1) $I=\int\limits_{}^{}\frac{dx}{cosx}=\int\limits_{}^{}\frac{cosxdx}{cos^2x}=\int\limits_{}^{}\frac{cosxdx}{1-sin^2x}$Đặt $t=sinx\Rightarrow I=\int\limits_{}^{}\frac{dt}{1-t^2}=\int\limits_{}^{}(\frac{1}{2(1-t)}+\frac{1}{2(1+t)})$$=\frac{1}{2}[ln(1+t)-ln(1-t)]+c=\frac{1}{2}.ln\frac{1+t}{1-t}+c=\frac{1}{2}.\frac{1+sinx}{1-sinx}+c$
1) $I=\int\limits_{}^{}\frac{dx}{cosx}=\int\limits_{}^{}\frac{cosxdx}{cos^2x}=\int\limits_{}^{}\frac{cosxdx}{1-sin^2x}$Đặt $t=sinx\Rightarrow I=\int\limits_{}^{}\frac{dt}{1-t^2}=\int\limits_{}^{}(\frac{1}{2(1-t)}+\frac{1}{2(1+t)})$$=\frac{1}{2}[ln(1+t)-ln(1-t)]=\frac{1}{2}.ln\frac{1+t}{1-t}=\frac{1}{2}.\frac{1+sinx}{1-sinx}$
1) $I=\int\limits_{}^{}\frac{dx}{cosx}=\int\limits_{}^{}\frac{cosxdx}{cos^2x}=\int\limits_{}^{}\frac{cosxdx}{1-sin^2x}$Đặt $t=sinx\Rightarrow I=\int\limits_{}^{}\frac{dt}{1-t^2}=\int\limits_{}^{}(\frac{1}{2(1-t)}+\frac{1}{2(1+t)})$$=\frac{1}{2}[ln(1+t)-ln(1-t)]
+c=\frac{1}{2}.ln\frac{1+t}{1-t}
+c=\frac{1}{2}.\frac{1+sinx}{1-sinx}
+c$