Question

If $\int \frac{\cos x}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx=\log \left|\frac{2f\left(x\right)-1}{2f\left(x\right)+1}\right|+C,$ then which of the following is/are true?

• A. $f\left(x\right)=2\cos x$
• B. $f\left(x\right)=\sin x$
• C. Maximum value of $f\left(x\right)=1$
• D. Maximum value of $f\left(x\right)=2$

Answer 1

Answer: (B), (C)

Maximum value of $f\left(x\right)=1$

Substitute $y=\sin x\Rightarrow dy=\cos xdx$
$I=\int \frac{dy}{y^2-{\left(\frac{1}{2}\right)}^2}$
$=\frac{1}{2\times \frac{1}{2}}\log \left|\frac{y-\frac{1}{2}}{y+\frac{1}{2}}\right|+C$
$=\log \left|\frac{2y-1}{2y+1}\right|+C=\log \left|\frac{2\sin x-1}{2\sin x+1}\right|+C$
Here $f\left(x\right)=\sin x$
Maximum value of $f\left(x\right)=1$