Tính [TEX]\int_0^{\frac{\pi }{2}} {\frac{1}{{{{\cos }^3}x}}dx} [/TEX]
[TEX]I= \int_0^{\frac{\pi }{2}} {\frac{1}{{{{\cos }^3}x}}dx} = \int_0^{\frac{\pi}{2}}\frac{cosxdx}{(1-sin^2x)^2}[/TEX]
[TEX]t=sinx \Rightarrow dt=cosxdx[/TEX]
[TEX]I=\int_0^1\frac{dt}{(1-t^2)^2}[/TEX]
[TEX] = \frac{-1}{4}\int_0^1\frac{d(t-1)}{(t-1)^2}+\frac{1}{4}\int_0^1\frac{d(t-1)}{t-1}+\frac{1}{4}\int_0^1\frac{d(t+1)}{t+1}-\frac{1}{4}\int_0^1\frac{d(t+1)}{(t+1)^2[/TEX]
[TEX] = (\frac{t}{2(t^2-1)}+\frac{1}{4}ln|\frac{t-1}{t+1}|)|^1_0[/TEX] :|