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Question
Evaluate:
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Solution
\[\int\left( \frac{2 \cos^2 x - \cos 2x}{\cos^2 x} \right)dx\]
\[ = \int\left( \frac{2 \cos^2 x - \left( 2 \cos^2 x - 1 \right)}{\cos^2 x} \right)dx \left[ \because \cos 2x = 2 \cos^2 x - 1 \right]\]
\[ = \int \sec^2\text{ x dx}\]
\[ = \ \text{tan x} + C\]
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