Question

Evaluate $\int (\sqrt{\tan x}+\sqrt{\cot x})dx$.

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

Answer 1

The correct option is C $\sqrt{2}{\tan }^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+C$

Let $I=\int (\sqrt{\tan x}+\sqrt{\cot x})dx$
$=\int \frac{\tan x+1}{\sqrt{\tan x}}dx$
Put $\tan x=t^2$
$\Rightarrow {\sec }^{2}xdx=2tdt$
$\Rightarrow dx=\frac{2tdt}{1+{\tan }^{2}t}=\frac{2t}{1+t^4}dt$
Therefore, $I=\int \frac{t^2+1}{\sqrt{t^2}}\cdot \frac{2t}{t^4+1}dt$
$=2\int \frac{t^2+1}{t^4+1}dt$
$=2\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-2+2}dt$
$=2\int \frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+(\sqrt{2}{)}^2}dt$
$=2\int \frac{du}{u^2+(\sqrt{2}{)}^2}$
where, $u=t-\frac{1}{t}\Rightarrow du=(1+\frac{1}{t^2})dt$
$\Rightarrow I=\frac{2}{\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)+C$
$=\sqrt{2}{\tan }^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)$.