fond-the-value-of-0-2pi-dt-a-cost-2-with-a-gt-1-

Question Number 37350 by math khazana by abdo last updated on 12/Jun/18

f o n d t h e v a l u e o f \int_{0}^{2 π} \frac{d t}{{(a + c o s t)}^{2}} w i t h a > 1 .

Commented by math khazana by abdo last updated on 12/Jun/18

I = \int_{0}^{π} \frac{d t}{{(a + c o s t)}^{2}} + \int_{π}^{2 π} \frac{d t}{{(a + c o s t)}^{2}} = K + H

c h a n g e m e n t t a n (\frac{t}{2}) = x g i v e

k = \int_{0}^{\infty} \frac{1}{{a + \frac{1 - x^{2}}{1 + x^{2}}}^{2}} \frac{2 d x}{1 + x^{2}} = 2 \int_{0}^{\infty} \frac{1 + x^{2}}{{a + {a x}^{2} + 1 - x^{2}}} d x

= 2 \int_{0}^{\infty} \frac{1 + x^{2}}{{a + 1 + (a - 1) x^{2}}^{2}} d x

= \frac{2}{{(a + 1)}^{2}} \int_{0}^{\infty} \frac{1 + x^{2}}{{1 + \frac{a - 1}{a + 1} x^{2}}^{2}} d x

=_{\sqrt{\frac{a - 1}{a + 1}} x = u} \frac{2}{{(a + 1)}^{2}} \int_{0}^{\infty} \frac{1 + \frac{a + 1}{a - 1} u^{2}}{{1 + u^{2}}^{2}} \sqrt{\frac{a + 1}{a - 1}} d u

= \frac{1}{(a - 1) \sqrt{\frac{a + 1}{a - 1}}} \frac{2}{{(a + 1)}^{2}} \int_{0}^{\infty} \frac{a - 1 + (a + 1) u^{2}}{{1 + u^{2}}^{2}} d u

= \frac{2}{{(a + 1)}^{2} \sqrt{a^{2} - 1}} \int_{0}^{\infty} \frac{a - 1 + (a + 1) u^{2}}{{1 + u^{2}}^{2}} d u b u t

\int_{0}^{\infty} \frac{a - 1 + (a + 1) u^{2}}{{(1 + u^{2})}^{2}} d u

= (a - 1) \int_{0}^{\infty} \frac{d u}{{(1 + u^{2})}^{2}} + (a + 1) \int_{0}^{\infty} \frac{1 + u^{2} - 1}{{(1 + u^{2})}^{2}} d u

= (a - 1) \int_{0}^{\infty} \frac{d u}{{(1 + u^{2})}^{2}} + (a + 1) \frac{π}{2} - (a + 1) \int_{0}^{\infty} \frac{d u}{{(1 + u^{2})}^{2}}

= - 2 \int_{0}^{\infty} \frac{d u}{{(1 + u^{2})}^{2}} + (a + 1) \frac{π}{2} c h a n g . u = t a n θ

- \frac{π}{2} + \frac{a π}{2} + \frac{π}{2} = \frac{π a}{2} b e c a u s e

\int_{0}^{+ \infty} \frac{d u}{{(1 + u^{2})}^{2}} = \int_{0}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} θ) d θ}{{(1 + {t a n}^{2} θ)}^{2}}

= \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + {t a n}^{2} θ} = \int_{0}^{\frac{π}{2}} {c o s}^{2} θ d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + c o s (2 θ)) d θ = \frac{π}{4} \Rightarrow

K = \frac{2}{{(a + 1)}^{2} \sqrt{a^{2} - 1}} \frac{π a}{2} = \frac{π a}{{(a + 1)}^{2} \sqrt{a^{2} - 1}} .

Commented by math khazana by abdo last updated on 12/Jun/18

H = \int_{π}^{2 π} \frac{d t}{{(a + c o s t)}^{2}} c h a n g . t = π + x g i v e

H = \int_{0}^{π} \frac{d x}{{(a - c o s x)}^{2}} =_{t a n (\frac{x}{2}) = u} \int_{0}^{+ \infty} \frac{1}{{(a - \frac{1 - u^{2}}{1 + u^{2}})}^{2}} \frac{2 d u}{1 + u^{2}}

= 2 \int_{0}^{+ \infty} \frac{1 + u^{2}}{{a + {a u}^{2} - 1 + u^{2}}^{2}} d u \dots w e f o l l o w t h e

s a m e m e t h o d t o f i n d H \dots . b e c o n t i n u e d . .