find-A-n-0-pi-2-1-cos-n-1-x-2sin-x-2-dx-

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} = \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 \dots .