Question

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

• A. $x-\log \left|\sin \right(x-\frac{\pi }{4}\left)\right|+c$
• B. $x+\log \left|\sin \right(x+\frac{\pi }{4}\left)\right|+c$
• C. $x-\log \left|\cos \right(x-\frac{\pi }{4}\left)\right|+c$
• D. $x+\ln |\sin x-\cos x|+c$

Answer 1

The correct option is D $x+\ln |\sin x-\cos x|+c$

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

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

$=2\int \frac{\sin x}{\sin x-\cos x}dx$

$=\int \frac{\sin x+\sin x+\cos x-\cos x}{\sin x-\cos x}dx$

$=\int \frac{\sin x-\cos x}{\sin x-\cos x}dx+\int \frac{\sin x+\cos x}{\sin x-\cos x}dx$

$=x+\int \frac{\sin x+\cos x}{\sin x-\cos x}dx$

$=x+\ln |\sin x-\cos x|+c$