Question

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{4} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) d\ x . Then, \]
\[I = \int_0^\frac{\pi}{4} \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) d\ x \]
\[ \Rightarrow I = \int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx\]
\[ \Rightarrow I = \sqrt{2} \int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} dx\]
\[ \Rightarrow I = \sqrt{2} \int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{\sqrt{1 - \left( \sin x - \cos x \right)^2}} dx\]
\[Let\ \sin x - \cos\ x = t . Then, \cos x\ + \sin x\ dx\ = dt\]
\[When\ x = 0, t = 1\ and\ x\ = \frac{\pi}{4}, t = 0\]
\[ \therefore I = \sqrt{2} \int_{- 1}^0 \frac{dt}{\sqrt{1 - t^2}}\]
\[ \Rightarrow I = \sqrt{2} \left[ \sin^{- 1} t \right]_{- 1}^0 \]

\[ \Rightarrow I =\sqrt{2}\left[\sin^{-1}(0)-\sin^{-1}(-1)\right]\]

\[ \Rightarrow I = \frac{\pi}{\sqrt{2}}\]