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Question

{1+2 tan x(tan x+sec x)}12 dx is equal to

A
ln |sec x(sec xtan x)|+c
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B
ln |cosec x(sec x+tan x)|+c
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C
ln |sec x(sec x+tan x)|+c
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D
ln |(sec x+tan x)|+c
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Solution

The correct option is C ln |sec x(sec x+tan x)|+c
{1+2 tan x(tan x+sec x)}12 dx
{1+2 tan2 x+2 tan x sec x}12 dx

usng trigonometric identity sec2 xtan2 x=1
{sec2 x+tan2 x+2 tan x sec x}12 dx
=(sec x+tan x) dx
=ln sec x+ln (sec x+tan x)+c
=ln |sec x(sec x+tan x)|+c

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