Question
$\int\limits_0^{\pi/4}\Bigg(\sqrt{\text{tan x}}+\sqrt{{\text{cot x}}}\Bigg)\text{ dx}=\sqrt{2}\cdot\frac{\pi}{2}$
Putting sin x – cos x = t, to get (cos x + sin x) dx = dt
and sin x cos x = $\frac{\text{1 - t}^{2}}{2}$
$\therefore 1=\sqrt{2}\int\limits_{-1}^{0}\frac{\text{dt}}{\sqrt{\text{1 - t}^{2}}}$= $\sqrt{2}\cdot[\sin^{-1}\text{t}]^{0}_{-1}$
$=\sqrt{2}(\sin^{-1}\text{0}-\sin^{-1}(-1)=\sqrt{2}\cdot\frac{\pi}{2}$.
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