The-value-of-the-integral-0-pi-1-a-2-2a-cos-x-1-dx-a-gt-1-is-

Question Number 57138 by mustakim420 last updated on 30/Mar/19

The value of the integral

\underset{0}{\int^{π}} \frac{1}{a^{2} - 2 a \cos x + 1} d x (a > 1) is

Commented by maxmathsup by imad last updated on 30/Mar/19

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

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

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

c h a n g e m e n t (a + 1) t = (a - 1) u g i v e

f (a) = 2 \int_{0}^{\infty} \frac{1}{{(a - 1)}^{2} (1 + u^{2})} \frac{a - 1}{a + 1} d u = \frac{2}{a^{2} - 1} {[a r c t a n (u)]}_{0}^{+ \infty}

= \frac{2}{a^{2} - 1} \frac{π}{2} = \frac{π}{a^{2} - 1} \Rightarrow f (a) = \frac{π}{a^{2} - 1} w i t h a > 1 .

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Mar/19

\int \frac{d x}{a^{2} + 1 - 2 a c o s x}

\frac{1}{2 a} \int \frac{d x}{k - c o s x} [k = \frac{a^{2} + 1}{2 a}]

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

d x = \frac{2 d t}{1 + t^{2}}

\frac{1}{2 a} \int \frac{2 d t}{(1 + t^{2}) (k - \frac{1 - t^{2}}{1 + t^{2}})}

\frac{1}{a} \int \frac{d t}{k + {k t}^{2} - 1 + t^{2}}

\frac{1}{a} \int \frac{d t}{(k - 1) + t^{2} (k + 1)}

\frac{1}{a (k + 1)} \int \frac{d t}{\frac{k - 1}{k + 1} + t^{2}}

\frac{1}{a (\frac{a^{2} + 1}{2 a} + 1)} \times \frac{1}{\sqrt{\frac{k - 1}{k + 1}}} {t a n}^{- 1} (\frac{t}{\sqrt{\frac{k - 1}{k + 1}}})

k = \frac{a^{2} + 1}{2 a}

\frac{k - 1}{k + 1} = \frac{a^{2} + 1 + 2 a}{a^{2} + 1 - 2 a} = \frac{{(a + 1)}^{2}}{{(a - 1)}^{2}}

\sqrt{\frac{k - 1}{k + 1}} = \frac{a + 1}{a - 1}

\frac{2}{{(a + 1)}^{2}} \times \frac{1}{\frac{a - 1}{a + 1}} ∣ {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{\frac{a - 1}{a + 1}}) ∣_{0}^{π}

\frac{2}{a^{2} - 1} \times [{t a n}^{- 1} (\infty) - 0]

\frac{2}{a^{2} - 1} \times \frac{π}{2} \to \frac{π}{a^{2} - 1}