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Question Number 46639 by Tinkutara last updated on 29/Oct/18

Commented by maxmathsup by imad last updated on 30/Oct/18
$let A =∫_0 ^π ((sin(2x))/(x+1))dx ⇒A=_(2x=t) ∫_0 ^(2π) ((sin(t))/((t/2)+1)) (dt/2) = ∫_0 ^(2π) ((sint)/(t+2)) =_(byparts) [−(1/(t+2)) cost]_0 ^(2π) −∫_0 ^(2π) ((−1)/((t+2)^2 )) (−cost)dt =(1/2) −(1/(2π +2)) −∫_0 ^(2π) ((cost)/((t+2)^2 )) dt for that i think that I = ∫_0 ^(2π) ((cosx)/((x+2)^2 ))dx ⇒ A =(1/2){1−(1/(π +1))}−I =(π/(2π+2)) −I .$
$l e t A = \int_{0}^{π} \frac{s i n (2 x)}{x + 1} d x \Rightarrow A =_{2 x = t} \int_{0}^{2 π} \frac{s i n (t)}{\frac{t}{2} + 1} \frac{d t}{2}$ $= \int_{0}^{2 π} \frac{s i n t}{t + 2} =_{b y p a r t s} {[- \frac{1}{t + 2} c o s t]}_{0}^{2 π} - \int_{0}^{2 π} \frac{- 1}{{(t + 2)}^{2}} (- c o s t) d t$ $= \frac{1}{2} - \frac{1}{2 π + 2} - \int_{0}^{2 π} \frac{c o s t}{{(t + 2)}^{2}} d t f o r t h a t i t h i n k t h a t I = \int_{0}^{2 π} \frac{c o s x}{{(x + 2)}^{2}} d x \Rightarrow$ $A = \frac{1}{2} {1 - \frac{1}{π + 1}} - I = \frac{π}{2 π + 2} - I .$
Commented by Tinkutara last updated on 31/Oct/18
Thanks Sir! Maybe it's a mistake in question.
	Answered by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18

	$\int \frac{s i n 2 x}{x + 1} d x$ $\frac{1}{x + 1} \int s i n 2 x - \int [\frac{d}{d x} (\frac{1}{x + 1}) \int s i n 2 x d x] d x$ $\frac{1}{x + 1} \times \frac{- c o s 2 x}{2} - \int \frac{- 1}{{(x + 1)}^{2}} \times \frac{- c o s 2 x}{2} d x$ $\frac{- c o s 2 x}{2 (x + 1)} - \int \frac{c o s 2 x}{2 {(x + 1)}^{2}} d x$ $s o \int_{0}^{π} \frac{s i n 2 x}{x + 1} d x$ $=∣ \frac{- c o s 2 x}{2 (x + 1)} ∣_{0}^{π} - \frac{1}{2} \int_{0}^{π} \frac{c o s 2 x}{{(x + 1)}^{2}} d x$ $= \frac{- 1}{2} (\frac{c o s 2 π}{π + 1} - \frac{c o s 0}{0 + 1}) - \frac{1}{2} j$ $\frac{- 1}{2} (\frac{1}{π + 1} - 1) - \frac{1}{2} j$ $c o n t d . . .$
	Commented by Tinkutara last updated on 30/Oct/18

	$W h y \int_{0}^{2 π} \frac{c o s t}{{(t + 2)}^{2}} d t = 2 \int_{0}^{π} \frac{c o s t}{{(t + 2)}^{2}} d t = 4 I ?$ $S i r h o w f (2 a - x) = f (x) h e r e ?$
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