Question

Let $f\left(x\right)=\int \frac{\cot x-\tan x}{\cos 4x+1}dx.$ If $f\left(\frac{\pi }{8}\right)=0,$ then the value of $4f\left(\frac{\pi }{6}\right)$ is

• A. $\ln 3$
• B. $\ln \sqrt{3}$
• C. $\ln 9$
• D. $0$

Answer 1

The correct option is A $\ln 3$

$f\left(x\right)=\int \frac{\cot x-\tan x}{\cos 4x+1}dx=\int \frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{2{\cos }^{2}2x-1+1}dx$

$=\int \frac{\frac{{\cos }^{2}x-{\sin }^{2}x}{\sin x.\cos x}}{2{\cos }^{2}2x}dx$
$=\int \frac{\cos 2x}{2\sin x.\cos x.{\cos }^{2}2x}dx$
$=\int \frac{2}{\sin 4x}dx$
$=2\int \text{cosec}4xdx$

$f\left(x\right)=\frac{1}{2}\ln |\text{cosec}4x-\cot 4x|+C$
$=\frac{1}{2}\ln |\frac{1}{\sin 4x}-\frac{\cos 4x}{\sin 4x}|+C$
$=\frac{1}{2}\ln |\tan 2x|+C$

$f\left(\frac{\pi }{8}\right)=0\Rightarrow C=0$
$\Rightarrow f\left(x\right)=\frac{1}{2}\ln |\tan 2x|$

$\therefore 4f\left(\frac{\pi }{6}\right)=4\times \frac{1}{2}\ln \sqrt{3}=2\times \frac{1}{2}\ln 3=\ln 3$