The correct option is D π/4 I=∫_0^π/2dx1+tan #160; #160;x ∫_o^af(x)=∫_o^af(a x)dx I=∫_0^π/2dx1+tan #160; #160;x=∫_0^π/2dx1+ #16 ...

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Property 1

∫0π/2dx/1+tan...

Question

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

\frac{π}{4}

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\frac{π}{2}

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Solution

The correct option is D $\frac{π}{4}$
$I = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + tan x}$
$\int_{o}^{a} f (x) = \int_{o}^{a} f (a - x) d x$
$I = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + tan x} = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + cot x} = \int_{0}^{\frac{π}{2}} \frac{tan x}{1 + tan x} d x$
$2 I = \int_{0}^{\frac{π}{2}} 1 \cdot d x$
$I = (\frac{π}{2}) \cdot \frac{1}{2}$
$I = \frac{π}{4}$
$\int_{0}^{\frac{π}{2}} \frac{d x}{1 + tan x} = \frac{π}{4}$

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