+(積分定数), \frac{2}{1-\sqrt{3}} = -\sqrt{3}-1 $$
に注意して積分を行います。

\begin{align*} & \left(\frac{\sqrt{3}}{3} t + \frac{\sqrt{3}-2}{3} \right)\frac{1}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} \\ =& \frac{\frac{\sqrt{3}}{6}(2t + (1-\sqrt{3})) + \frac{-1+\sqrt{3}}{6}}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} \\ =& \frac{\frac{\sqrt{3}}{6}(2t + (1-\sqrt{3}))}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} + \frac{\frac{-1-\sqrt{3}}{6}}{\left(t - \frac{1-\sqrt{3}}{2} \right)^2 + \frac{2-\sqrt{3}}{2}} \\ =& \frac{\frac{\sqrt{3}}{6}(2t + (1-\sqrt{3}))}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} + \frac{\frac{-1+\sqrt{3}}{6}}{\left(t - \frac{1-\sqrt{3}}{2} \right)^2 + \left(\frac{1-\sqrt{3}}{2}\right)^2} \\ \end{align*}

であるから

\begin{align*} & \int_0^a \left(\frac{\sqrt{3}}{3} t + \frac{\sqrt{3}-2}{3} \right)\frac{1}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} dt \\ =& \int_0^a \left( \frac{\frac{\sqrt{3}}{6}(2t + (1-\sqrt{3}))}{t^2 + (1-\sqrt{3})t + (2-\sqrt{3})} + \frac{\frac{-1+\sqrt{3}}{6}}{\left(t - \frac{1-\sqrt{3}}{2} \right)^2 + \left(\frac{1-\sqrt{3}}{2}\right)^2} \right) dt \\ =& \frac{\sqrt{3}}{6} \left[ \log({t^2 + (1-\sqrt{3})t + (2-\sqrt{3})}) \right]_0^a + \frac{-1+\sqrt{3}}{6} \cdot \frac{2}{1-\sqrt{3}} \left[ \arctan\left(t(-\sqrt{3}-1) + 1 \right) \right]_0^a \\ =& \frac{\sqrt{3}}{6}\left(\log{(4-\sqrt{2}-\sqrt{6})} - \log{(2-\sqrt{3})} \right) - \frac{1}{3}\left( \arctan\left(a(-\sqrt{3}-1) + 1 \right) - \frac{\pi}{4} \right) \end{align*}

$\sqrt{3}$に関して第3項と「共役」の関係になっています。

足し合わせる

上で求めた4つの値を全部足せば積分が求まります。第1項と第2項の和は

$$ -\frac{\sqrt{2}}{6} \log\frac{2}{1+\sqrt{2}} + \frac{\sqrt{2}}{6} \log{2} = \frac{\sqrt{2}}{6}\log{(1+\sqrt{2})} $$

で与えられます。

第3項と第4項を結果の項別に足すと、

\begin{align*} & -\frac{\sqrt{3}}{6}\log{(4-\sqrt{2}+\sqrt{6})} + \frac{\sqrt{3}}{6}\log{(4-\sqrt{2}-\sqrt{6})} \\ =& \frac{\sqrt{3}}{6}\log{(5-2\sqrt{6})(2-\sqrt{3})} \end{align*}

\begin{align*} & \frac{\sqrt{3}}{6}\log{(2+\sqrt{3})} - \frac{\sqrt{3}}{6}\log{(2-\sqrt{3})} \\ =& \frac{\sqrt{3}}{6}\log{\frac{2+\sqrt{3}}{2-\sqrt{3}}} \end{align*}

よって最初の2項の合計は$ \frac{\sqrt{3}}{6}\log(2+\sqrt{3})(5-2\sqrt{6}) $

$$ S = a(\sqrt{3}-1) + 1, T = a(-\sqrt{3}-1) + 1 $$

とすると

$$ S + T = -2a + 2 = 4 - 2\sqrt{2}, ST = -2a^2 - 2a + 1 = 2\sqrt{2} - 3 $$

であるから

\begin{align*} & -\frac{1}{3} \arctan{S} -\frac{1}{3} \arctan{T} \\ =& -\frac{1}{3}\arctan{\frac{S+T}{1-ST}} \\ =& -\frac{1}{3}\arctan{\frac{4-2\sqrt{2}}{4 - 2\sqrt{2}}} \\ =& -\frac{1}{3}\arctan{1} \\ =& -\frac{\pi}{12} \end{align*}

また

$$ \frac{1}{3} \left( \frac{\pi}{4} + \frac{\pi}{4} \right) = \frac{\pi}{6} $$

よって、求める積分値は

\begin{align*} & \frac{\sqrt{2}}{6}\log{(1+\sqrt{2})} + \frac{\sqrt{3}}{6}\log(2+\sqrt{3})(5-2\sqrt{6}) - \frac{\pi}{12} + \frac{\pi}{6} \\ =& \frac{\sqrt{2}}{6}\log{(1+\sqrt{2})} + \frac{\sqrt{3}}{6}\log(2+\sqrt{3})(5-2\sqrt{6}) + \frac{\pi}{12} \end{align*}

となります！！！

教訓

積分に王道なし。