数学作問1 解説

まず$t=\sqrt{x}$と置換します.すると,
$$I=\int_{0}^{1}2t\sqrt{t^2+t}\,dt=\int_{0}^{1}2t\sqrt{\biggl(t+\dfrac{1}{2}\biggl)^2-\dfrac{1}{4}}\,dt$$

$u=t+\dfrac{1}{2}$と置換.
$$\dfrac{1}{2}I=\int_{\frac{1}{2}}^{\frac{3}{2}}\biggl(u-\dfrac{1}{2}\biggl)\sqrt{u^2-\dfrac{1}{4}}\,du$$

$u=\dfrac{1}{2\cos{\theta}}$と置換.

$\tan{\theta}=2\sqrt{u^2-\dfrac{1}{4}}$

$du=\dfrac{\tan{\theta}}{2\cos{\theta}}d\theta$となる.ただしここで,

$\cos{\theta}=\dfrac{1}{3}$となるような

$\theta$を$\theta_0$とする.

$$\dfrac{1}{2}I=\int_{0}^{\theta_0}\dfrac{1}{2}\biggl(\dfrac{1}{\cos{\theta}}-1\biggl)\cdot \dfrac{1}{2}\tan{\theta}\cdot \dfrac{\tan{\theta}}{2\cos{\theta}}d\theta$$

$$=\dfrac{1}{8}\int_{0}^{\theta_0}\biggl(\dfrac{1}{\cos^2{\theta}}-1\biggl)\biggl(\dfrac{1}{\cos^2{\theta}}-\dfrac{1}{\cos^3{\theta}}\biggl)\,d\theta$$
$$4I=\int_{0}^{\theta_0}\dfrac{1}{\cos^4{\theta}}-\dfrac{1}{\cos^3{\theta}}-\dfrac{1}{\cos^2{\theta}}+\dfrac{1}{\cos{\theta}}\,d\theta$$


ここで次のようにおく.
$$J_n=\int_{0}^{\theta_0}\dfrac{1}{\cos^n{\theta}}\,d\theta$$
$4I=J_4-J_3-J_2+J_1$である.
$J_{n+2}$と$J_n$の関係を示しましょう.

$$J_{n+2}=\int_{0}^{\theta_0}\dfrac{1}{\cos^{n+2}{\theta}}\,d\theta=\int_{0}^{\theta_0}\dfrac{1}{\cos^n{\theta}}\cdot\dfrac{1}{\cos^2{\theta}}\,d\theta$$
部分積分をすると,
$$J_{n+2}=\biggl[\tan{\theta}\cdot\dfrac{1}{\cos^n{\theta}} \biggl]_{0}^{\theta_0}-n\int_{0}^{\theta_0}\tan{\theta}\cdot\dfrac{\sin{\theta}}{\cos^{n+1}{\theta}}\,d\theta$$
$$=\biggl[\dfrac{\sin{\theta}}{\cos^{n+1}{\theta}} \biggl]_{0}^{\theta_0}-n\int_{0}^{\theta_0}\dfrac{\sin^2{\theta}}{\cos^{n+2}{\theta}}\,d\theta$$
$$=\biggl[\dfrac{\sqrt{1-\cos^2{\theta}}}{\cos^{n+1}{\theta}} \biggl]_{0}^{\theta_0}-n\int_{0}^{\theta_0}\dfrac{1-\cos^2{\theta}}{\cos^{n+2}{\theta}}\,d\theta$$
$\cos{\theta_0}=\dfrac{1}{3}$なので,
$$=\dfrac{\frac{2\sqrt{2}}{3}}{(\frac{1}{3})^{n+1}}-n(J_{n+2}-J_n)$$
よって,
$$(n+1)J_{n+2}=2\sqrt{2}\cdot3^{n}+nJ_n$$
$$J_{n+2}=\dfrac{2\sqrt{2}\cdot3^{n}}{n+1}+\dfrac{n}{n+1}J_n$$


次に$J_1$を求める.
$$J_1=\int_{0}^{\theta_0}\dfrac{1}{\cos{\theta}}\,d\theta=\int_{0}^{\theta_0}\dfrac{\cos{\theta}}{\cos^2{\theta}}\,d\theta$$
$$ \int_{0}^{\theta_0}\dfrac{\cos{\theta}}{1-\sin^2{\theta}}\,d\theta=\dfrac{1}{2}\int_{0}^{\theta_0}\dfrac{\cos{\theta}}{1-\sin{\theta}}+\dfrac{\cos{\theta}}{1+\sin{\theta}}\,d\theta$$
$$=\dfrac{1}{2}\biggl[-\log{|1-\sin{x}|}+\log{|1+\sin{x}|}\biggl]_0^{\theta_0}$$
$$=\dfrac{1}{2}\log{\Biggl|\dfrac{1+\sin{\theta_0}}{1-\sin{\theta_0}}\Biggl|}$$
$\sin{\theta_{0}}=\dfrac{2\sqrt{2}}{3}$より,代入して整理すると,
$$J_1=2\log{(1+\sqrt{2})}$$

$J_2$を解く.
$$J_2=\int_{0}^{\theta_0}\dfrac{1}{\cos^{2}{\theta}}\,d\theta=\tan{\theta_0}=2\sqrt{2}$$

$J_3$は関係式から求める.
$$J_3=3\sqrt{2}+\dfrac{1}{2}J_1$$
$$=3\sqrt{2}+\log{(1+\sqrt{2})}$$

$J_4$も関係式より,
$$J_4=6\sqrt{2}+\dfrac{2}{3}J_2$$
$$=6\sqrt{2}+\dfrac{4\sqrt{2}}{3}=\dfrac{22\sqrt{2}}{3}$$
まとめると,
$$J_1=2\log{(1+\sqrt{2})}$$
$$J_2=2\sqrt{2}$$
$$J_3=3\sqrt{2}+\log{(1+\sqrt{2})}$$
$$J_4=\dfrac{22\sqrt{2}}{3}$$
$4I=J_4-J_3-J_2+J_1$であるから,
$$4I=\dfrac{7\sqrt{2}}{3}+\log{(1+\sqrt{2})}$$
よって,
$$\int_{0}^{1}\sqrt{x+\sqrt{x}}\,dx=\dfrac{7\sqrt{2}}{12}+\dfrac{1}{4}\log{(1+\sqrt{2})}$$