_ ^ 1 + x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\sqrt {1 + \sin x} \;dx = } $

A
$\frac{1}{2}\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right) + c$
B
$\frac{1}{2}\left( {\sin \frac{x}{2} - \cos \frac{x}{2}} \right) + c$
C
$2\sqrt {1 + \sin x} + c$
✓
$ - 2\sqrt {1 - \sin x} + c$

✓

Answer

Correct option: D.

$ - 2\sqrt {1 - \sin x} + c$

d
(d)$\int_{}^{} {\sqrt {1 + \sin x} \,dx} = \int_{}^{} {\sqrt {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}} } dx$
$ = \int_{}^{} {\sin \frac{x}{2}\,dx} + \int_{}^{} {\cos \frac{x}{2}\,dx} = - 2\cos \frac{x}{2} + 2\sin \frac{x}{2} + c$
$ = - 2\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right) + c = - 2\sqrt {(1 - \sin x)} + c.$

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