Answer
$$\int\frac{dt}{\cos^2 t\sqrt{1+\tan t}}=2\sqrt{1+\tan t}+C$$
Work Step by Step
$$A=\int\frac{dt}{\cos^2 t\sqrt{1+\tan t}}$$
Let $u=1+\tan t $
Then we have $du=\frac{1}{\cos^2 t}dt$.
Also, $\sqrt{1+\tan t}=\sqrt u=u^{1/2}$
Substitute into $A$: $$A=\int\frac{1}{u^{1/2}}du$$ $$A=\int(u^{-1/2})du$$ $$A=\frac{u^{1/2}}{\frac{1}{2}}+C$$ $$A=2\sqrt u+C$$ $$A=2\sqrt{1+\tan t}+C$$
