Answer:[tex]\frac{\tan^2(x)}{2}+C[/tex]Step-by-step explanation:Since I know [tex](\tan(x))'=\sec^2(x)[/tex], then I will let [tex]u=\tan(x)[/tex] which will give me [tex]du=\sec^2(x) dx[/tex].[tex]\int \tan(x)\sec^2(x) dx[/tex][tex]\int u du[/tex] I replaced tan(x) with u and sec^2(x) dx with du.[tex]\int u^1 du[/tex]Use the integration power rule:[tex]\frac{u^{1+1}}{1+1}+C[/tex][tex]\frac{u^2}{2}+C[/tex]Now put this back in terms of x with u=tan(x):[tex]\frac{\tan^2(x)}{2}+C[/tex]Check:To check we will find the derivative of our antiderivative and see it it gives us the integrand.[tex]\frac{d}{dx}(\frac{\tan^2(x)}{2}+C)[/tex]By sum rule for derivatives:[tex]\frac{d}{dx}(\frac{\tan^2(x)}{2})+\frac{d}{dx}(C)[/tex]Constant multiple rule will be applied to the first term while the constant rule will be applied to the second term:[tex]\frac{1}{2} \frac{d}{dx}(\tan^2(x))+0[/tex]Power rule+Chain rule:[tex]\frac{1}{2}\cdot 2\tan(x) \cdot sec^2(x)[/tex]Simplify the 1/2 times 2 part:[tex]\tan(x)\sec^2(x)[/tex]This is what our integrand was so we have confirmed are integration was correct.