91Ó°ÊÓ

Contour integral:Contour integration is a method of calculating integrals along paths in the complex plane.

The integration is given by,

$I={∫}_0^{\mathrm{π}}\frac{\cos \left(\mathrm{θ}\right)d\mathrm{θ}}{5-4\cos \left(\mathrm{θ}\right)}.....\left(1\right)$

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Convert (1) to contour integral, but first, we shall assume that:

$$\begin{gathered} \cos \mathrm{θ}=\frac{z+{\displaystyle \frac{1}{z}}}{2} \\ -d\mathrm{θ}=\frac{dz}{iz} \end{gathered}$$

The contour is the unit circle.

So, substitution in the above equation (1) yields as follows:

$$\begin{gathered} I_1=âˆ\circledR \frac{\left({\displaystyle \frac{z+{\displaystyle \frac{1}{z}}}{2}}\right){\displaystyle \frac{dz}{iz}}}{5-4\left(\frac{z+{\displaystyle \frac{1}{z}}}{2}\right)} \\ =\frac{-1}{i}âˆ\circledR \frac{1+z^{2}dz}{z\left(4z^2-10z+4\right)} \\ =\frac{-1}{i}âˆ\circledR \frac{1+z^{2}dz}{z\left(2z-4\right)\left(2z-1\right)} \\ =\frac{-1}{4i}âˆ\circledR \frac{1+z^{2}dz}{z\left(z-2\right)\left(z-{\displaystyle \frac{1}{2}}\right)}....\left(2\right) \end{gathered}$$

The value of integral (2) is given as follows:

$$\begin{gathered} I_1=2\mathrm{Ï€¾\pm }\underset{}{∑}\mathrm{R}\left({\mathrm{Z}}_{\mathrm{i}}\right) \\ =2\mathrm{Ï€¾\pm }\left(-\frac{1}{4\mathrm{i}}\right)\left[{\left(\frac{1+{\mathrm{z}}^2}{\mathrm{z}\left(\mathrm{z}-2\right)}\right)}_{\mathrm{z}→0.5}+{\left(\frac{1+{\mathrm{z}}^2}{\left(\mathrm{z}-2\right)\left(\mathrm{z}-{\displaystyle \frac{1}{2}}\right)}\right)}_{\mathrm{z}→0}\right] \\ =2\mathrm{Ï€¾\pm }\left(-\frac{1}{4\mathrm{i}}\right)\left(1-\frac{5}{3}\right) \\ =\frac{\mathrm{Ï€}}{3} \end{gathered}$$

Because the function $\frac{\cos \left(\mathrm{θ}\right)}{5-4\cos \left(\mathrm{θ}\right)}$ is even, the value from $0→2\mathrm{π}$ is twice the value form $$" width="9" height="19" role="math">0→π, so the value of I is given as follows:

$$\begin{gathered} I=\left(\frac{1}{2}\right)I_1 \\ I=\left(\frac{\mathrm{Ï€}}{3}\right)\left(\frac{1}{2}\right) \\ =\frac{\mathrm{Ï€}}{6} \end{gathered}$$

Therefore, the answer is $\frac{\mathrm{Ï€}}{6}$.