Question

\( \int_0^\pi \frac{x \tan(x)}{\sec(x) + \cos(x)} \, dx = ? \)

Answer 1

1. Simplify the Integrand:
We are given the integral: \[ \int_0^\pi \frac{x \tan(x)}{\sec(x) + \cos(x)} \, dx \] We know that \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) and \( \sec(x) = \frac{1}{\cos(x)} \). Substituting these into the integral: \[ \int_0^\pi \frac{x \left( \frac{\sin(x)}{\cos(x)} \right)}{\left( \frac{1}{\cos(x)} \right) + \cos(x)} \, dx \] To simplify the denominator, we find a common denominator: \[ \frac{1}{\cos(x)} + \cos(x) = \frac{1 + \cos^2(x)}{\cos(x)} \] Substituting this back into the integral: \[ \int_0^\pi \frac{x \left( \frac{\sin(x)}{\cos(x)} \right)}{\frac{1 + \cos^2(x)}{\cos(x)}} \, dx \] Simplifying by multiplying by the reciprocal of the denominator: \[ \int_0^\pi \frac{x \sin(x)}{\cos(x)} \cdot \frac{\cos(x)}{1 + \cos^2(x)} \, dx \] The \( \cos(x) \) terms cancel, leaving: \[ \int_0^\pi \frac{x \sin(x)}{1 + \cos^2(x)} \, dx \]

2. Apply the Property of Definite Integrals:
We can use the property of definite integrals: \[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \] Let: \[ I = \int_0^\pi \frac{x \sin(x)}{1 + \cos^2(x)} \, dx \] Then, applying the property: \[ I = \int_0^\pi \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx \] Since \( \sin(\pi - x) = \sin(x) \) and \( \cos(\pi - x) = -\cos(x) \), we have: \[ I = \int_0^\pi \frac{(\pi - x) \sin(x)}{1 + \cos^2(x)} \, dx \]

3. Add the Two Integrals:
Now, add the original integral and the transformed integral: \[ 2I = \int_0^\pi \frac{x \sin(x)}{1 + \cos^2(x)} \, dx + \int_0^\pi \frac{(\pi - x) \sin(x)}{1 + \cos^2(x)} \, dx \] Simplifying: \[ 2I = \int_0^\pi \frac{(x + \pi - x) \sin(x)}{1 + \cos^2(x)} \, dx \] \[ 2I = \int_0^\pi \frac{\pi \sin(x)}{1 + \cos^2(x)} \, dx \] \[ 2I = \pi \int_0^\pi \frac{\sin(x)}{1 + \cos^2(x)} \, dx \]

4. Solve the Integral:
Let \( u = \cos(x) \), so that \( du = -\sin(x) \, dx \). When \( x = 0 \), \( u = \cos(0) = 1 \), and when \( x = \pi \), \( u = \cos(\pi) = -1 \). Substituting: \[ 2I = \pi \int_1^{-1} \frac{-du}{1 + u^2} \] Simplifying: \[ 2I = \pi \int_{-1}^1 \frac{du}{1 + u^2} \] The integral of \( \frac{1}{1 + u^2} \) is \( \arctan(u) \), so: \[ 2I = \pi \left[ \arctan(u) \right]_{-1}^1 \] Substituting the limits: \[ 2I = \pi \left[ \arctan(1) - \arctan(-1) \right] \] Using \( \arctan(1) = \frac{\pi}{4} \) and \( \arctan(-1) = -\frac{\pi}{4} \), we get: \[ 2I = \pi \left[ \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) \right] \] \[ 2I = \pi \cdot \frac{\pi}{2} \] \[ 2I = \frac{\pi^2}{2} \]

5. Find \( I \):
Dividing both sides by 2: \[ I = \frac{\pi^2}{4} \]

Final Answer:
Therefore, the value of the integral is: \[ \int_0^\pi \frac{x \tan(x)}{\sec(x) + \cos(x)} \, dx = \frac{\pi^2}{4} \]