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

$$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}=2\sqrt{\frac{t^2}{(1+t^2{)}^2}}=\frac{2t}{1+t^2}(\because 0\le x\le \frac{\pi }{4}で\sin x\ge 0,t\ge 0)$$

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

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

と求まりました。

計算

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

$$\begin{array}{rl}& {\int }_0^{\frac{\pi }{4}}\frac{{\sin }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}dx\\ =& {\int }_0^a\frac{{\left(\frac{2t}{1+t^2}\right)}^2}{{\left(\frac{2t}{1+t^2}\right)}^3+{\left(\frac{1-t^2}{1+t^2}\right)}^3}\cdot \frac{2}{1+t^2}dt\\ =& {\int }_0^a\frac{(2t{)}^2(1+t^2)}{(2t{)}^3+(1-t^2{)}^3}\cdot \frac{2}{1+t^2}dt\\ =& {\int }_0^a\frac{2(2t{)}^2}{(2t{)}^3+(1-t^2{)}^3}dt\\ =& {\int }_0^a\frac{8t^2}{-t^6+3t^4+8t^3-3t^2+1}dt\end{array}$$

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。

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

$t-\frac{1}{t}=2$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)

$$\begin{array}{rl}t-\frac{1}{t}& =-1{\pm }_1\sqrt{3}i\\ t^2+(1{\mp }_1\sqrt{3}i)t-1& =0\\ {(t+\frac{1{\mp }_1\sqrt{3}i}{2})}^2-\frac{-1{\mp }_1\sqrt{3}i}{2}-1& =0\\ {(t+\frac{1{\mp }_1\sqrt{3}i}{2})}^2& =\frac{-1{\mp }_1\sqrt{3}i}{2}+1\\ {(t+\frac{1{\mp }_1\sqrt{3}i}{2})}^2& =\frac{1{\mp }_1\sqrt{3}i}{2}\\ t+\frac{1{\mp }_1\sqrt{3}i}{2}& ={\pm }_2\frac{\sqrt{3}{\mp }_{1}i}{2}\\ t& =-\frac{1{\mp }_1\sqrt{3}i}{2}{\pm }_2\frac{\sqrt{3}{\mp }_{1}i}{2}\\ t& =-\frac{1{\pm }_2\sqrt{3}}{2}{\pm }_1\left(\frac{\sqrt{3}{\pm }_{2}1}{2}\right)i\end{array}$$

(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、

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

と実数の範囲で因数分解できます。そこで、

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

となる実数$A$～$F$を求めます。

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

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

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

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

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

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

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、

$$\begin{array}{rl}& \frac{8t^2}{-t^6+3t^4+8t^3-3t^2+1}\\ =& -\frac{\sqrt{2}}{6}\cdot \frac{1}{t-(1+\sqrt{2})}\\ +& \frac{\sqrt{2}}{6}\cdot \frac{1}{t-(1-\sqrt{2})}\\ +& (-\frac{\sqrt{3}}{3}t+\frac{-\sqrt{3}-2}{3})\frac{1}{t^2+(1+\sqrt{3})t+(2+\sqrt{3})}\\ +& (\frac{\sqrt{3}}{3}t+\frac{\sqrt{3}-2}{3})\frac{1}{t^2+(1-\sqrt{3})t+(2-\sqrt{3})}\end{array}$$

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、

$$\begin{array}{rl}\frac{2x}{1-x^2}& =1\\ 2x& =1-x^2\\ x^2+2x-1& =0\\ x& =-1\pm \sqrt{2}\end{array}$$

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

$$\begin{array}{rl}& {\int }_0^a-\frac{\sqrt{2}}{6}\cdot \frac{1}{t-(1+\sqrt{2})}dt\\ =& -\frac{\sqrt{2}}{6}{\int }_0^a\frac{1}{t-(1+\sqrt{2})}dt\\ =& -\frac{\sqrt{2}}{6}[\log |t-(1+\sqrt{2})|{]}_0^a\\ =& -\frac{\sqrt{2}}{6}\log \frac{2}{1+\sqrt{2}}\end{array}$$

第2項

$$\begin{array}{rl}& {\int }_0^a\frac{\sqrt{2}}{6}\cdot \frac{1}{t-(1-\sqrt{2})}dt\\ =& \frac{\sqrt{2}}{6}{\int }_0^a\frac{1}{t-(1-\sqrt{2})}dt\\ =& \frac{\sqrt{2}}{6}[\log |t-(1-\sqrt{2})|{]}_0^a\\ =& \frac{\sqrt{2}}{6}\log \frac{2(\sqrt{2}-1)}{\sqrt{2}-1}\\ =& \frac{\sqrt{2}}{6}\log 2\end{array}$$

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。

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

であるから

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

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。

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

であるから

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

$\sqrt{3}$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、

$$\begin{array}{rl}& -\frac{\sqrt{3}}{6}\log (4-\sqrt{2}+\sqrt{6})+\frac{\sqrt{3}}{6}\log (4-\sqrt{2}-\sqrt{6})\\ =& \frac{\sqrt{3}}{6}\log (5-2\sqrt{6})(2-\sqrt{3})\end{array}$$

$$\begin{array}{rl}& \frac{\sqrt{3}}{6}\log (2+\sqrt{3})-\frac{\sqrt{3}}{6}\log (2-\sqrt{3})\\ =& \frac{\sqrt{3}}{6}\log \frac{2+\sqrt{3}}{2-\sqrt{3}}\end{array}$$

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

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

また

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

よって、求める積分値は

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

となります！！！

教訓

積分に王道なし。