1. The integral we start with: \[ \int x \sin x \sec^3 x \, dx \] You simplify the expression by recognizing that \(\sin x \sec^3 x = \tan x \sec^2 x\), so the integral becomes: \[ \int x \tan x \sec^2 x \, dx \] Now we apply integration by parts. \[\] 2. Applying integration by parts: \[\] The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] Here, let: \[ u = x \quad \text{and} \quad dv = \tan x \sec^2 x \, dx \] Then: \[ du = dx \quad \text{and} \quad v = \frac{\tan^2 x}{2} \] 3. Substituting in the integration by parts formula: \[ \int x \tan x \sec^2 x \, dx = \left[x \frac{\tan^2 x}{2}\right] - \int \frac{\tan^2 x}{2} \, dx \] So the integral simplifies to: \[ \frac{1}{2} \left[ x \tan^2 x - \int (\sec^2 x - 1) \, dx \right] \] 4. Simplifying the remaining integral: \[\] We can split the integral: \[ \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx = \tan x - x \] Substituting this back into the expression: \[ \frac{1}{2} \left[ x \tan^2 x - (\tan x - x) \right] = \frac{1}{2} \left[ x \tan^2 x - \tan x + x \right] + c \] This simplifies to: \[ \frac{1}{2} \left[ x \sec^2 x - \tan x \right] + c \] 5. The result simplifies to: \[ \frac{1}{2} \left[ f(x) \sec^2 x + g(x) \left( \frac{\tan x}{x} \right) \right] + c \] Based on this, we deduce that: \[ f(x) = x \quad \text{and} \quad g(x) = -x \] hence: \[ f(x) + g(x) = 0 \]