We have, I=π/2∫0dx1+tan√tanα(π/2−x)
⇒I=π/2∫0dx1+cot√tanαx=π/2∫0sin√tanαx.dxsin√tanαx+cos√tanαx ........(A)
Also, I=π/2∫0dx1+tan√tanαx=π/2∫0cos√tanαx.dxcos√tanαx+sin√tanαx ........(B)
(A) + (B) gives, 2I=π/2∫0dx=π2
⇒I=π4
∴dIdα=0
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