The given function is \(\displaystyle \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x \).
Solution:
\( \begin{aligned} &\int \frac{\cos 2x + 2 \sin^2 x}{\cos^2 x} \, dx \\ &= \int \frac{\cos^2 x \,- \sin^2 x + 2 \sin^2 x}{\cos^2 x} \, dx \quad \because \cos 2x = \cos^2 x \,- \sin^2 x \\ &= \int \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \, dx \\ &= \int \frac{1}{\cos^2 x} \, dx \quad \because \cos^2 x + \sin^2 x = 1 \\ &= \int \sec^2 x \, dx \\ &= \tan x + C, \quad \text{C is a constant} \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \frac{\cos 2x + 2 \sin^2 x}{\cos^2 x} \, dx = \tan x + C} \\ \)Please let me know in the comments if you find any errors in this solution.