Em tự chứng minh đẳng thức sau
$\dfrac{\tan\left(x-\dfrac{\pi}{4}\right)}{\cos2x} = - \frac{1}{(\sin x +\cos x)^2}$
Suy ra
$I=\int\limits_{0}^{\frac{\pi}{6}}\dfrac{\tan\left(x-\dfrac{\pi}{4}\right)}{\cos2x}dx=-\int\limits_{0}^{\frac{\pi}{6}}\frac{1}{(\sin x +\cos x)^2}dx $
$=-\int\limits_{0}^{\frac{\pi}{6}}\frac{\sin^2 x +\cos^2 x }{(\sin x +\cos x)^2}dx=-\int\limits_{0}^{\frac{\pi}{6}}\frac{(\sin x+ \cos x)(\sin x)' - (\sin x +\cos x)'\sin x }{(\sin x +\cos x)^2}dx$
$=-\left[ {\frac{\sin x}{\sin x+ \cos x}} \right]_{0}^{\frac{\pi}{6}}$