Question

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

• A. ${\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2}\tan x}\right)+c$
• B. $2{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2}\tan x}\right)+c$
• C. $\sqrt{2}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+c$
• D. $\sqrt{3}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2}\tan x}\right)+c$

Answer 1

Answer: (C)

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

Given, $\int (\sqrt{\cot x}+\sqrt{\tan x})dx$
Put $\tan x=t^2$ $\Rightarrow$ ${\sec }^{2}xdx=2tdt$ . Then the equation becomes:
$\Rightarrow$ $\int (t+\frac{1}{t})\frac{2tdt}{t^4+1}$
$\Rightarrow$ $2\int \frac{t^2+1}{t^4+1}dt$
Now dividing the numerator and denominator by $t^2$ we get :
$\Rightarrow$ $2\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$
$\Rightarrow$ $2\int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t}{)}^2+2}dt$
Putting $(t-\frac{1}{t})=k$ $\Rightarrow$ $(1+\frac{1}{t^2})dt=dk$
$\Rightarrow$ $2\int \frac{1}{k^2+(\sqrt{2}{)}^2}dk$
$\Rightarrow$ $2\int \frac{1}{k^2+(\sqrt{2}{)}^2}dk=\frac{2}{\sqrt{2}}{\tan }^{-1}\left(\frac{k}{\sqrt{2}}\right)+c=\sqrt{2}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+c=\sqrt{2}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+c$
Hence, option 'C' is correct.