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Question Number 30178 by abdo imad last updated on 17/Feb/18

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+cosxcos\theta }with-\pi <\theta <\pi .$$

Commented by prof Abdo imad last updated on 22/Feb/18

$$letputcos\theta =tI={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+tcosx}thech.tan\left(\frac{x}{2}\right)=u$$$$giveI={\int }_0^{\mathrm{\infty }}\frac{1}{1+t\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$$$$I={\int }_0^{\mathrm{\infty }}\frac{2du}{1+u^2+t(1-u^2)}du$$$$={\int }_0^{\mathrm{\infty }}\frac{2du}{1+t+(1-t)u^2}=\frac{2}{1+t}{\int }_0^{\mathrm{\infty }}\frac{du}{1+\frac{1-t}{1+t}{u}^2}$$$$lettbenusethech.\sqrt{\frac{1-t}{1+t}}u=\alpha \Rightarrow$$$$I=\frac{2}{1+t}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\alpha }^2}\sqrt{\frac{1+t}{1-t}}d\alpha$$$$=\frac{2}{\sqrt{1-t^2}}{\int }_0^{\mathrm{\infty }}\frac{d\alpha }{1+{\alpha }^2}=\frac{\pi }{\sqrt{1-t^2}}=\frac{\pi }{\sqrt{1-{cos}^2\theta }}\Rightarrow$$$$I=\frac{\pi }{\mid sin\theta \mid }if\theta \ne 0alsowemuststudythecase\theta =0$$$$....$$

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