- Evaluate the integral
\[ \int _ { 0 } ^ { 2 \pi } { \frac { d x } { a ^ { 2 } + \cos ^ { 2 } x } } . \]
Where \(a > 1\)
Solution
The value is \[ \boxed{\frac{2\pi}{a\sqrt{a^2+1}}}. \]
1. Reduce the problem to the standard integral \(\int_0^{2\pi}(c+\cos t)^{-1}dt\).
Using \(\cos^2x=(1+\cos2x)/2\) and setting \[ c=2a^2+1>1, \] we obtain \[ \frac1{a^2+\cos^2x}=\frac{2}{c+\cos2x}. \] Hence, with \(t=2x\), \[ \int_0^{2\pi}\frac{dx}{a^2+\cos^2x} =\int_0^{4\pi}\frac{dt}{c+\cos t} =2\int_0^{2\pi}\frac{dt}{c+\cos t}, \] because the last integrand is \(2\pi\)-periodic.
2. The unit-circle contour gives\(\int_0^{2\pi}(c+\cos t)^{-1}dt=2\pi/\sqrt{c^2-1}\).
Put \(z=e^{it}\). Then \(dt=dz/(iz)\) and \(\cos t=(z+z^{-1})/2\), so \[ \int_0^{2\pi}\frac{dt}{c+\cos t} =\frac{2}{i}\int_{|z|=1}\frac{dz}{z^2+2cz+1}. \] The two roots of the denominator are \[ z_\pm=-c\pm\sqrt{c^2-1}. \] Since \(c>1\), \(z_-<-1\). Moreover \[ |z_+|=c-\sqrt{c^2-1} =\frac1{c+\sqrt{c^2-1}}<1, \] so exactly \(z_+\) lies inside the unit circle. It is simple, and the residue of \(2/[i(z^2+2cz+1)]\) there is \[ \frac{2}{i(2z_++2c)} =\frac1{i\sqrt{c^2-1}}. \] The residue theorem therefore yields \[ \int_0^{2\pi}\frac{dt}{c+\cos t} =2\pi i\frac1{i\sqrt{c^2-1}} =\frac{2\pi}{\sqrt{c^2-1}}. \]
3. Substitute \(c=2a^2+1\).
By steps 1 and 2, \[ \int_0^{2\pi}\frac{dx}{a^2+\cos^2x} =\frac{4\pi}{\sqrt{(2a^2+1)^2-1}} =\frac{4\pi}{\sqrt{4a^2(a^2+1)}}. \] Since \(a>1\), the square root is \(2a\sqrt{a^2+1}\), giving \[ \frac{2\pi}{a\sqrt{a^2+1}}. \]