Answer:
[tex]2(\tan(x))^\frac{1}{2}+C[/tex]
Explanation:
Let [tex]u=\tan(x)[/tex].
Therefore, [tex]du=\sec^2(x)dx[/tex] or [tex]du=(\sec(x))^2dx[/tex].
[tex]\int \frac{(\sec(x))^2 dx}{\sqrt{\tan(x)}}[/tex]
Putting the integral in terms of [tex]u[/tex] now:
[tex]\int \frac{du}{\sqrt{u}}[/tex]
or
[tex]\int \frac{1}{\sqrt{u}} du[/tex]
or
[tex]\int \frac{1}{u^\frac{1}{2}}du[/tex]
or
[tex]\int u^\frac{-1}{2} du[/tex]
Let's integrate now:
[tex]\frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+C[/tex]
[tex]\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C[/tex]
[tex]\frac{2}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C[/tex]
[tex]\frac{2u^{\frac{1}{2}}}{1}+C[/tex]
[tex]2u^{\frac{1}{2}}+C[/tex]
Putting back in terms of [tex]x[/tex]:
[tex]2(\tan(x))^\frac{1}{2}+C[/tex]
