The value of the definite integral -0-tan x- x-tan x d x is equal toA- -2-B- -1- -1-D- -2

The correct option is D π(π 2)tan x x+tan x=tan x ( x tan x) Let I= ∫_0^ππtan x x+ tan xdx ⇒ I=∫_0^ππ( xtan x tan^2 x)dx =∫_0^ππ ( xtan x+1 ^2 x)dx =π [ x ...

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Standard XII

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

The value of the definite integral $π \int 0 \frac{π tan x}{sec x + tan x} d x$ is equal to

π (1 - π)

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π (π - 2)

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π (2 - π)

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π (π - 1)

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Solution

The correct option is B $π (π - 2)$
$\frac{tan x}{sec x + tan x} = tan x (sec x - tan x)$
Let $I = π \int 0 \frac{π tan x}{sec x + tan x} d x$
$\Rightarrow I = π \int 0 π (sec x tan x - {tan}^{2} x) d x$
$= π \int 0 π (sec x tan x + 1 - {sec}^{2} x) d x$
$= π [sec x + x - tan x]_{0}^{π}$
$= π (π - 2)$

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