Question

If $\int (\sin 2x-\cos 2x)dx=\frac{1}{\sqrt{2}}\sin (2x-a)+C$, then

• A. $a=\frac{-5\pi }{4},C\in R$
• B. $a=\frac{5\pi }{4},C\in R$
• C. $a=\frac{\pi }{4},CϵR$
• D. $a=\frac{\pi }{2},CϵR$

Answer 1

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

Consider, $I=\int (\sin 2x-\cos 2x)dx$

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

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

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

$\because \cos (A+B)=\cos A\cos B-\sin A\sin B$

$\Rightarrow I=-\sqrt{2}\int \left(\cos (2x+\frac{\pi }{4})\right)dx$

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

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

According to question

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

$\Rightarrow a=\frac{-5\pi }{4},C\in R$