\(\int\displaystyle\frac{\cos 2x+2\sin^2 x}{\cos^2 x}\,dx =\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\tan x+C\)
Using \(\cos 2x = 1-2\sin^2 x\), the numerator is 1:
\(\cos 2x+2\sin^2 x = 1-2\sin^2 x+2\sin^2 x\)
\(= 1\)
So
\( \displaystyle \int\frac{\cos 2x+2\sin^2 x}{\cos^2 x}\,dx = \int\frac{1}{\cos^2 x}\,dx \)
\( \displaystyle = \int\sec^2 x\,dx \)
\(= \tan x+C\)
NCERT (2026-27 edition):
Integration using Trigonometric Identities
