Question

The value of the integral
\[ I = \int_{0}^{2\pi} \frac{\cos 2\theta \, d\theta}{5 + 4\cos\theta} \]
is:

• A. \( \dfrac{\pi}{2} \)
• B. \( \dfrac{\pi}{4} \)
• C. \( \pi \)
• D. \( \dfrac{\pi}{6} \)

Hint:

Set \( z=e^{i\theta} \) and use residues, or recall the standard integral \( \int_0^{2\pi}\frac{\cos n\theta}{a+b\cos\theta}d\theta \). With \( a=5,\,b=4 \), note \( \sqrt{a^2-b^2}=3 \).

Answer 1

The Correct Option is D

Step 1: Convert to a contour integral on the unit circle. Put \( z = e^{i\theta} \), so that \( d\theta = \dfrac{dz}{iz} \), \( \cos\theta = \dfrac{1}{2}\left(z + \dfrac{1}{z}\right) \) and \( \cos 2\theta = \dfrac{1}{2}\left(z^{2} + \dfrac{1}{z^{2}}\right) \).

Step 2: The denominator becomes
\[ 5 + 4\cos\theta = 5 + 2\left(z + \tfrac{1}{z}\right) = \frac{2z^{2} + 5z + 2}{z} = \frac{(2z+1)(z+2)}{z}. \]

Step 3: Substitute everything. The numerator \( \cos 2\theta = \dfrac{z^{4}+1}{2z^{2}} \), so
\[ I = \oint_{|z|=1} \frac{\dfrac{z^{4}+1}{2z^{2}}}{\dfrac{(2z+1)(z+2)}{z}} \, \frac{dz}{iz} = \frac{1}{2i}\oint_{|z|=1} \frac{z^{4}+1}{z^{2}(2z+1)(z+2)} \, dz. \]

Step 4: Poles inside \( |z|=1 \) are \( z=0 \) (order 2) and \( z=-\tfrac{1}{2} \) (simple); \( z=-2 \) lies outside.

Residue at \( z=-\tfrac12 \): with \( f(z)=\dfrac{z^{4}+1}{z^{2}(2z+1)(z+2)} \),
\[ \operatorname{Res}_{z=-1/2} = \frac{z^{4}+1}{z^{2}\cdot 2\,(z+2)}\Big|_{z=-1/2} = \frac{\tfrac{1}{16}+1}{\tfrac14 \cdot 2 \cdot \tfrac32} = \frac{17/16}{3/4} = \frac{17}{12}. \]

Step 5: Residue at the double pole \( z=0 \). Let \( g(z)=\dfrac{z^{4}+1}{(2z+1)(z+2)}=\dfrac{z^{4}+1}{2z^{2}+5z+2} \). Then
\[ \operatorname{Res}_{z=0} = g'(0). \]
At \( z=0 \): numerator \(=1\), denominator \(=2\), numerator derivative \(=0\), denominator derivative \(=5\), so
\[ g'(0) = \frac{0\cdot 2 - 1\cdot 5}{2^{2}} = -\frac{5}{4}. \]

Step 6: Sum of residues \( = \dfrac{17}{12} - \dfrac{5}{4} = \dfrac{17}{12} - \dfrac{15}{12} = \dfrac{2}{12} = \dfrac{1}{6}. \)
\[ I = \frac{1}{2i}\cdot 2\pi i \cdot \frac{1}{6} = \frac{\pi}{6}. \]
Hence the integral equals \( \dfrac{\pi}{6} \).
\[ \boxed{\, I = \dfrac{\pi}{6} \,} \]