le give A_n = ∫_0 ^(π/2) ((sin((2n−1)x))/(sinx))dx and B_n =∫_0 ^(π/2) ((sin^2 (nx))/(sin^2 x))dx 1)calculate A_n 2)prove that B_(n+1) −B_n = A_(n+1) .then find B


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 29506 by abdo imad last updated on 09/Feb/18

$l e g i v e A_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n ((2 n - 1) x)}{s i n x} d x a n d B_{n} = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (n x)}{{s i n}^{2} x} d x$ $1) c a l c u l a t e A_{n}$ $2) p r o v e t h a t B_{n + 1} - B_{n} = A_{n + 1} . t h e n f i n d B_{n} .$
Commented by abdo imad last updated on 11/Feb/18

$A_{n + 1} - A_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 n + 1) x - s i n (2 n - 1) x}{s i n x} d x b u t$ $s i n (2 n + 1) x = s i n (2 n x) c o s x + c o s (2 n x) s i n x$ $s i n (2 n - 1) x = s i n (2 n x) c o s x - c o s (2 n x) s i n x \Rightarrow$ $s i n (2 n + 1) x - s i n (2 n - 1) x = 2 c o s (2 n x) s i n x \Rightarrow$ $A_{n + 1} - A_{n} = \int_{0}^{\frac{π}{2}} 2 c o s (2 n x) d x = {[\frac{1}{n} s i n (2 n x)]}_{0}^{\frac{π}{2}} = 0 \forall n$ $A_{n} = A_{1} = \frac{π}{2} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum