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Q. `int` `(sec x - tan x)/(sec x + tan x)` `dx` = `?` A) `log ( sec x + tan x) + c` B)...

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user8235304 | Student, Grade 11 | Valedictorian

Posted June 30, 2013 at 1:51 AM via web

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Q. `int` `(sec x - tan x)/(sec x + tan x)` `dx` = `?`

A) `log ( sec x + tan x) + c`

B) `2 tan x - 2 sec x - x + c`

C) `tan x + 2 sec x - x + c`

D)`2 tan x - 2 sec x + x + c`

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llltkl | College Teacher | Valedictorian

Posted June 30, 2013 at 2:27 AM (Answer #1)

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`int(secx-tanx)/(secx+tanx)dx`

multiplying both numerator and denominator by (secx-tanx)

=`int(secx-tanx)^2/((secx+tanx)(secx-tanx))dx`

=`int(sec^2x-2secxtanx+tan^2x)/(sec^2x-tan^2x)dx` (`sec^2x=tan^2x+1` )

=`int(2sec^2x-2secxtanx-1)dx`

=`2intsec^2xdx-2intsecxtanxdx-intdx`

=`2tanx-2secx-x+c`

where, c is the constant of integration.

Therefore the correct answer is option B).

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