神鳥奈紗さんの積分問題10を解いてみました

神鳥奈紗 さんの 積分問題10 を勝手に解いてみました。

$$ \int_{0}^{\pi/2}\frac{dx}{1+\cos^4 x}\\ =\int_{0}^{\infty}\frac{1}{1+(\frac{1}{1+\tan^2x})^2}\frac{1}{1+\tan^2 x}d(\tan x)\\ =\int_{0}^{\infty}\frac{x^2+1}{x^4+2x^2+2}dx\\ =\int_{0}^{\infty}\frac{x^2+1}{(x^2-2px+\sqrt{2})(x^2+2px+\sqrt{2})}dx,\ (p=2^{1/4}\cos(3\pi/8))\\ =\int_{0}^{\infty}\left( \frac{ax+b}{x^2-2px+\sqrt{2}}+\frac{-ax+b}{x^2+2px+\sqrt{2}}\right)dx\\ =\int_{0}^{\infty}\left( \frac{(x^2-2px+\sqrt{2})'a/2+ap+b}{x^2-2px+\sqrt{2}}+\frac{(x^2+2px+\sqrt{2})'(-a)/2+ap+b}{x^2+2px+\sqrt{2}}\right)dx\\ =\left[ \frac{a}{2}\log(x^2-2px+\sqrt{2})- \frac{a}{2}\log(x^2+2px+\sqrt{2})\right]_{0}^{\infty}+\int_{0}^{\infty}\frac{ap+b}{x^2-2px+\sqrt{2}}dx+\int_{0}^{\infty}\frac{ap+b}{x^2+2px+\sqrt{2}}dx\\ =\frac{a}{2}\left[\log\frac{x^2-2px+\sqrt{2}}{x^2+2px+\sqrt{2}}\right]_{0}^{\infty}+\int_{0}^{\infty}\frac{ap+b}{x^2-2px+\sqrt{2}}dx+\int_{0}^{\infty}\frac{ap+b}{x^2+2px+\sqrt{2}}dx\\ =\int_{0}^{\infty}\frac{ap+b}{(x-p)^2-p^2+\sqrt{2}}dx+\int_{0}^{\infty}\frac{ap+b}{(x+p)^2-p^2+\sqrt{2}}dx\\ =(ap+b)\frac{\pi/2-(-\pi/8)}{\sqrt{\sqrt{2}-p^2}}+(ap+b)\frac{\pi/2-\pi/8}{\sqrt{\sqrt{2}-p^2}}\\ =(ap+b)\frac{\pi}{\sqrt{\sqrt{2}-p^2}}\\ =\frac{1}{2}\left(\frac{1-1/\sqrt{2}}{2}+\frac{1}{\sqrt{2}}\right)\frac{\pi}{\sqrt{\sqrt{2}-\sqrt{2}\cos^2(3\pi/8)}}\\ =\frac{1}{2}\frac{2+\sqrt{2}}{4}\frac{\pi}{\sqrt{\sqrt{2}}\sin\frac{3}{8}\pi}\\ =\frac{1}{2}\sqrt{\frac{2+\sqrt{2}}{4\sqrt{2}}}\pi\\ =\frac{\sqrt{\sqrt{2}+1}}{4}\pi $$