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Question Number 35687 by prof Abdo imad last updated on 22/May/18

c a l c u l a t e f (a) = \int_{0}^{π} \frac{d x}{1 - a c o s x} a f r o m R .

2) a p p l i c a t i o n c a l c u l a t e \int_{0}^{π} \frac{d x}{1 - 2 c o s x}

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

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

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

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

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

c a s e 1 \frac{1 + a}{1 - a} > 0 a n d a \neq 1 \Rightarrow \frac{(1 + a) (1 - a)}{{(1 - a)}^{2}} > 0 \Rightarrow

1 - a^{2} > 0 \Rightarrow ∣ a ∣< 1 w e g e t

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

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

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

c a s e 2 \frac{1 + a}{1 - a} < 0 \Rightarrow \frac{1 + a}{a - 1} > 0 a n d

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

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

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

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

= \frac{1}{\sqrt{a^{2} - 1}} {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{0}^{+ \infty} = 0 .

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

2) w e h a v e a = 2 a n d ∣ a ∣> 1 \Rightarrow

\int_{0}^{π} \frac{d x}{1 - 2 c o s x} = 0

Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18

\int_{0}^{Π} \frac{d x}{1 - a c o s x}

= \frac{1}{a} \int_{0}^{Π} \frac{d x}{\frac{1}{a} - c o s x}

l e t k = 1 / a

= k \int_{0}^{Π} \frac{d x}{k - \frac{1 - {t a n}^{2} (\frac{x}{2})}{1 + {t a n}^{2} (\frac{x}{2})}}

= k \int_{0}^{Π} \frac{(1 + {t a n}^{2} \frac{x}{2})}{k + {k t a n}^{2} \frac{x}{2} - 1 + {t a n}^{2} \frac{x}{2}} d x

t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} \times \frac{1}{2} d x

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

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

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

= \frac{2 k}{k + 1} \times \frac{1}{\sqrt{\frac{k - 1}{k + 1}}} \times ∣ {t a n}^{- 1} (\frac{t}{\sqrt{\frac{k - 1}{k + 1}}}) ∣_{0}^{\infty}

= \frac{2 k}{k + 1} \times \frac{\sqrt{k + 1}}{\sqrt{k - 1}} \times (\frac{Π}{2})

= \frac{2}{1 + \frac{1}{k}} \times \frac{\sqrt{1 + \frac{1}{k}}}{\sqrt{1 - \frac{1}{k}}} \times (\frac{Π}{2})

= \frac{2}{1 + a} \times \frac{\sqrt{1 + a}}{\sqrt{1 - a}} \times (\frac{Π}{2})