10) \( \int_{0}^{\pi / 2} \frac{1-\sin x}{x+\cos x} d x \)

To evaluate the integral: \[ I = \int_{0}^{\pi / 2} \frac{1 - \sin x}{x + \cos x} \, dx \] we can recognize that the integrand has a structure that suggests a straightforward substitution. Let's proceed step-by-step: 1. **Identify the Derivative:** Notice that the numerator \(1 - \sin x\) is the derivative of the denominator \(x + \cos x\) with respect to \(x\): \[ \frac{d}{dx}(x + \cos x) = 1 - \sin x \] 2. **Rewrite the Integrand:** This observation allows us to rewrite the integrand as: \[ \frac{1 - \sin x}{x + \cos x} = \frac{d}{dx} \ln(x + \cos x) \] Therefore, the integral becomes: \[ I = \int_{0}^{\pi / 2} \frac{d}{dx} \ln(x + \cos x) \, dx \] 3. **Integrate Using Fundamental Theorem of Calculus:** By the Fundamental Theorem of Calculus, we can evaluate the integral directly: \[ I = \ln(x + \cos x) \bigg|_{0}^{\pi / 2} \] 4. **Evaluate the Limits:** Substitute the limits \(x = \pi/2\) and \(x = 0\) into the expression: \[ I = \ln\left(\frac{\pi}{2} + \cos\left(\frac{\pi}{2}\right)\right) - \ln(0 + \cos 0) \] Simplifying further: \[ I = \ln\left(\frac{\pi}{2} + 0\right) - \ln(1) = \ln\left(\frac{\pi}{2}\right) \] Since \(\ln(1) = 0\), the expression reduces to: \[ I = \ln\left(\frac{\pi}{2}\right) \] **Final Answer:** \[ \ln\left(\frac{\pi}{2}\right) \]