Question

Solve :

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

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

Answer 1

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

Write $\sqrt{tanx}=\frac{\sqrt{\sin x}}{\sqrt{\cos x}}$,

and $\sqrt{\cot x}=\frac{\sqrt{\cos x}}{\sqrt{\sin x}}$

Let $I=\int \frac{(\sin x+\cos x)}{\sqrt{\sin x\cdot \cos x}}dx$
$=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2\sin x\cdot \cos x}}dx$
$=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$
Put $\sin x-\cos x=t$
$\Rightarrow (\cos x+\sin x)dx=dt$
Also, $\sin 2x=(1-t^2)$
$\therefore I=\sqrt{2}\int \frac{dt}{\sqrt{1-t^2}}$
$=\sqrt{2}{\sin }^{-1}t+C$
$=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C$
$=\sqrt{2}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+C$