Question

$\int_{}^{}\left( \sqrt{\tan x} + \sqrt{\cot x} \right)dx$

• A. $\sqrt{2}\tan^{- 1}\left( \frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) + c$
• B. $\sqrt{2}\tan^{- 1}\left( \frac{\sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{2}} \right) + c$
• C. $\tan^{- 1}\left( \frac{\sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{2}} \right) + c$
• D. None of these

Answer 1

Answer: (A)

$\sqrt{2}\tan^{- 1}\left( \frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) + c$

Answer 2

I = $\int_{}^{}{\left( \sqrt{\tan x} + \sqrt{\cot x} \right)dx}$

Put $\tan x = t^{2}$ $\Rightarrow \sec^{2}xdx = 2tdt$⇒ $dx = \frac{2t}{1 + t^{4}}dt$

I $= \int_{}^{}\left( t + \frac{1}{t} \right).\frac{2t}{1 + t^{4}} ⥂ dt$

$= 2\int_{}^{}\frac{\left( 1 + \frac{1}{t^{2}} \right)}{t^{2} + \frac{1}{t^{2}} + 2 - 2}dt$ $\mathbf{= 2}\int_{}^{}\frac{\left( \mathbf{1 +}\frac{\mathbf{1}}{\mathbf{t}^{\mathbf{2}}} \right)\mathbf{dt}}{\left( \mathbf{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{t}} \right)^{\mathbf{2}}\mathbf{+}\left( \sqrt{\mathbf{2}} \right)^{\mathbf{2}}}$

Put $t - \frac{1}{t} = p$ $\Rightarrow \left( 1 + \frac{1}{t^{2}} \right)dt = dp$

$= \int_{}^{}\frac{dp}{p^{2} + \left( \sqrt{2} \right)^{2}} = \frac{2}{\sqrt{2}}\tan^{- 1}\frac{p}{\sqrt{2}} + c$ $= \sqrt{2}\tan^{- 1}\frac{\left( t - \frac{1}{t} \right)}{\sqrt{2}} + c$

$= \sqrt{2}\tan^{- 1}\left( \frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) + c$