Integrate $\int\frac{\sqrt{\tan(x)}}{\cos^2x}dx$ I need help with this integral: $$\int\frac{\sqrt{\tan x}}{\cos^2x}dx$$ I tried substitution and other methods, but all have lead me to this expression: $$2\int\sqrt{\tan x}(1+\tan^2 x)dx$$ where I can't calculate anything... Any suggestions? Thanks!

As you have noted, your integral simplifies to
$$2\int\sqrt{\tan x}\ \sec^2x\ dx$$
If one makes the substitution $u=\tan x$, one gets $du=\sec^2x dx$, which reduces our integral to
$$2\int u^{1/2}du$$
$$=2\frac{u^{3/2}}{3/2}+C$$
$$=\frac{4u^{3/2}}{3}+C$$
$$=\frac{4\tan^{3/2}x}{3}+C$$
