Question

If $\int \frac{\sin x}{\sin (x-\pi /4)}dx=A\left(f\left(x\right)\right)+\log |\sin x-\cos x|+C$ then

• A. $A-\sqrt{2}$
• B. $A=1/\sqrt{2}$
• C. $f\left(x\right)=\sin x$
• D. $f\left(x\right)=x$

Answer 1

Answer: (B), (D)

The correct options are
B $A=1/\sqrt{2}$
D $f\left(x\right)=x$

Let $I=\int \frac{\sin x}{\sin (x-\pi /4)}dx$

Substitute $u=x-\pi /4$
$I=\int \frac{\sin (u+\pi /4)}{\sin u}du=\int \frac{\sin u\cos \left(\pi /4\right)+\cos u\sin \left(\pi /4\right)}{\sin u}du$
$=\frac{1}{\sqrt{2}}\int du+\frac{1}{\sqrt{2}}\int \cot udu=\frac{1}{\sqrt{2}}u+\frac{1}{\sqrt{2}}\log \left(\sin u\right)+c$
$=\frac{1}{\sqrt{2}}(x-\pi /4)+\log \left(\sin (x-\pi /4)\right)+c$
$=\frac{1}{\sqrt{2}}(x-\log (\sin x-\cos x)+\log \frac{1}{\sqrt{2}})+c$

$=\frac{1}{\sqrt{2}}(x-\log (\sin x-\cos x))+c$

Comparing with , $\int \frac{\sin x}{\sin (x-\pi /4)}dx=A\left(f\left(x\right)\right)+\log |\sin x-\cos x|+C$

We get, $A=\frac{1}{\sqrt{3}}andf\left(x\right)=x$