Question

The value of $\sqrt{2}\int \frac{\sin xdx}{\sin (x-\frac{\pi }{4})}$ is
(where $C$ is constant of integration)

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

Answer 1

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

Let $I=\sqrt{2}\int \frac{\sin xdx}{\sin (x-\frac{\pi }{4})}$
Put $x-\frac{\pi }{4}=t$
$\Rightarrow dx=dt$
$\therefore I=\sqrt{2}\int \frac{\sin (t+\frac{\pi }{4})}{\sin t}dt$
$\Rightarrow I=\frac{\sqrt{2}}{\sqrt{2}}\int \left(\frac{\sin t+\cos t}{\sin t}\right)dt$
$\Rightarrow I=\int (1+\cot t)dt$
$\Rightarrow I=t+\ln |\sin t|+c_1$
$\Rightarrow I=x-\frac{\pi }{4}+\ln \left|\sin (x-\frac{\pi }{4})\right|+c_1$
$\therefore I=x+\ln \left|\sin (x-\frac{\pi }{4})\right|+C$
$(\text{where}C=c_1-\frac{\pi }{4})$