I=∫ _ 0 ^π #160; 1 1+ 2 ^tan x #160; #160; .........(i) I=∫ _ 0 ^π #160; 1 1+ 2 ^tan (π x) #160; #160; I=∫ _ 0 ^π #160; 1 1+ ...

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First Fundamental Theorem of Calculus

∫0πdx1+2tan x

Question

$\int_{0}^{π} \frac{d x}{1 + 2^{t a n x}}$

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Solution

$I = \int_{0}^{π} \frac{1}{1 + 2^{tan x}} . . . . . . . . . (i) I = \int_{0}^{π} \frac{1}{1 + 2^{tan (π - x)}} I = \int_{0}^{π} \frac{1}{1 + 2^{- tan x}} I = \int_{0}^{π} d \frac{2^{tan x}}{1 + 2^{tan x}} . . . . . . . . . (i i)$
Adding $(i)$ and $(i i)$
$2 I = \int_{0}^{π} 1. d x 2 I = π I = \frac{π}{2}$

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