Question

The value of $\int \sqrt{2}\left(\frac{\sin x}{\sin (x-\frac{\pi }{4})}\right)dx$ is

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

Answer 1

The correct option is D $x+\log \left|\sin (x-\frac{\pi }{4})\right|+c$

$\sqrt{2}\int \frac{\sin xdx}{\sin (x-\frac{\pi }{4})}$
$=\sqrt{2}\int \frac{\sin x}{\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x}$
$=\int \frac{2\sin xdx}{\sin x-\cos x}$
$=\int \frac{\sin x-\cos x}{\sin x-\cos x}dx+\int \frac{\sin x+\cos x}{\sin x-\cos x}dx$
$=\int 1dx+\int \frac{\sin x+\cos x}{\sin x-\cos x}dx$
Put $t=\sin x-\cos x$
$\Rightarrow dt=\cos x+\sin x$

$=x+\int \frac{dt}{t}$

$=x+\log \left|t\right|+c_1$
$=x+\log \left|(\sin x-\cos x)\right|+c_1$
$=x+\log \left|\sin (x-\frac{\pi }{4})\right|+c$ where $c=c_1+\log \sqrt{2}$
Hence, option 'D' is correct.