find in terms of n the value of A_n = ∫_0 ^1 (1+x^2 )^(n/2) sin(narctanx)dx . ( n∈ N).


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 28268 by abdo imad last updated on 22/Jan/18

$f i n d i n t e r m s o f n t h e v a l u e o f$ $A_{n} = \int_{0}^{1} {(1 + x^{2})}^{\frac{n}{2}} s i n (n a r c t a n x) d x . (n \in N) .$
Commented by abdo imad last updated on 23/Jan/18

$l e t i n t r o d u c e t h e p o l y n o m e P (x) = \frac{1}{2 i} ({(1 + i x)}^{n} - {(1 - i x)}^{n})$ $l e t w r i t e P (x) i n g e o m e t r i c f o r m . w e h a v e$ $∣ 1 + i x ∣= \sqrt{1 + x^{2}} = {(1 + x^{2})}^{\frac{1}{2}} a n d$ $1 + i x = \sqrt{1 + x^{2}} (\frac{1}{\sqrt{1 + x^{2}}} + i \frac{x}{\sqrt{1 + x^{2}}}) = r e^{i θ} \Rightarrow$ $r = \sqrt{1 + x^{2}} a n d t a n θ = x \Rightarrow θ = a r t a n x s o$ ${(1 + i x)}^{n} = {(1 + x^{2})}^{\frac{n}{2}} e^{i n a r c t a n x}$ ${(1 - i x)}^{n} = {(1 + x^{2})}^{\frac{n}{2}} e^{- i n a r t a n x} a n d$ $P (x) = \frac{1}{2 i} (2 i I m ({(1 + i x)}^{n})) = {(1 + x^{2})}^{\frac{n}{2}} s i n (n a r c t a n x)$ $\Rightarrow A_{n} = \int_{0}^{1} P (x) d x b u t w e h a v e p r o v e d t h a t$ $P (x) = \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} x^{2 p + 1}$ $\int_{0}^{1} P (x) d x = \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} \frac{C_{n}^{2 p + 1}}{2 p + 2} = A_{n} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum