If _ ^ ( x - x) ;dx = 2 (x + ) + c , then = - Vidyadip

MCQ

If $\int_{}^{} {(\cos x - \sin x)\;dx = \sqrt 2 \sin (x + \alpha ) + c} $, then $\alpha = $

A
$\frac{\pi }{3}$
B
$ - \frac{\pi }{3}$
✓
$\frac{\pi }{4}$
D
$ - \frac{\pi }{4}$

✓

Answer

Correct option: C.

$\frac{\pi }{4}$

c
(c) Given that $\int_{}^{} {(\cos x - \sin x)\,dx} = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sin x + \cos x + c = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sqrt 2 \left( {\frac{{\sin x}}{{\sqrt 2 }} + \frac{{\cos x}}{{\sqrt 2 }}} \right) + c = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) + c = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin (x + \alpha ) \Rightarrow \alpha = \frac{\pi }{4}.$

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