Question

$\int \left(\sin 2x-\cos 2x\right)dx=\frac{1}{\sqrt{2}}\sin \left(2x-a\right)+b$ then

• A. $a=\frac{5\pi }{4},b\in R$
• B. $a=-\frac{5\pi }{4},b\in R$
• C. $a=\frac{\pi }{4},b\in R$
• D. None of these

Answer 1

The correct option is A $a=\frac{5\pi }{4},b\in R$

$\int (\sin 2x-\cos 2x)dx=-\sqrt{2}\int (\frac{-1}{\sqrt{2}}\sin 2x+\frac{1}{\sqrt{2}}\cos 2x)dx$

$=-\sqrt{2}\int (\sin 2x\sin \frac{5\pi }{4}-\cos \frac{5\pi }{4}\cos 2x)dx$

$=-\sqrt{2}\cos (2x-\frac{5\pi }{4})dx$

$=\frac{\sqrt{2}}{2}\sin (2x-\frac{5\pi }{4})+b$

$=\frac{1}{\sqrt{2}}\sin (2x-\frac{5\pi }{4})+b$

Comparing with $\frac{1}{\sqrt{2}}\sin \left(2x-a\right)+b$

$\therefore a=\frac{5\pi }{4},b\in R$