Tính tích phân: $I=\int \limits_{0}^{\frac{\pi}{6}} \dfrac{2\sin^2(\dfrac{\pi}{4}-x)}{\cos 2x}\mathrm{d}x.$

Tính tích phân: $I=\int \limits_{0}^{\frac{\pi}{6}} \dfrac{2\sin^2(\dfrac{\pi}{4}-x)}{\cos 2x}\mathrm{d}x.$

Giải
$\begin{aligned}I=\int_{0}^{\frac{\pi }{6}}\frac{2\sin^2(x-\frac{\pi }{4})}{\cos 2x}\mbox{dx}&=\int_{0}^{\frac{\pi }{6}}\frac{1-\cos (2x-\frac{\pi }{2})}{\cos^2 x-\sin^2 x}\mbox{dx}\\&=\int_{0}^{\frac{\pi }{6}}\frac{1-\sin 2x}{(\cos x-\sin x)(\cos x+\sin x)}\mbox{dx}\\&=\int_{0}^{\frac{\pi }{6}}\frac{(\cos x-\sin x)^2}{(\cos x-\sin x)(\cos x+\sin x)}\mbox{dx}\\&=\int_{0}^{\frac{\pi }{6}}\frac{\cos x-\sin x}{\cos x+\sin x}\mbox{dx}\\ &=\int_{0}^{\frac{\pi }{6}}\frac{d(\cos x+\sin x)}{\cos x+\sin x}\\ &=\ln|\cos x+\sin x| \Big|_{0}^{\frac{\pi }{6}}=\ln\dfrac{\sqrt{3}+1}{2}\end{aligned}$