∫((sin(x))/(1+x^2 ))dx


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Question Number 111023 by mohammad17 last updated on 01/Sep/20

$\int \frac{s i n (x)}{1 + x^{2}} d x$
Commented by mohammad17 last updated on 01/Sep/20

$h e l p m e s i r$
	Answered by mathdave last updated on 02/Sep/20

	$s o l u t i o n$ $s e r i e s f o r m o f \frac{1}{1 + x^{2}} = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} x^{2 n}$ $I = \int \frac{\sin x}{1 + x^{2}} d x = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int x^{2 n} \sin x d x$ $u s i n g I B P$ $l e t u = x^{2 n}, d u = 2 x^{2 n} a n d \int d v = \int \sin x d x, v = - \cos x$ $b u t \int u d v = u v - \int v d u$ $I = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} [(- x^{2 n} \cos x + 2 \int x^{2 n} \cos x d x]$ $I = - {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} x^{2 n} \cos x + 2 {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int x^{2 n} \cos x d x$ $I = - {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} x^{2 n} \cos x + 2 {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} [x^{2 n} \sin x - 2 \int x^{2 n} \sin x d x]$ $I = - {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} x^{2 n} \cos x + 2 {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} x^{2 n} \sin x - 4 I$ $I + 4 I = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} x^{2 n} [2 \sin x - \cos x]$ $∵ I = \int \frac{\sin x}{1 + x^{2}} d x = \frac{{(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} x^{2 n}}{5} [2 \sin x - \cos x]$ $m a t h d a v e (02 / 09 / 2020)$
	Commented by mohammad17 last updated on 02/Sep/20

	$y o u a r e a n a w e s o m e p e r s o n t h a n k y o u s i r$
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