The value of the integration int_{-pi / 4}^{p | vedclass

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(B) Let $I = \int_{-\pi/4}^{\pi/4} (\lambda|\sin x| + \frac{\mu \sin x}{1+\cos x} + \gamma) \, dx$. We can split the integral into three parts: $I = \

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is independent of $\mu$ only

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(B) Let $I = \int_{-\pi/4}^{\pi/4} (\lambda|\sin x| + \frac{\mu \sin x}{1+\cos x} + \gamma) \, dx$. We can split the integral into three parts: $I = \lambda \int_{-\pi/4}^{\pi/4} |\sin x| \, dx + \mu \int_{-\pi/4}^{\pi/4} \frac{\sin x}{1+\cos x} \, dx + \gamma \int_{-\pi/4}^{\pi/4} 1 \, dx$. Consider the second integral $J = \int_{-\pi/4}^{\pi/4} \frac{\sin x}{1+\cos x} \, dx$. Let $f(x) = \frac{\sin x}{1+\cos x}$. Then $f(-x) = \frac{\sin(-x)}{1+\cos(-x)} = \frac{-\sin x}{1+\cos x} = -f(x)$. Since $f(x)$ is an odd function and the interval $[-\pi/4, \pi/4]$ is symmetric about $0$, the integral $J = 0$. Thus, $I = \lambda \int_{-\pi/4}^{\pi/4} |\sin x| \, dx + \gamma \int_{-\pi/4}^{\pi/4} 1 \, dx$. Since the term involving $\mu$ vanishes, the value of the integral is independent of $\mu$.

The value of the integration $\int_{-\pi / 4}^{\pi / 4} (\lambda|\sin x| + \frac{\mu \sin x}{1+\cos x} + \gamma) \, dx$

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