Question

Find the value of : $\int (\sqrt{\tan x}+\sqrt{\cot x})dx=?$

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

Answer 1

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

$\int \sqrt{\tan x}+\int \sqrt{\cot x}dx$

$\int \sqrt{\frac{\sin x}{\cos x}}+\int \sqrt{\frac{\cos x}{\sin x}}dx$

$\int \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\int \frac{\sqrt{\cos x}}{\sqrt{\sin x}}dx$

Taking L.C.M

$\int \frac{\sqrt{{\sin }^{2}x}+\sqrt{{\cos }^{2}x}}{\sqrt{\cos x\sin x}}dx$

Multiply numerator and denominator by $\sqrt{2}$

$\sqrt{2}\int \frac{\sqrt{{\sin }^{2}x}+\sqrt{{\cos }^{2}x}}{\sqrt{2\sin x\cos x}}dx$

$\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$ $(\sin 2x=2\sin x\cos x)$

Let $\sin x-\cos x=t$

$\Rightarrow (\cos x+\sin x)dx=dt⟶\left(1\right)$

squaring both sides

$\Rightarrow {\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=t^2$

$1-\sin 2x=t^2$ $[{\sin }^{2}x+{\cos }^2=1,\sin 2x=2\sin x\cos x]$

$1-t^2=\sin 2x$ $⟶\left(2\right)$

Putting the values $\left(1\right)\&\left(2\right)$ in A

$=\sqrt{2}\int \frac{dt}{\sqrt{1-t^2}}$

$=\sqrt{2}{\sin }^{-1}t+c$

$=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+c$

Hence, the answer is $\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+c.$