0-pi-cos-2-1-2acos-a-2-d-a-2-lt-1-answer-

Question Number 65202 by arcana last updated on 26/Jul/19

\int_{0}^{π} \frac{\cos 2 θ}{1 - 2 a \cos θ + a^{2}} d θ, a^{2} < 1

answer ?

Answered by Tanmay chaudhury last updated on 26/Jul/19

f o r m u l a t a k e n f r o m S l l o n e y

\frac{1 - a^{2}}{1 - 2 a c o s θ + a^{2}}

= 1 + 2 a c o s θ + 2 a^{2} c o s 2 θ + 2 a^{3} c o s 3 θ + \dots

n o w

I (r) = \int_{0}^{π} \frac{c o s r θ}{1 - 2 a c o s θ + a^{2}} d θ

= \frac{1}{1 - a^{2}} \int_{0}^{π} \frac{1 - a^{2}}{1 - 2 a c o s θ + a^{2}} \times c o s r θ d θ

= \frac{1}{1 - a^{2}} \int_{0}^{π} (1 + 2 a c o s θ + 2 a^{2} c o s 2 θ + 2 a^{3} c o s 3 θ + \dots) c o s r θ d θ

\frac{1}{1 - a^{2}} \int_{0}^{π} (c o s r θ + 2 a c o s θ c o s r θ + . . + 2 a^{r} c o s r θ . c o s r θ + \dots) d θ

a g a i n f o r m u l a

\int_{0}^{π} c o s m θ c o s n θ d θ = \int_{0}^{π} s i n m θ s i n n θ d θ = 0

w h e n m \neq n

i f m = n t h e n \int_{0}^{π} c o s m θ c o s n θ d θ = \frac{π}{2}

s o

I (r) = \frac{1}{1 - a^{2}} \times (0 + 0 + \dots + 2 a^{r} \times \frac{π}{2})

I (r) = \frac{a^{r} π}{1 - a^{2}}

s o r e q u i r e d a n s w e r = I (2) = \frac{a^{2} π}{1 - a^{2}}

Commented by arcana last updated on 26/Jul/19

gracias!

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