If f(x) = int left[ tan^2 x + cot^2 x + frac{ | vedclass

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(A) Given $f(x) = \int \left[ 	an^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{(2 \sin x \cos x)^2} ight] dx$. Since $\sin^2 2x = 4 \sin^2 x \cos

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(A) Given $f(x) = \int \left[ 	an^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{(2 \sin x \cos x)^2} ight] dx$. Since $\sin^2 2x = 4 \sin^2 x \cos^2 x$, the expression becomes $\int \left[ 	an^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{4 \sin^2 x \cos^2 x} ight] dx$. $= \int \left[ 	an^2 x + \cot^2 x + \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} ight] dx$. $= \int \left[ (\sec^2 x - 1) + (\csc^2 x - 1) + \sec x 	an x + \csc x \cot x ight] dx$. $= 	an x - x - \cot x - x + \sec x - \csc x + C$. $f(x) = 	an x - \cot x + \sec x - \csc x - 2x + C$. Given $f\left(\frac{\pi}{4}ight) = 0$: $	an\frac{\pi}{4} - \cot\frac{\pi}{4} + \sec\frac{\pi}{4} - \csc\frac{\pi}{4} - 2\left(\frac{\pi}{4}ight) + C = 0$. $1 - 1 + \sqrt{2} - \sqrt{2} - \frac{\pi}{2} + C = 0 \implies C = \frac{\pi}{2}$. So $f(x) = 	an x - \cot x + \sec x - \csc x - 2x + \frac{\pi}{2}$. Now $f\left(\frac{\pi}{6}ight) = 	an\frac{\pi}{6} - \cot\frac{\pi}{6} + \sec\frac{\pi}{6} - \csc\frac{\pi}{6} - 2\left(\frac{\pi}{6}ight) + \frac{\pi}{2}$. $= \frac{1}{\sqrt{3}} - \sqrt{3} + \frac{2}{\sqrt{3}} - 2 - \frac{\pi}{3} + \frac{\pi}{2} = \frac{3}{\sqrt{3}} - \sqrt{3} - 2 + \frac{\pi}{6} = \sqrt{3} - \sqrt{3} - 2 + \frac{\pi}{6} = \frac{\pi}{6} - 2$. Then $3 \left[ f\left(\frac{\pi}{6}ight) + 2 ight] = 3 \left[ \frac{\pi}{6} - 2 + 2 ight] = 3 \left( \frac{\pi}{6} ight) = \frac{\pi}{2}$.

If $f(x) = \int \left[ \tan^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{\sin^2 2x} \right] dx$ and $f\left(\frac{\pi}{4}\right) = 0$, then $3 \left[ f\left(\frac{\pi}{6}\right) + 2 \right] = $

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