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！

$$\int_{0}^{\frac{\pi}{2}}\sqrt{\sin x-\sin^2{x}}dx \\ =\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x-\cos^2{x}}dx \\ =\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x(1-\cos x)}dx \\ =\int_{0}^{\frac{\pi}{2}}\sqrt{2\cos x\cdot\sin^2\frac{x}{2}}dx \\ =\int_{0}^{\frac{\pi}{2}}\sqrt{2\cos x}\cdot\sin\frac{x}{2}dx \\ =\int_{0}^{\frac{\pi}{2}}\sqrt{2(2\cos^2{\frac{x}{2}-1})}\cdot\sin\frac{x}{2}dx \\ =2\sqrt{2}\int_{\frac{1}{\sqrt{2}}}^{1}\sqrt{2t^2-1}dt \cdots(*1) \\ =2\sqrt{2}\int_{0}^{\frac{\pi}{4}}\sqrt{\frac{1}{\cos^2x}-1}\cdot\frac{1}{\sqrt{2}}\cdot\frac{\sin x}{\cos^2x}dx \\ =2\int_{0}^{\frac{\pi}{4}}\tan x\cdot\frac{\sin x}{\cos^2x}dx \\ =2\int_{0}^{\frac{\pi}{4}}\frac{1-\cos^2x}{\cos^3x}dx \\ =2(\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^3x}dx-\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos x}dx)$$
ここで、$$\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^3x}dx, \int_{0}^{\frac{\pi}{4}}\frac{1}{\cos x}dx$$
をそれぞれ計算していく。
$$\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos x}dx \\ =\int_{0}^{\frac{\pi}{4}}\frac{\cos x}{\cos^2x}dx \\ =\int_{0}^{\frac{\pi}{4}}\frac{\cos x}{1-\sin^2x}dx \\ =\int_{0}^{\frac{1}{\sqrt{2}}}\frac{1}{1-t^2}dt \\ =\frac{1}{2}\int_{0}^{\frac{1}{\sqrt{2}}}(\frac{1}{1+t}+\frac{1}{1-t})dt \\ =\frac{1}{2}\left[\log(1+t)-\log(1-t)\right]_{0}^{\frac{1}{\sqrt{2}}} \\ =\frac{1}{2}\log\frac{\sqrt{2}+1}{\sqrt{2}-1} \\ =\frac{1}{2}\log(3+2\sqrt{2}) \\ =\log(\sqrt{2}+1)$$

$$\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^3x}dx \\ =\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^2x}\cdot\frac{1}{\cos x}dx \\ =\left[\tan x\cdot\frac{1}{\cos x}\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\tan x\cdot\frac{\sin x}{\cos^2x}dx \\ =\sqrt{2}-\int_{0}^{\frac{\pi}{4}}\frac{\sin^2x}{\cos^3x}dx \\ =\sqrt{2}-\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^3x}dx+\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos x}dx$$

よって
$$\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^3x}dx=\frac{1}{2}(\sqrt{2}+\log(\sqrt{2}+1))$$

以上から、
$$2(\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^3x}dx-\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos x}dx)\\ =\sqrt{2}-\log(\sqrt{2}+1) \\ =\sqrt{2}+\log(\frac{1}{\sqrt{2}+1}) \\ =\sqrt{2}+\log(\sqrt{2}-1)$$

$$*1\cdots t=\frac{1}{\sqrt2\tan x}$$