1. We are asked to find the integral $$\int \frac{\tan x}{\cos x} \, dx$$ and identify which of the given options matches the result. 2. Recall that $$\tan x = \frac{\sin x}{\cos x}$$, so substitute this into the integral: $$\int \frac{\tan x}{\cos x} \, dx = \int \frac{\frac{\sin x}{\cos x}}{\cos x} \, dx = \int \frac{\sin x}{\cos^2 x} \, dx$$ 3. Rewrite the integral as: $$\int \sin x \cdot \sec^2 x \, dx$$ 4. Use substitution: let $$u = \cos x$$, then $$du = -\sin x \, dx$$ or $$-du = \sin x \, dx$$. 5. Substitute into the integral: $$\int \sin x \cdot \sec^2 x \, dx = \int \sin x \cdot \frac{1}{\cos^2 x} \, dx = \int \frac{\sin x}{u^2} \, dx$$ Since $$\sin x \, dx = -du$$, the integral becomes: $$\int \frac{\sin x}{u^2} \, dx = \int \frac{-du}{u^2} = -\int u^{-2} \, du$$ 6. Integrate: $$-\int u^{-2} \, du = -\left(-u^{-1}\right) + C = \frac{1}{u} + C$$ 7. Substitute back $$u = \cos x$$: $$\frac{1}{\cos x} + C = \sec x + C$$ 8. Therefore, the integral evaluates to $$\sec x + C$$. 9. Comparing with the options: (a) $$\sec^2 x$$ (b) $$\sec x$$ (c) $$\tan x \sec x$$ (d) $$\ln \cos x$$ The correct answer is (b) $$\sec x$$. Final answer: $$\int \frac{\tan x}{\cos x} \, dx = \sec x + C$$.