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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} .