$ \left(\int \frac{2 \cos x+1}{(2+\cos x)^2} d x\right)-\frac{\sin x}{2+\cos x}= $

Question

Aicharya · Accepted Answer

Let $f(x)=\frac{\sin x}{2+\cos x}$ $ \begin{aligned} f^{\prime}(x) & =\frac{(2+\cos x) \cos x-\sin x(-\sin x)}{(2+\cos x)^2} \ & =\frac{2 \cos x+1}{(2+\cos x)^2} \end{aligned} $ $\frac{\sin x}{2+\cos x}$ is the antiderivatives of $\frac{2 \cos x+1}{(2+\cos x)^2}$ $ \begin{aligned} & \Rightarrow \int\left(\frac{2 \cos x+1}{(2+\cos x)^2}ight) d x=\frac{\sin x}{2+\cos x}+C \ & \Rightarrow \int \frac{(2 \cos x+1)}{(2+\cos x)^2} d x-\frac{\sin x}{2+\cos x}=C \end{aligned} $

$ \left(\int \frac{2 \cos x+1}{(2+\cos x)^2} d x\right)-\frac{\sin x}{2+\cos x}= $

Solution