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Question
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
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Solution
`(cos2x + 2sin^2 x)/ cos^2x`
= `(cos 2x + (1 - cos 2x))/cos^2` [cos2x = 1-2sin2 x]
=` 1/(cos2 x)`
= `sec^2 x`
∴ \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] = \[\int\]`sec^2x dx = tan x +c`
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