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Let $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}}}$

$\frac{{dI}}{{d\alpha }} = ?$

We have, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}(\pi /2 - x)}}} $

$ \Rightarrow I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\cot }^{\sqrt {\tan \alpha } }}x}}}  = \int\limits_0^{\pi /2} {\frac{{{{\sin }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\sin }^{\sqrt {\tan \alpha } }}x + {{\cos }^{\sqrt {\tan \alpha } }}x}}} $ ........(A)

Also, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}} = } \int\limits_0^{\pi /2} {\frac{{{{\cos }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\cos }^{\sqrt {\tan \alpha } }}x + {{\sin }^{\sqrt {\tan \alpha } }}x}}}$ ........(B)

(A) + (B) gives, $2I = \int\limits_0^{\pi /2} {dx}  = \frac{\pi }{2}$

$ \Rightarrow I = \frac{\pi }{4}$

$\therefore \frac{{dI}}{{d\alpha }} = 0$

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