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Question Number 78271 by msup trace by abdo last updated on 15/Jan/20

$$calculate{A}_{\theta }={\int }_0^{\frac{\pi }{2}}\frac{dx}{2+cos\theta sinx}$$$$-\pi <\theta <\pi$$

Commented by mathmax by abdo last updated on 19/Jan/20

$$A\left(\theta \right)={\int }_0^{\frac{\pi }{2}}\frac{dx}{2+cos\theta sinx}\Rightarrow A\left(\theta \right){=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^1\frac{2dt}{(1+t^2)(2+cos\theta \frac{2t}{1+t^2})}$$$$={\int }_0^1\frac{2dt}{2+2t^2+2tcos\theta }={\int }_0^1\frac{dt}{t^2+tcos\theta +1}={\int }_0^1\frac{dt}{t^2+2t\frac{cos\theta }{2}+\frac{{cos}^2\theta }{4}+1-\frac{{cos}^2\theta }{4}}$$$$={\int }_0^1\frac{dt}{{(t+\frac{cos\theta }{2})}^2+\frac{4-{cos}^2\theta }{4}}{=}_{t+\frac{cos\theta }{2}=\frac{\sqrt{4-{cos}^2\theta }}{2}u\Rightarrow u=\frac{2t+cos\theta }{\sqrt{4-{cos}^2\theta }}}$$$$={\int }_{\frac{cos\theta }{\sqrt{4-{cos}^2\theta }}}^{\frac{2+cos\theta }{\sqrt{4-{cos}^2\theta }}}\frac{1}{\frac{4-{cos}^2\theta }{4}(1+u^2)}\times \frac{\sqrt{4-{cos}^2\theta }}{2}du$$$$=2{\int }_{\frac{cos\theta }{\sqrt{4-{cos}^2\theta }}}^{\frac{2+cos\theta }{\sqrt{4-{cos}^2\theta }}}\frac{du}{1+u^2}=2{\left[arctan\left(u\right)\right]}_{\frac{cos\theta }{\sqrt{4-{cos}^2\theta }}}^{\frac{2+cos\theta }{\sqrt{4-{cos}^2\theta }}}$$$$=2\{arctan\left(\frac{2+cos\theta }{\sqrt{4-{cos}^2\theta }}\right)-arctan\left(\frac{cos\theta }{\sqrt{4-{cos}^2\theta }}\right)\}.$$

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