Question:
$\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}$
Solution:
$\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}$
$=\frac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x}$ $\left[\cos 2 x=1-2 \sin ^{2} x\right]$
$=\frac{1}{\cos ^{2} x}$
$=\sec ^{2} x$
$\therefore \int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x=\int \sec ^{2} x d x=\tan x+\mathrm{C}$
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