Was it right for my book to assume that $\sin x$ and $\cos x$ is positive in finding $\int\frac{\sqrt{\tan x}}{\sin x\cos x}dx$?

Question

Question:

Determine: $$\int\frac{\sqrt{\tan x}}{\sin x\cos x}dx$$

My book's attempt:

$$\int\frac{\sqrt{\tan x}}{\sin x\cos x}dx$$

$$=\int\frac{\sqrt{\frac{\sin x}{\cos x}}}{\sin x\cos x}dx$$

$$=\int\frac{\frac{\sqrt{\sin x}}{\sqrt{\cos x}}}{\sin x\cos x}dx\tag{1}$$

$$...$$

My book assumed in $(1)$ that $\sin x$ and $\cos x$ are both positive. However, the question didn't mention that we are in the first quadrant. From the question, we can only understand that $\tan x$ is positive (because of $\sqrt{\tan x}$): $x$ could be in the 1st or the third quadrant. So, isn't my book making an invalid move in $(1)$?

Answer 1

You may avoid worrying about the domains of validity by integrating as follows

$$\int\frac{\sqrt{\tan x}}{\sin x\cos x}dx = \int\frac{\sqrt{\tan x}}{\tan x}\sec^2 x \>dx = \int \frac1{\sqrt{\tan x} }d({\tan x})=2 \sqrt{\tan x} $$ which is valid for either domains.