Question

If \[ \int \frac{1}{\cos 4x \cos 2x}\,dx = \frac{1}{4\sqrt{2}}\log\left|\frac{1+f(x)}{1-f(x)}\right| -\frac{1}{2}\log g(x)+C, \] then \[ g\left(\frac{\pi}{6}\right)-\sqrt{2}f\left(\frac{\pi}{6}\right)= \]

• A. \(\frac{\pi}{2\sqrt{2}}\)
• B. \(\pi+3\)
• C. \(2\)
• D. \(1\)

Hint:

For integrals involving \(\cos 2x\) and \(\cos 4x\), the substitution \(t=\tan x\) often converts the expression into a rational function of \(t\), which can then be simplified using partial fractions.

Answer 1

The Correct Option is C

Step 1: Put \(t=\tan x\).
We have \[ \int \frac{1}{\cos 4x \cos 2x}\,dx \] Let \[ t=\tan x \] Then, \[ dx=\frac{dt}{1+t^2} \] Also, \[ \cos 2x=\frac{1-t^2}{1+t^2} \] and \[ \cos 4x=\frac{t^4-6t^2+1}{(1+t^2)^2} \] Therefore, the integral becomes \[ \int \frac{(1+t^2)^2}{(1-t^2)(t^4-6t^2+1)}\,dt \]

Step 2: Use partial fraction form.
After simplifying by partial fractions, the integral can be written in the form \[ \frac{1}{4\sqrt{2}}\log\left|\frac{1+f(x)}{1-f(x)}\right| -\frac{1}{2}\log g(x)+C \] where \[ f(x)=\sqrt{2}\sin 2x \] and \[ g(x)=\left|\frac{\tan x+1}{\tan x-1}\right| \]

Step 3: Find \(f\left(\frac{\pi}{6}\right)\).
Now, \[ f\left(\frac{\pi}{6}\right) = \sqrt{2}\sin\left(\frac{\pi}{3}\right) \] Since \[ \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}, \] we get \[ f\left(\frac{\pi}{6}\right) = \frac{\sqrt{6}}{2} \] Therefore, \[ \sqrt{2}f\left(\frac{\pi}{6}\right) = \sqrt{2}\cdot \frac{\sqrt{6}}{2} = \sqrt{3} \]

Step 4: Find \(g\left(\frac{\pi}{6}\right)\).
We know \[ \tan\frac{\pi}{6}=\frac{1}{\sqrt{3}} \] Thus, \[ g\left(\frac{\pi}{6}\right) = \left|\frac{\frac{1}{\sqrt{3}}+1}{\frac{1}{\sqrt{3}}-1}\right| \] \[ = \left|\frac{1+\sqrt{3}}{1-\sqrt{3}}\right| \] \[ = \left|\frac{(1+\sqrt{3})^2}{1-3}\right| \] \[ = \left|\frac{1+2\sqrt{3}+3}{-2}\right| \] \[ = 2+\sqrt{3} \]

Step 5: Calculate the required value.
Now, \[ g\left(\frac{\pi}{6}\right)-\sqrt{2}f\left(\frac{\pi}{6}\right) = (2+\sqrt{3})-\sqrt{3} \] \[ =2 \]

Step 6: Final conclusion.
Hence, \[ \boxed{2} \]