Question

Solve: $\int \frac{\sin 2x}{\sin (x-\frac{\pi }{4}).\sin (x+\frac{\pi }{4})}dx$

• A. $\log |{\sin }^{2}x+\frac{1}{2}|$
• B. $\log |\sin x-\frac{1}{2}|$
• C. $\log |{\sin }^{2}x-\frac{1}{2}|$
• D. $\log |{\sin }^{2}x+\frac{1}{\sqrt{2}}|$

Answer 1

The correct option is C $\log |{\sin }^{2}x-\frac{1}{2}|$

Consider, $I=\int \frac{\sin 2x}{\sin (x-\frac{\pi }{4})\sin (x+\frac{\pi }{4})}dx$

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

$=\int \frac{\sin 2x}{\frac{1}{2}({\sin }^{2}x-{\cos }^{2}x)}dx$

$=\int \frac{\sin 2x}{\frac{1}{2}(2{\sin }^{2}x-1)}dx$

$=\int \frac{\sin 2x}{({\sin }^{2}x-\frac{1}{2})}dx$

Take $z={\sin }^{2}x-\frac{1}{2}$

$\Rightarrow dz=2\sin x\cos xdx=\sin 2xdx$

Hence integration becomes-

$\int \frac{dz}{z}=\log |z|$

$I=\log |{\sin }^{2}x-\frac{1}{2}|$