Question

If $I=\int \sqrt{\tan x}dx,$ then I eqals

• A. $\frac{1}{\sqrt{2}}\left[{\sin }^{-1}(\sin x-\cos x)\right]+\log |\sin x+\cos x+\sqrt{\sin 2x1}|+C$
• B. $\sqrt{2}[{\sin }^{-1}(\sin x+\cos x)+\log \sin x-\cos x+\sqrt{\sin 2x}]+C$
• C. $\sqrt{2}[{\sin }^{-1}(\sin x+\cos x)+\log |\sin x-\cos x+\sqrt{\sin 2x}|]+C$
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{\sqrt{2}}\left[{\sin }^{-1}(\sin x-\cos x)\right]+\log |\sin x+\cos x+\sqrt{\sin 2x1}|+C$

Let $I=\int \sqrt{\tan x}dx=\frac{1}{2}(I_1-I_2)$
Where
$I_1=\int (\tan x+\cot x)dx=\int \frac{\sin x+\cos x}{\sqrt{\cos x\sin x}}dx$
Put $\sin x+\cos x=u$
$I_1=\sqrt{2}\int \frac{dt}{\sqrt{{1-t}^2}}=\sqrt{2}{\sin }^{-1}t+c$
$=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+c$
And
$I_2=\int (\sqrt{\cot x}-\sqrt{\tan x})dx=\int \frac{\cos x-\sin x}{\sqrt{\cos x\sin x}}dx$
Put $\sin x+\cos x=t\Rightarrow 2\sin x\cos x=t^2-1$
$I_2=\sqrt{2}\int \frac{dt}{\sqrt{t^2-1}}=\sqrt{2}\log (1+\sqrt{t^2-1})+c$
$=\sqrt{2}\log (\sin x+\cos x+\sqrt{\sin 2x})+c$