calculate-0-4pi-d-x-2-2x-cos-1-2-

Question Number 141222 by mathmax by abdo last updated on 16/May/21

calculate \int_{0}^{4 π} \frac{d θ}{{(x^{2} + 2 x \cos θ + 1)}^{2}}

Answered by Ar Brandon last updated on 16/May/21

Ω = \int_{0}^{4 π} \frac{d θ}{{(x^{2} + 2 xcos θ + 1)}^{2}} = 4 \int_{0}^{π} \frac{d θ}{{(x^{2} + 2 xcos θ + 1)}^{2}}, t = \tan \frac{θ}{2}

= 4 \int_{0}^{\infty} \frac{2 dt}{{(x^{2} + 2 x \cdot \frac{1 - t^{2}}{1 + t^{2}} + 1)}^{2}} \cdot \frac{1}{1 + t^{2}}

= 8 \int_{0}^{\infty} \frac{t^{2} + 1}{{(x^{2} (1 + t^{2}) + 2 x (1 - t^{2}) + 1 + t^{2})}^{2}} dt

= 8 \int_{0}^{\infty} \frac{t^{2} + 1}{{((x^{2} - 2 x + 1) t^{2} + (x^{2} + 2 x + 1))}^{2}} dt

= 8 \int_{0}^{\infty} \frac{t^{2} + 1}{{({(x - 1)}^{2} t^{2} + {(x + 1)}^{2})}^{2}} = \frac{8}{{(x - 1)}^{4}} \int_{0}^{\infty} \frac{t^{2} + 1}{{(t^{2} + {(\frac{x + 1}{x - 1})}^{2})}^{2}}

= \frac{8}{{(x - 1)}^{4}} {\int_{0}^{\infty} \frac{t^{2} + {(\frac{x + 1}{x - 1})}^{2}}{{(t^{2} + {(\frac{x + 1}{x - 1})}^{2})}^{2}} dt + \int_{0}^{\infty} \frac{1 - {(\frac{x + 1}{x - 1})}^{2}}{{(t^{2} + {(\frac{x + 1}{x - 1})}^{2})}^{2}} dt}

f (a) = \int_{0}^{\infty} \frac{dt}{t^{2} + a^{2}} = \frac{π}{2 a} \Rightarrow f^{'} (a) = - \int_{0}^{\infty} \frac{2 a}{{(t^{2} + a^{2})}^{2}} dt = - \frac{π}{{2 a}^{2}}

\Rightarrow \int_{0}^{\infty} \frac{dt}{{(t^{2} + a^{2})}^{2}} = \frac{π}{{4 a}^{3}}

Ω = \frac{8}{{(x - 1)}^{4}} {\frac{π}{2 (\frac{x + 1}{x - 1})} + (1 - {(\frac{x + 1}{x - 1})}^{2}) \times \frac{π}{4 {(\frac{x + 1}{x - 1})}^{3}}}