Question

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

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

Answer 1

Answer: (B)

a = –$\frac{5\pi}{4}$, b Ī R

Answer 2

ņ (sin 2x – cos 2x) dx = $\frac{1}{\sqrt{2}}$ sin (2x – a) + b

Ž $\sqrt { 2 }$ $\int_{}^{}\left( \frac{1}{\sqrt{2}}\sin 2x - \frac{1}{\sqrt{2}}\cos 2x \right)$dx

= $\frac { 1 } { \sqrt { 2 } }$ sin (2x – a) + b

Ž –$\sqrt { 2 }$ $\int_{}^{}\left( \frac{1}{\sqrt{2}}\cos 2x - \frac{1}{\sqrt{2}}\sin 2x \right)$dx

= $\frac { 1 } { \sqrt { 2 } }$ sin (2x – a) + b

Ž –$\sqrt { 2 }$ $\int_{}^{\int}\cos$ (2x + p/4) dx =$\frac{1}{\sqrt{2}}$sin (2x – a) + b

Ž –$\frac{\sqrt{2}}{2}$sin (2x + p/4) + c = $\frac{1}{\sqrt{2}}$sin (2x – a) + b

Ž $\frac { 1 } { \sqrt { 2 } }$sin $\left( 2x + \frac{5\pi}{4} \right)$ + c =$\frac{1}{\sqrt{2}}$sin (2x – a) + b

\ a = – $\frac{5\pi}{4}$, b Ī R.