$\int \frac{1 - {tan}^{2} 3 x}{1 + {tan}^{2} 3 x} d x$

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Solution

$\int \frac{1 - {tan}^{2} 3 x}{1 + {tan}^{2} 3 x} d x = \int cos 6 x d x = \frac{1}{6} sin 6 x + C$

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