Question

If $I=\int (\sqrt{\cot x}-\sqrt{\tan x})dx$, then I equals

• A. $\sqrt{2}\log (\sqrt{\tan x}-\sqrt{\cot x})+C$
• B. $\sqrt{2}\log |\sin x+\cos x+\sqrt{\sin 2x}|+C$
• C. $\sqrt{2}\log |\sin x-\cos x+\sqrt{2}\sin x\cos x|+C$
• D. $\sqrt{2}\log \left|\sin \right(x+\frac{\pi }{4})+\sqrt{2}\sin x\cos x|+C$

Answer 1

The correct option is B $\sqrt{2}\log |\sin x+\cos x+\sqrt{\sin 2x}|+C$

$I=\int (\sqrt{\cot x}-\sqrt{\tan x})dx$
$I=\int \frac{\cos x-\sin x}{\sqrt{\cos x\sin x}}dx$
$I=\sqrt{2}\int \frac{\cos x-\sin x}{\sqrt{1+\sin 2x-1}}dx$
$I=\int \frac{\cos x-\sin x}{\sqrt{(\cos x+\sin x{)}^2-1}}dx$
Put $\cos x+\sin x=t$
$\Rightarrow (\cos x-\sin x)dx=dt$
$\therefore I=\sqrt{2}\int \frac{dt}{\sqrt{t^2-1}}$
$=\sqrt{2}\log |t+\sqrt{t^2-1}|+C$
$\Rightarrow I=\sqrt{2}\log |\sin x+\cos x+\sqrt{\sin 2x}|+C$