最難積分解いてみた！

ヨビノリさんの最難積分を別の解法で解いてみました！

今回解く積分はこちら！

$${\int }_0^{\frac{\pi }{2}}\sqrt{\sin x-{\sin }^{2}x}dx$$

模範解答とは全く違う解法ですが、初見での解法を載せてみました。
では、実際に解いていきましょう！

$$\begin{gathered} {\int }_0^{\frac{\pi }{2}}\sqrt{\sin x-{\sin }^{2}x}dx \\ ={\int }_0^{\frac{\pi }{2}}\sqrt{\cos x-{\cos }^{2}x}dx \\ ={\int }_0^{\frac{\pi }{2}}\sqrt{\cos x(1-\cos x)}dx \\ ={\int }_0^{\frac{\pi }{2}}\sqrt{2\cos x\cdot {\sin }^2\frac{x}{2}}dx \\ ={\int }_0^{\frac{\pi }{2}}\sqrt{2\cos x}\cdot \sin \frac{x}{2}dx \\ ={\int }_0^{\frac{\pi }{2}}\sqrt{2(2{\cos }^2\frac{x}{2}-1)}\cdot \sin \frac{x}{2}dx \\ =2\sqrt{2}{\int }_{\frac{1}{\sqrt{2}}}^1\sqrt{2t^2-1}dt\cdots (\ast 1) \\ =2\sqrt{2}{\int }_0^{\frac{\pi }{4}}\sqrt{\frac{1}{{\cos }^{2}x}-1}\cdot \frac{1}{\sqrt{2}}\cdot \frac{\sin x}{{\cos }^{2}x}dx \\ =2{\int }_0^{\frac{\pi }{4}}\tan x\cdot \frac{\sin x}{{\cos }^{2}x}dx \\ =2{\int }_0^{\frac{\pi }{4}}\frac{1-{\cos }^{2}x}{{\cos }^{3}x}dx \\ =2({\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{3}x}dx-{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos x}dx) \end{gathered}$$
ここで、$${\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{3}x}dx,{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos x}dx$$
をそれぞれ計算していく。
$$\begin{gathered} {\int }_0^{\frac{\pi }{4}}\frac{1}{\cos x}dx \\ ={\int }_0^{\frac{\pi }{4}}\frac{\cos x}{{\cos }^{2}x}dx \\ ={\int }_0^{\frac{\pi }{4}}\frac{\cos x}{1-{\sin }^{2}x}dx \\ ={\int }_0^{\frac{1}{\sqrt{2}}}\frac{1}{1-t^2}dt \\ =\frac{1}{2}{\int }_0^{\frac{1}{\sqrt{2}}}(\frac{1}{1+t}+\frac{1}{1-t})dt \\ =\frac{1}{2}{[\log (1+t)-\log (1-t)]}_0^{\frac{1}{\sqrt{2}}} \\ =\frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1} \\ =\frac{1}{2}\log (3+2\sqrt{2}) \\ =\log (\sqrt{2}+1) \end{gathered}$$

$$\begin{gathered} {\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{3}x}dx \\ ={\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{2}x}\cdot \frac{1}{\cos x}dx \\ ={[\tan x\cdot \frac{1}{\cos x}]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\tan x\cdot \frac{\sin x}{{\cos }^{2}x}dx \\ =\sqrt{2}-{\int }_0^{\frac{\pi }{4}}\frac{{\sin }^{2}x}{{\cos }^{3}x}dx \\ =\sqrt{2}-{\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{3}x}dx+{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos x}dx \end{gathered}$$

よって
$${\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{3}x}dx=\frac{1}{2}(\sqrt{2}+\log (\sqrt{2}+1))$$

以上から、
$$\begin{gathered} 2({\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{3}x}dx-{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos x}dx) \\ =\sqrt{2}-\log (\sqrt{2}+1) \\ =\sqrt{2}+\log \left(\frac{1}{\sqrt{2}+1}\right) \\ =\sqrt{2}+\log (\sqrt{2}-1) \end{gathered}$$

$$\ast 1\cdots t=\frac{1}{\sqrt{2}\tan x}$$