# 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!

• Hint: what's the derivative of $\tan x$? – Simon S Feb 9 '15 at 19:55
• Hint: rewrite as $$\int \sec^2 x\sqrt{\tan x}dx$$ – John Joy Feb 10 '15 at 1:09
• This forum would be so much better if people only offered hints like you have done. Much more beneficial than full solutions. – mmgro27 Aug 13 '15 at 14:07

Hint: Let $u = \tan x$ then $du = \sec^2 x\ \ dx = \frac{1}{\cos^2 x} dx$
$$\int \dfrac{\sqrt{\tan x}}{\cos^2 x} dx = \int \sqrt{\tan x} \; d(\tan x) = \dfrac{2}{3}\sqrt{(\tan x)^3} +C$$
Set $t=\tan x$; then $\,\mathrm d\mkern1.5mu t=\dfrac1{\cos^2x}\mathrm d\mkern1.5mu x$ so the integral becomes $$\int\sqrt t\,\mathrm d\mkern1.5mu t = \frac 23 (t)^{\frac32}=\frac23\tan t\sqrt{\tan t}.$$
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 $$2\int u^{1/2}du$$ $$=2\frac{u^{3/2}}{3/2}+C$$ $$=\frac{4u^{3/2}}{3}+C$$ $$=\frac{4\tan^{3/2}x}{3}+C$$