The correct option is D π /2 I= ∫_0^πdx/1+2^tan ( π x )= ∫_0^πdx/1+2^ tan x or I= ∫_0^π2^tan x/1+2^tan xdx. Adding 2I= ∫_0^π1+2^tan x/1+2 ...

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Property 1

∫0πdx/1+2tan ...

Question

$\int_{0}^{π} \frac{d x}{1 + 2^{tan x}} =$

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π / 4

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π / 2

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π

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Solution

The correct option is D $π / 2$
$I = \int_{0}^{π} \frac{d x}{1 + 2^{tan (π - x)}} = \int_{0}^{π} \frac{d x}{1 + 2^{- tan x}}$
or $I = \int_{0}^{π} \frac{2^{tan x}}{1 + 2^{tan x}} d x .$ Adding
$2 I = \int_{0}^{π} \frac{1 + 2^{tan x}}{1 + 2^{tan x}} d x = \int_{0}^{π} 1 d x = {[x]}_{0}^{π} = π$

$∴ I = \frac{π}{2} .$

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