91Ó°ÊÓ

Recall the identity \(\cos t = \frac{1}{2} (e^{it} + e^{-it})\), where \(i\) is the imaginary unit. Using this identity, we can rewrite the integrand as: $$ \frac{1}{a+b(\frac{1}{2} (e^{it} + e^{-it}))}=\frac{2}{2a+be^{it}+be^{-it}} $$

Now, let's consider the contour integral of the above complex function over a circle with a unit radius and centered at the origin in the complex plane: $$ \oint_C \frac{2\mathrm{d} z}{(2a+bz+b\frac{1}{z})} $$ where the contour \(C\) is defined by \(z = e^{it}\) and \(0 \leq t \leq 2\pi\). By substituting \(z = e^{it}\) and \(\mathrm{d}z = ie^{it}\mathrm{d}t\), we have $$ \oint_C \frac{2\mathrm{d} z}{(2a+bz+b\frac{1}{z})} = \oint_C \frac{2ie^{it}\mathrm{d}t}{(2a+be^{it}+be^{-it})} $$ Notice that the integral on the right-hand side is exactly the given integral multiplied by \(i\). Therefore, the integral we want to find is simply the imaginary part of the contour integral on the left-hand side.

Now we will apply Cauchy's residue theorem to find the value of the contour integral. To find the residue, we should first find the poles of our function. The function \(f(z) = \frac{2}{(2a+bz+b\frac{1}{z})}\) has a pole at \(z=0\), and the remaining poles come from the zeros of the denominator \(2a+bz+b\frac{1}{z}\), i.e. the zeros of the quadratic equation \(b^2z^2+2abz-2a^2=0\). By solving this quadratic equation, we obtain two zeros: $$ z_\pm=\frac{-2a\pm\sqrt{(2a)^2+4(b^2)(-2a^2)}}{2b^2}=\frac{-a\pm\sqrt{a^2-b^2}}{b} $$ Since \(|z_+|>1\) and \(|z_-|<1\), the pole \(z=z_-\) is the pole that lies inside the contour \(C\). The residue at the pole \(z=z_-\) can be calculated as follows: $$ \text{Res}_{z=z_-}(f(z))=\lim_{z\rightarrow z_-}(z-z_-)f(z) = \lim_{z\rightarrow z_-}\left(\frac{2}{\frac{b}{z_-}+\frac{b}{z}}\right) = \frac{1}{2}\frac{4a}{b(z_+ - z_-)}=-\frac{2a}{b\sqrt{a^2-b^2}} $$ Now we can use Cauchy's residue theorem to find the value of the contour integral: $$ \oint_C f(z)dz=2\pi i\sum\text{Res} = 2\pi i\left(-\frac{2a}{b\sqrt{a^2-b^2}}\right) $$

As mentioned in Step 2, the original integral is the imaginary part of the contour integral. Therefore, we have $$ I=\int_{0}^{2 \pi} \frac{\mathrm{d} t }{a+b \cos t}= \Im\left[\oint_C f(z)dz\right]=\frac{4\pi a}{b\sqrt{a^2-b^2}} $$