If _ ^ dx 1 + x = ( x 2 + a ) + b , then - Vidyadip

MCQ

If $\int_{}^{} {\frac{{dx}}{{1 + \sin x}} = \tan \left( {\frac{x}{2} + a} \right) + b} $, then

A
$a = \frac{\pi }{4},\;b = 3$
B
$a = - \frac{\pi }{4},\;b = 3$
C
$a = \frac{\pi }{4},\;b = $arbitrary constant
✓
$a = - \frac{\pi }{4},\;b = $arbitrary constant

✓

Answer

Correct option: D.

$a = - \frac{\pi }{4},\;b = $arbitrary constant

d
(d)$\int_{}^{} {\frac{{dx}}{{1 + \sin x}}} = \tan x - \sec x + c = - \frac{{1 - \sin x}}{{\cos x}}$
$ = - \frac{{{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}}}{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}} + c = - \frac{{1 - \tan \frac{x}{2}}}{{1 + \tan \frac{x}{2}}} + c$
$ = \frac{{\tan \frac{x}{2} - 1}}{{1 + \tan \frac{x}{2}}} + c = \tan \left( {\frac{x}{2} - \frac{\pi }{4}} \right) + c$
$ \Rightarrow a = - \frac{\pi }{4},\,\,b = $ arbitrary constant.

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