find A_n =∫_0 ^(π/2) ((1−cos(n+1)x)/(2sin((x/2))))dx .


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Question Number 48494 by maxmathsup by imad last updated on 24/Nov/18

$f i n d A_{n} = \int_{0}^{\frac{π}{2}} \frac{1 - c o s (n + 1) x}{2 s i n (\frac{x}{2})} d x .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18
	$A_n −A_(n−1) =∫_0 ^(π/2) ((cos(n)x−cos(n+1)x)/(2sin((x/2))))dx =∫_0 ^(π/2) ((2sin(2n+1)(x/2).sin((x/2)))/(2sin((x/2))))dx =∫_0 ^(π/2) sin(2n+1)(x/2)dx =(((−1)/(2n+1)))∣cos(2n+1)(x/2)∣_0 ^(π/2) =(((−1)/(2n+1))){cos(2n+1)(π/4)−1} =(1/(2n+1)){1−cos(n(π/2)+(π/4))} if n=even =(1/(2n+1))(1+(1/(√2))) or (1/(2n+1))(1−(1/(√2))) if n odd =(1/(2n+1))(1+(1/(√2))) or (1/(2n+1))(1−(1/(√2))) now considering A_n −A_(n−1) =(1/(2n+1))(1+(1/(√2))) pls wait....$
	$A_{n} - A_{n - 1} = \int_{0}^{\frac{π}{2}} \frac{c o s (n) x - c o s (n + 1) x}{2 s i n (\frac{x}{2})} d x$ $= \int_{0}^{\frac{π}{2}} \frac{2 s i n (2 n + 1) \frac{x}{2} . s i n (\frac{x}{2})}{2 s i n (\frac{x}{2})} d x$ $= \int_{0}^{\frac{π}{2}} s i n (2 n + 1) \frac{x}{2} d x$ $= (\frac{- 1}{2 n + 1}) ∣ c o s (2 n + 1) \frac{x}{2} ∣_{0}^{\frac{π}{2}}$ $= (\frac{- 1}{2 n + 1}) {c o s (2 n + 1) \frac{π}{4} - 1}$ $= \frac{1}{2 n + 1} {1 - c o s (n \frac{π}{2} + \frac{π}{4})}$ $i f n = e v e n$ $= \frac{1}{2 n + 1} (1 + \frac{1}{\sqrt{2}}) o r \frac{1}{2 n + 1} (1 - \frac{1}{\sqrt{2}})$ $i f n o d d$ $= \frac{1}{2 n + 1} (1 + \frac{1}{\sqrt{2}}) o r \frac{1}{2 n + 1} (1 - \frac{1}{\sqrt{2}})$ $n o w c o n s i d e r i n g$ $A_{n} - A_{n - 1} = \frac{1}{2 n + 1} (1 + \frac{1}{\sqrt{2}})$ $p l s w a i t . . . .$
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