Question

Find the value of the integral $\frac{\left(1+2\cos x\right)}{{\left(2+\cos x\right)}^2}dx$ from $0$ to $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

Answer 1

Solve the above integral.

The given integral is $\frac{\left(1+2\cos x\right)}{{\left(2+\cos x\right)}^2}dx$

Assume that, $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(1+2\cos x\right)}{{\left(2+\cos x\right)}^2}dx$

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(1+2\cos x\right)}{{\left(2+\cos x\right)}^2}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left({\sin }^{2}x+{\cos }^{2}x+2\cos x\right)}{{\left(2+\cos x\right)}^2}dx[\because {\mathbf{sin}}^{\mathbf{2}}\mathit{x}\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}] \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left({\sin }^{2}x+\cos x\left(\cos x+2\right)\right)}{{\left(2+\cos x\right)}^2}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(\cos x\right)}{\left(2+\cos x\right)}dx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{2}x}{{\left(2+\cos x\right)}^2}dx \end{gathered}$$

Assume that, $I_1={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos x}{\cos x+2}dx$

$$\begin{gathered} I_1={\left[\frac{1}{\cos x+2}\right]}_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}\cos xdx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{d\left({\displaystyle \frac{1}{\cos x+2}}\right)}{dx}\int \cos xdx\right]dx \\ \Rightarrow I_1={\left[\frac{\sin x}{\cos x+2}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{-\left(-\sin x\right)}{{\left(2+\cos x\right)}^2}\sin xdx \\ \Rightarrow I_1={\left[\frac{\sin x}{\cos x+2}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{{\left(2+\cos x\right)}^2}\sin xdx \end{gathered}$$

So, $I=I_1+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{2}x}{{\left(2+\cos x\right)}^2}dx$

$$\begin{gathered} I={\left[\frac{\sin x}{\cos x+2}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{{\left(2+\cos x\right)}^2}\sin xdx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{{\left(2+\cos x\right)}^2}\sin xdx \\ \Rightarrow I={\left[\frac{\sin x}{\cos x+2}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=\left[\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)+2}-\frac{\sin \left({\displaystyle 0}\right)}{\cos \left({\displaystyle 0}\right)+2}\right] \\ \Rightarrow I=\left[\frac{1}{0+2}-\frac{0}{1+2}\right] \\ \Rightarrow I=\frac{1}{2} \end{gathered}$$

Hence, the required final answer is $\frac{1}{2}$.