Question

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

• A. $a=\frac{3\pi }{4},C\in R$
• B. $a=\frac{5\pi }{4},C\in R$
• C. $a=\frac{\pi }{4},C\in R$
• D. $a=\frac{\pi }{3},C\in R$

Answer 1

Answer: (C)

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

Consider the given integral.

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

Now,

$I=\int (\sin 2x+\cos 2x)dx$

$I=-\frac{\cos 2x}{2}+\frac{\sin 2x}{2}+C$

$I=\frac{1}{2}[\sin 2x-\cos 2x]+C$

$I=\frac{\sqrt{2}}{2}[\frac{1}{\sqrt{2}}\sin 2x-\frac{1}{\sqrt{2}}\cos 2x]+C$

$I=\frac{1}{\sqrt{2}}[\cos \frac{\pi }{4}\sin 2x-\sin \frac{\pi }{4}\cos 2x]+C$

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

On comparing above equation, we get

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

Hence, this is the answer.