Given I_n =∫_(nπ) ^((n+1)π) e^(−x) sinx dx , n∈N. 1. Find a relation between I_(n+1) and I


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Question Number 155496 by mathocean1 last updated on 01/Oct/21

$G i v e n I_{n} = \underset{n π}{\int^{(n + 1) π}} e^{- x} s i n x d x, n \in N .$ $1 . F i n d a r e l a t i o n b e t w e e n I_{n + 1} a n d I_{n} .$
	Answered by ArielVyny last updated on 01/Oct/21

	$I_{n} = \int_{n π}^{(n + 1) π} e^{- x} s i n x d x$ $d u = e^{- x} \to u = - e^{- x}$ $v = s i n x \to d v = c o s x$ $I_{n} = {[- e^{- x} s i n x]}_{n π}^{(n + 1) π} + \int_{n π}^{(n + 1)} e^{- x} c o s x d x$ $d u = e^{- x} \to u = - e^{- x}$ $v = c o s x \to d v = - s i n x$ $2 I_{n} = - {[e^{- x} s i n x + e^{- x} c o s x]}_{n π}^{(n + 1) π}$ $2 I_{n} = - [e^{- π (n + 1)} s i n ((n + 1) π) + e^{- π (n + 1)} c o s ((n + 1) π) - e^{- n π} s i n (n π) - e^{- n π} c o s (n π)]$ $2 I_{n} = - [e^{- π (n + 1)} (s i n (n π) c o s π + c o s (n π) s i n π) + e^{- π (n + 1)} (c o s n π c o s π + s i n (n π) s i n (π) - e^{- n π} s i n (n π) - e^{- n π} c o s (n π)]$ $2 I_{n} = - [- e^{- π (n + 1)} {(- 1)}^{n} - e^{- n π} {(- 1)}^{n}]$ $2 I_{n} = {(- 1)}^{n} e^{- π (n + 1)} + {(- 1)}^{n} e^{- n π}$ $2 I_{n + 1} = {(- 1)}^{n + 1} e^{- π (n + 2)} + {(- 1)}^{n} e^{- (n + 1) π}$ $2 I_{n + 1} - 2 I_{n} = {(- 1)}^{n + 1} e^{- π (n + 2)} + {(- 1)}^{n + 1} e^{- n π}$ $I_{n + 1} - I_{n} = \frac{{(- 1)}^{n + 1} e^{- π (n + 2)} + {(- 1)}^{n + 1} e^{- n π}}{2}$
	Commented by mathocean1 last updated on 22/Oct/21

	$t h a n k y o u$
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