∫_0 ^π ((sin((21x)/2))/(sin(x/2)))dx


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Question Number 91018 by M±th+et+s last updated on 27/Apr/20

$\int_{0}^{π} \frac{s i n \frac{21 x}{2}}{s i n \frac{x}{2}} d x$
Commented by mathmax by abdo last updated on 29/Apr/20
$let take atry changement (x/2) =t give ∫_0 ^π ((sin(((21x)/2)))/(sin((x/2))))dx =2∫_0 ^(π/2) ((sin(21t))/(sint))dt let w(t) =(1/(sint)) we have w(t) =(2/(e^(it) −e^(−it) )) =(2/(z−z^(−1) )) ( z=e^(it) ) =((2z)/(z^2 −1)) =−2z((1/(1−z^2 ))) =−2z Σ_(n=0) ^∞ z^(2n) =−2e^(it) Σ_(n=0) ^∞ e^(2int) ⇒ I =−4 ∫_0 ^(π/2) sin(21t)Σ_(n=0) ^∞ e^((2n+1)it) dt =−4 Σ_(n=0) ^∞ ∫_0 ^(π/2) sin(21t)e^((2n+1)t) dt =−4 Σ_(n=0) ^∞ A_n A_n =∫_0 ^(π/2) sin(21t)(cos(2n+1)t +isin(2n+1)t)dt =∫_0 ^(π/2) sin(21t)cos(2n+1)t +i ∫_0 ^(π/2) sin(21t)sin(2n+1)t dt sina cosb =cos((π/2)−a)cosb =(1/2)(cos((π/2)−a+b)+cos((π/2)−a−b)) =(1/2){sin(a−b)+sin(a+b)} ⇒ ∫_0 ^(π/2) sin(21t)cos(2n+1)t dt =(1/2)∫_0 ^(π/2) {sin(20−2n)t +sin(2n+22)t}dt =(1/2)[−(1/(20−2n))cos(20−2n)t+(1/(2n+22))cos(2n+22)t]_0 ^(π/2) =(1/2){−(1/(20−2n))cos(10−n)π+(1/(2n+22))cos(n+11)π +(1/(20−2n))−(1/(2n+22))}....be continued...$
$l e t t a k e a t r y c h a n g e m e n t \frac{x}{2} = t g i v e$ $\int_{0}^{π} \frac{s i n (\frac{21 x}{2})}{s i n (\frac{x}{2})} d x = 2 \int_{0}^{\frac{π}{2}} \frac{s i n (21 t)}{s i n t} d t l e t w (t) = \frac{1}{s i n t}$ $w e h a v e w (t) = \frac{2}{e^{i t} - e^{- i t}} = \frac{2}{z - z^{- 1}} (z = e^{i t})$ $= \frac{2 z}{z^{2} - 1} = - 2 z (\frac{1}{1 - z^{2}}) = - 2 z \sum_{n = 0}^{\infty} z^{2 n} = - 2 e^{i t} \sum_{n = 0}^{\infty} e^{2 i n t} \Rightarrow$ $I = - 4 \int_{0}^{\frac{π}{2}} s i n (21 t) \sum_{n = 0}^{\infty} e^{(2 n + 1) i t} d t$ $= - 4 \sum_{n = 0}^{\infty} \int_{0}^{\frac{π}{2}} s i n (21 t) e^{(2 n + 1) t} d t = - 4 \sum_{n = 0}^{\infty} A_{n}$ $A_{n} = \int_{0}^{\frac{π}{2}} s i n (21 t) (c o s (2 n + 1) t + i s i n (2 n + 1) t) d t$ $= \int_{0}^{\frac{π}{2}} s i n (21 t) c o s (2 n + 1) t + i \int_{0}^{\frac{π}{2}} s i n (21 t) s i n (2 n + 1) t d t$ $s i n a c o s b = c o s (\frac{π}{2} - a) c o s b = \frac{1}{2} (c o s (\frac{π}{2} - a + b) + c o s (\frac{π}{2} - a - b))$ $= \frac{1}{2} {s i n (a - b) + s i n (a + b)} \Rightarrow$ $\int_{0}^{\frac{π}{2}} s i n (21 t) c o s (2 n + 1) t d t = \frac{1}{2} \int_{0}^{\frac{π}{2}} {s i n (20 - 2 n) t + s i n (2 n + 22) t} d t$ $= \frac{1}{2} {[- \frac{1}{20 - 2 n} c o s (20 - 2 n) t + \frac{1}{2 n + 22} c o s (2 n + 22) t]}_{0}^{\frac{π}{2}}$ $= \frac{1}{2} {- \frac{1}{20 - 2 n} c o s (10 - n) π + \frac{1}{2 n + 22} c o s (n + 11) π$ $+ \frac{1}{20 - 2 n} - \frac{1}{2 n + 22}} . . . . b e c o n t i n u e d . . .$
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