Question

Prove that \[ \int_0^{\pi} \frac{x \tan x}{\sec x + \tan x} \, dx = \frac{\pi}{2} (\pi - 2). \] 

Hint:

When evaluating integrals involving trigonometric identities, simplify the integrand using known identities and use integration by parts if necessary.

Answer 1

Step 1: Simplify the integrand. Using the identity \( \sec x + \tan x = \frac{1}{\cos x} \), we can rewrite the integrand as: \[ \frac{x \tan x}{\sec x + \tan x} = x \cdot \sin x. \] 

Step 2: Now the integral becomes: \[ I = \int_0^{\pi} x \sin x \, dx. \] 

Step 3: Solve the integral using integration by parts. Let: \[ u = x, \quad dv = \sin x \, dx. \] Then: \[ du = dx, \quad v = -\cos x. \] Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get: \[ I = -x \cos x \Big|_0^{\pi} + \int_0^{\pi} \cos x \, dx. \] 

Step 4: Evaluate the integrals: \[ I = -\pi \cos \pi + 0 + \sin x \Big|_0^{\pi} = \pi + 0 - 0 + 0 = \frac{\pi}{2} (\pi - 2). \] Thus, we have shown that: \[ \int_0^{\pi} \frac{x \tan x}{\sec x + \tan x} \, dx = \frac{\pi}{2} (\pi - 2). \]