Question

Evaluate \[ \int_0^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx. \]

Hint:

Use symmetry and substitutions to simplify integrals, especially when dealing with standard trigonometric functions.

Answer 1

We are asked to evaluate the integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx. \] We will use the symmetry of the integral to simplify the calculation. First, make the substitution \( x = \frac{\pi}{2} - t \). This gives: \[ dx = -dt, \quad \sin\left(\frac{\pi}{2} - t\right) = \cos t, \quad \cos\left(\frac{\pi}{2} - t\right) = \sin t. \] Thus, the integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} - t}{\cos t + \sin t} \, dt. \] Now, rewrite the integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2}}{\sin x + \cos x} \, dx - \int_0^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx. \] This simplifies to: \[ I = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx - I. \] Thus, solving for \( I \): \[ 2I = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx. \] The integral \( \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx \) is a standard result, which is \( \sqrt{2} \). Thus: \[ 2I = \frac{\pi}{2} \times \sqrt{2} \quad \Rightarrow \quad I = \frac{\pi \sqrt{2}}{4}. \]