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

MCQ

If $\int_{}^{} {(\sin 2x - \cos 2x)} \;dx = \frac{1}{{\sqrt 2 }}\sin (2x - a) + b$, then

A
$a = \frac{\pi }{4},\;b = 0$
B
$a = - \frac{\pi }{4},\;b = 0$
C
$a = \frac{{5\pi }}{4},\;b = $any constant
✓
$a = - \frac{{5\pi }}{4},\;b = $any constant

✓

Answer

Correct option: D.

$a = - \frac{{5\pi }}{4},\;b = $any constant

d
(d)$\int_{}^{} {(\sin 2x - \cos 2x)\,dx = \frac{1}{{\sqrt 2 }}\sin (2x - a) + b} $
$ \Rightarrow - \frac{1}{2}(\sin 2x + \cos 2x) = \frac{1}{{\sqrt 2 }}\sin (2x - a) + b$
$ \Rightarrow - \left[ {\frac{1}{{\sqrt 2 }}\sin 2x + \frac{1}{{\sqrt 2 }}\cos 2x} \right] = \sin (2x - a) + b\sqrt 2 $
$ \Rightarrow \sin \left( {2x + \frac{{5\pi }}{4}} \right) = \sin (2x - a) + b\sqrt 2 $
$ \Rightarrow b$ is any constant and $a = \frac{{ - 5\pi }}{4}$.

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