If int frac{1}{cos 4x cos 2x} dx = frac{1}{2s | vedclass

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(C) We have $I = \int \frac{1}{\cos 4x \cos 2x} dx$. Multiply and divide by $\sin 2x$: $I = \int \frac{\sin 2x}{\sin 2x \cos 4x \cos 2x} dx = \int \fr

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$2$

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(C) We have $I = \int \frac{1}{\cos 4x \cos 2x} dx$. Multiply and divide by $\sin 2x$: $I = \int \frac{\sin 2x}{\sin 2x \cos 4x \cos 2x} dx = \int \frac{\sin 2x}{\sin 2x \cos 2x (2\cos^2 2x - 1)} dx$. Using $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$,this is not the simplest path. Alternatively,use $\frac{1}{\cos 4x \cos 2x} = \frac{1}{\sin 2x} \left( \frac{\sin(4x-2x)}{\cos 4x \cos 2x} ight) = \frac{1}{\sin 2x} (	an 4x - 	an 2x)$. Actually,the standard identity is $\frac{1}{\cos 4x \cos 2x} = \frac{1}{\sin 2x} \frac{\sin(4x-2x)}{\cos 4x \cos 2x} = \frac{1}{\sin 2x} (	an 4x - 	an 2x)$. Integrating,we get $I = \int \frac{	an 4x}{\sin 2x} dx - \int \frac{	an 2x}{\sin 2x} dx$. Using the partial fraction decomposition provided in the prompt: $I = \frac{1}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2}\sin 2x}{1-\sqrt{2}\sin 2x} ight| - \frac{1}{2} \log |\sec 2x + 	an 2x| + C$. Comparing with the given form,$f(x) = \sqrt{2}\sin 2x$ and $g(x) = |\sec 2x + 	an 2x|$. At $x = \frac{\pi}{6}$: $f\left(\frac{\pi}{6}ight) = \sqrt{2} \sin\left(\frac{\pi}{3}ight) = \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}}$. $g\left(\frac{\pi}{6}ight) = \sec\left(\frac{\pi}{3}ight) + 	an\left(\frac{\pi}{3}ight) = 2 + \sqrt{3}$. Thus,$g\left(\frac{\pi}{6}ight) - \sqrt{2} f\left(\frac{\pi}{6}ight) = (2 + \sqrt{3}) - \sqrt{2} \left(\frac{\sqrt{3}}{\sqrt{2}}ight) = 2 + \sqrt{3} - \sqrt{3} = 2$.

If $\int \frac{1}{\cos 4x \cos 2x} dx = \frac{1}{2\sqrt{2}} \log \left(\frac{1+f(x)}{1-f(x)}\right) - \frac{1}{2} \log g(x) + C$,then find the value of $g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right)$.

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