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

$\begin{array}{l} \int(\sqrt{\tan x}+\sqrt{\cot x}) d x=\int\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\right) d x=\int\left(\frac{\sin x+\cos x}{\sqrt{\sin x \cdot \cos x}}\right) d x \\ =\int \frac{\sqrt{2}(\sin x+\cos x) d x}{\sqrt{2 \sin x \cos x}}=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{1-(1-2 \sin x \cos x)}} d x=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)}} \\ =\int \frac{\sqrt{2}(\sin x+\cos x) d x}{\sqrt{1-(\sin x-\cos x)^{2}}} \mid \text { ধরি, }-\cos x+\sin x=z \Rightarrow \frac{d z}{d x}=(\sin x+\cos x) \Rightarrow(\sin x+\cos x) d x=d z \\ \text { তাহলে, } \int \frac{\sqrt{2} d z}{\sqrt{1-z^{2}}}=\sqrt{2} \sin ^{-1} z+c=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+c \text { (Ans.) }\end{array}$