[TEX]1) I=\int_{0}^{\frac{\pi}{4}}\frac{-d(cosx)}{cos^2x\sqrt{1+cos^2x}}=\int_{0}^{\frac{\pi}{4}} \frac{-d(cosx)}{cos^3x\sqrt{1+\frac{1}{cos^2x}}}=\int_{0}^{\frac{\pi}{4}} \frac{d(1+\frac{1}{cos^2x})}{2 \sqrt{1+\frac{1}{cos^2x}}}=\sqrt{1+\frac{1}{cos^2t}}|_0^{\frac{\pi}{4}}[/TEX]
2) Đặt x=-t
[TEX]3) \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{-d(cosx)}{sin^2x.cos^2x}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{-d(cosx)}{(1-cos^2x)cos^2x}=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{dt}{(1-t^2)t^2}[/TEX]