91Ó°ÊÓ

Let's make the substitution \(t = \tan\left(\frac{x}{2}\right)\). With this substitution, we have \(\sin x = \frac{2t}{1+t^2}\), and \(\cos x = \frac{1-t^2}{1+t^2}\). Furthermore, we need to compute the new limits of integration in terms of \(t\). When \(x=-\pi\), \(t=\tan\left(\frac{-\pi}{2}\right)=-1\); when \(x=0\), \(t=\tan\left(\frac{0}{2}\right)=0\). We also need to compute the differential \(dx\) in terms of \(dt\): \((1+\tan^{2}(x/2))\frac{d x}{2}=d(\tan (x/2))=\frac{2 d t}{1+t^{2}}\) Therefore, \(dx = \frac{2dt}{1+t^2}\). Now we can rewrite the integral in terms of \(t\): $$\int_{-\pi}^{0} \frac{\sin x}{2+\cos x} d x = \int_{-1}^{0} \frac{\left(\frac{2t}{1+t^2}\right)}{2+\left(\frac{1-t^2}{1+t^2}\right)} \cdot \frac{2dt}{1+t^2}$$

Simplify the integral: $$\int_{-1}^{0} \frac{\left(\frac{2t}{1+t^2}\right)}{2+\left(\frac{1-t^2}{1+t^2}\right)} \cdot \frac{2dt}{1+t^2} = \int_{-1}^{0} \frac{4t^2 dt}{(2+t^2)(1+t^2)}$$ Now, we integrate the expression with respect to \(t\) using partial fraction decomposition. The partial fraction decomposition is given by: $$\frac{4t^2}{(2+t^2)(1+t^2)} = A\frac{1}{1+t^2} + B\frac{1}{2+t^2}$$ Solving for \(A\) and \(B\), we obtain \(A=2\) and \(B=-2\). Now we can write the integral as: $$\int_{-1}^{0} \left(2\frac{1}{1+t^2} - 2\frac{1}{2+t^2}\right) dt$$

By evaluating the integral using antiderivatives, we have: $$\int_{-1}^{0} \left(2\frac{1}{1+t^2} - 2\frac{1}{2+t^2}\right) dt = 2\left[\arctan t\right]_{-1}^0 - 2\left[\frac{1}{\sqrt{2}}\arctan \frac{t}{\sqrt{2}}\right]_{-1}^0$$ Plugging in the limits of integration, we have: $$2(\arctan 0 - \arctan(-1)) - 2\left[\frac{1}{\sqrt{2}}(\arctan 0 - \arctan(-\frac{1}{\sqrt{2}}))\right] = 2\left(\frac{\pi}{4}\right) - \left(\frac{\pi}{2}\right) = \frac{\pi}{2}$$

Thus, the value of the integral is: $$\int_{-\pi}^{0} \frac{\sin x}{2+\cos x} d x = \frac{\pi}{2}$$