Question

$\int (\sin 2x-\cos 2x)dx=\frac{1}{\sqrt{\left(k\right)}}\sin (2x-a)+b.$ Find the value of $k$.

Answer 1

Let $I=\int (\sin 2x-\cos 2x)dx$
$=\sqrt{2}\int (\frac{1}{\sqrt{\left(2\right)}}\sin 2x-\frac{1}{\sqrt{\left(2\right)}}\cos 2x)dx$
$=-\sqrt{2}.\int \cos (2x+\frac{\pi }{4})dx=-\sqrt{2}.\frac{1}{2}\sin (2x+\frac{\pi }{4})+c$
$=\frac{1}{\sqrt{\left(2\right)}}\sin (\pi +2x+\frac{\pi }{4})+b$ $(\because \sin (\pi +\theta )=-\sin \theta )$
$=\frac{1}{\sqrt{\left(2\right)}}\sin (2x-(-\frac{5\pi }{4}))+b$
Therefore $a=-\frac{5\pi }{4},k=2,b$ is any constant