By contour techniques∫\smallint∫ 1/(2+cosθ\thetaθ) limit 0 to 2π\piπ .
∫1cos(θ)+2 dθ{\displaystyle\int}\dfrac{1}{\cos\left({\theta}\right)+2}\,\mathrm{d}{\theta}∫cos(θ)+21dθ
=∫sec2(θ2)tan2(θ2)+3 dθ={\displaystyle\int}\dfrac{\sec^2\left(\frac{{\theta}}{2}\right)}{\tan^2\left(\frac{{\theta}}{2}\right)+3}\,\mathrm{d}{\theta}=∫tan2(2θ)+3sec2(2θ)dθ ...simplify using trigonometric
Substitute
u=tan(θ2)3u=\dfrac{\tan\left(\frac{{\theta}}{2}\right)}{\sqrt{3}}u=3tan(2θ), dudθ=sec2(θ2)23\dfrac{\mathrm{d}u}{\mathrm{d}{\theta}} = \dfrac{\sec^2\left(\frac{{\theta}}{2}\right)}{2\sqrt{3}}dθdu=23sec2(2θ)
dθ=23sec2(θ2) du\mathrm{d}{\theta}=\dfrac{2\sqrt{3}}{\sec^2\left(\frac{{\theta}}{2}\right)}\,\mathrm{d}udθ=sec2(2θ)23du
=∫233u2+3 du={\displaystyle\int}\dfrac{2\sqrt{3}}{3u^2+3}\,\mathrm{d}u=∫3u2+323du
=23∫1u2+1 du={{\dfrac{2}{\sqrt{3}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u=32∫u2+11du
=2arctan(u)3=\dfrac{2\arctan\left(u\right)}{\sqrt{3}}=32arctan(u) 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}=[32arctan(3tan(2θ))]02π
=2π3=\dfrac{2{\pi}}{\sqrt{3}}=32π
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