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Question Number 65768 by mathmax by abdo last updated on 03/Aug/19

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{dx}{x^2-2\left(cos\theta \right)x+1}$$

Commented by mathmax by abdo last updated on 04/Aug/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{dx}{x^2-2xcos\theta +1}\Rightarrow I={\int }_0^{\mathrm{\infty }}\frac{dx}{x^2-2xcos\theta +{cos}^2\theta +{sin}^2\theta }$$$$={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x-cos\theta )}^2+{sin}^2\theta }{=}_{x-cos\theta =\mid sin\theta \mid u}{\int }_{-\frac{cos\theta }{\mid sin\theta \mid }}^{+\mathrm{\infty }}\frac{\mid sin\theta \mid du}{{sin}^2\theta (1+u^2)}$$$$=\frac{1}{\mid sin\theta \mid }{\left[arctanu\right]}_{-\frac{cos\theta }{\mid sin\theta \mid }}^{+\mathrm{\infty }}=\frac{1}{\mid sin\theta \mid }\{\frac{\pi }{2}+arctan\left(\frac{cos\theta }{\mid sin\theta \mid }\right)\}$$$$case1sin\theta >0\Rightarrow I=\frac{1}{sin\theta }\{\frac{\pi }{2}+arctan\left(\frac{1}{tan\theta }\right)\}$$$$=\frac{1}{sin\theta }\{\frac{\pi }{2}\stackrel{-}{+}\frac{\pi }{2}-\theta \}$$$$case2sin\theta <0\Rightarrow I=-\frac{1}{sin\theta }\{\frac{\pi }{2}-arctan\left(\frac{cos\theta }{sn\theta }\right)\}$$$$=\frac{1}{sin\theta }\{-\frac{\pi }{2}\stackrel{-}{+}\frac{\pi }{2}-\theta \}$$

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