Question

\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]

Answer 1

\[\int\frac{\sqrt{\tan x}}{\sin x \cos x}dx\]

\[ = \int\frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \times \cos^2 x}dx\]

\[ = \int\frac{\sqrt{\tan x}}{\tan x} \times \sec^2 \text{x dx}\]

\[ = \int\frac{1}{\sqrt{\tan x}} \times \sec^2 \text{x dx}\]

\[ = \int \left( \tan x \right)^{- \frac{1}{2}} \sec^2 x dx\]

\[Let \tan x = t\]

\[ \Rightarrow \sec^2 x = \frac{dt}{dx}\]

\[ \Rightarrow \sec^2 \text{x dx} = dt\]

\[Now, \int \left( \tan x \right)^{- \frac{1}{2}} \sec^2\text{ x dx}\]

\[ = \int t^{- \frac{1}{2}} dt\]

\[ = \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]

\[ =\text{2}    \sqrt{t} + C\]

\[ = 2 \sqrt{\tan x} + C\]