Question

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

• A. $\sqrt{2}{\sin }^{-1}(\sin x+\cos x)+C$
• B. $\sqrt{2}{\cos }^{-1}(\sin x-\cos x)+C$
• C. $\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C$
• D. $\sqrt{2}{\cos }^{-1}(\sin x+\cos x)+C$

Answer 1

The correct option is C $\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C$

Let $I=\int (\sqrt{\tan x}+\sqrt{cotx})dx=\int \frac{\tan x+1}{\sqrt{\tan x}}dx$

Substituting $\tan x=t^2\Rightarrow {\sec }^{2}xdx=2tdt$

$I=\int \frac{t^2+1}{\sqrt{t^2}}.\frac{2t}{t^4+1}dt=2\int \frac{t^2+1}{t^4+1}dt=2\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-2+2}dt=2\int \frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+{\left(\sqrt{2}\right)}^2}dt$

Put $t-\frac{1}{t}=u\Rightarrow (1+\frac{1}{t^2})dt=du$

$I=2\int \frac{du}{u^2+{\left(\sqrt{2}\right)}^2}=\frac{2}{\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)+c=\sqrt{2}{\tan }^{-1}\left(\frac{(\sqrt{\tan x}-\sqrt{cotx})}{\sqrt{2}}\right)+c=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C$