The given function is \(\displaystyle \frac{\sec x}{\sec x + \tan x} \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \frac{\sec x}{\sec x + \tan x} \, dx \).
Solution:
\( \begin{aligned} &\int \frac{\sec x}{\sec x + \tan x} \, dx \\ &= \int \frac{\sec x \cdot \cos x}{(\sec x + \tan x) \cdot \cos x} \, dx \quad \text{multiply the numerator and the denominator by } \cos x \\ &= \int \frac{1}{1 + \sin x} \, dx \\ &= \int \frac{1 \,- \sin x}{(1 \,- \sin x) \cdot (1 + \sin x)} \, dx \\ &= \int \frac{1 \,- \sin x}{1 \,- \sin^2 x} \, dx \\ &= \int \frac{1 \,- \sin x}{\cos^2 x} \, dx \\ &= \int \frac{1}{\cos^2 x} \, dx \,- \int \frac{\sin x}{\cos^2 x} \, dx \\ &= \int \sec^2 x \, dx \,- \int \frac{\sin x}{\cos^2 x} \, dx \\ &= \tan x + C_1 \,- \int \frac{\sin x}{\cos^2 x} \, dx \\ \end{aligned} \)Let \( u = \cos x \), then \( du = -\sin x \, dx \). Therefore,
\( \begin{aligned} &\int \frac{\sin x}{\cos^2 x} \, dx \\ &= -\int \frac{1}{u^2} \, du \\ &= \frac{1}{u} + C_2 = \sec x + C_2 \\ \end{aligned} \)Combining the results, the integral is:
\( \begin{aligned} \boxed{\int \frac{\sec x}{\sec x + \tan x} \, dx = \tan x \,- \sec x + C} \end{aligned} \)where \(C = C_1 – C_2 \) is the constant of integration.
Please let me know in the comments if you find any errors in this solution.