The given function is \( \displaystyle \frac{2\, – 3 \sin x}{\cos^2 x} \) and we need to find the integration of this function with respect to \( x \).
Solution:
\( \begin{aligned} &\int \frac{2\, – 3 \sin x}{\cos^2 x} \, dx \\ &= \int \frac{2}{\cos^2 x} \, dx \, – \int \frac{3 \sin x}{\cos^2 x} \, dx \\ &= 2 \int \sec^2 x \, dx \, – 3 \int \frac{\sin x}{\cos^2 x} \, dx \\ &= 2 \int \sec^2 x \, dx \, – 3 \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \, dx \\ &= 2 \int \sec ^{2} x \, dx \, -3 \int \tan x \cdot \sec x \, dx \\ &= 2 \tan x \, – 3 \sec x + C , \quad \text{C is a constant}\\ \end{aligned} \)Thus, the integration of the given function is
\( \boxed{\int \frac{2\, – 3 \sin x}{\cos^2 x} \, dx = 2 \tan x \, – 3 \sec x + C} \\ \)The following part shows how the integral of \( \displaystyle \int \frac{3 \sin x}{\cos^2 x} \, dx \) is calculated.
\( \begin{aligned} \text{Let } u = \cos x \implies du = -\sin x \, dx \\ \implies -du = \sin x \, dx \\ \end{aligned} \)Therefore, by substitution:
\( \begin{aligned} &\int \frac{3 \sin x}{\cos^2 x} \, dx = 3 \int \frac{\sin x \, dx}{\cos^2 x} \\ &= 3 \int \frac{-du}{u^2} = -3 \int u^{-2} \, du = -3 \left( \frac{u^{-1}}{-1} \right) + c \\ &= \frac{3}{u} + c = \frac{3}{\cos x} + c \\ &= 3 \sec x + c, \quad \text{c is a constant} \\ \end{aligned} \)Please let me know in the comments if you find any errors in this solution.