Question #233409

By contour techniques\smallint 1/(2+cosθ\theta) limit 0 to 2π\pi .


1
Expert's answer
2021-09-17T03:44:49-0400

1cos(θ)+2dθ{\displaystyle\int}\dfrac{1}{\cos\left({\theta}\right)+2}\,\mathrm{d}{\theta}


=sec2(θ2)tan2(θ2)+3dθ={\displaystyle\int}\dfrac{\sec^2\left(\frac{{\theta}}{2}\right)}{\tan^2\left(\frac{{\theta}}{2}\right)+3}\,\mathrm{d}{\theta} ...simplify using trigonometric

Substitute 

u=tan(θ2)3u=\dfrac{\tan\left(\frac{{\theta}}{2}\right)}{\sqrt{3}}, dudθ=sec2(θ2)23\dfrac{\mathrm{d}u}{\mathrm{d}{\theta}} = \dfrac{\sec^2\left(\frac{{\theta}}{2}\right)}{2\sqrt{3}}


dθ=23sec2(θ2)du\mathrm{d}{\theta}=\dfrac{2\sqrt{3}}{\sec^2\left(\frac{{\theta}}{2}\right)}\,\mathrm{d}u


=233u2+3du={\displaystyle\int}\dfrac{2\sqrt{3}}{3u^2+3}\,\mathrm{d}u


=231u2+1du={{\dfrac{2}{\sqrt{3}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u


=2arctan(u)3=\dfrac{2\arctan\left(u\right)}{\sqrt{3}} substitution u


=[2arctan(tan(θ2)3)3]02π=[\dfrac{2\arctan\left(\frac{\tan\left(\frac{{\theta}}{2}\right)}{\sqrt{3}}\right)}{\sqrt{3}}]_0^{2\pi}


=2π3=\dfrac{2{\pi}}{\sqrt{3}}




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