L(sin^n (x))=?


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Question Number 174508 by Gbenga last updated on 03/Aug/22

$L (\sin^{n} (x)) = ?$
	Answered by Mathspace last updated on 03/Aug/22
	$sin^n x=(((e^(ix) −e^(−ix) )/(2i)))^n =(1/((2i)^n ))Σ_(k=0) ^n C_n ^k (e^(ix) )^k (−e^(−ix) )^(n−k) =(1/((2i)^n ))Σ_(k=0) ^n C_n ^k (−1)^(n−k) e^(ikx−i(n−k)x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k e^(i(2k−n)x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k {cos(2k−n)x +isin(2k−n)x} ⇒L(sin^n x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k L(cos(2k−n)x +i(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k L(sin(2k−n)x L(cos(2k−n)x) =∫_0 ^∞ cos(2k−n)t e^(−xt) dt =Re(∫_0 ^∞ e^(−xt+i(2k−n)t) dt) and ∫_0 ^∞ e^((−x+i(2k−n))t) dt =[(1/(−x+i(2k−n)))e^((−x+i(2k−n))t) ]_0 ^∞ =−(1/(−x+i(2k−n))) =(1/(x−i(2k−n)))=((x+i(2k−n))/(x^2 +(2k−n)^2 )) ⇒Re(....)=(x/(x^2 +(2k−n)^2 )) ⇒ L(sin^n x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k (x/(x^2 +(2k−n)^2 )) +i(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k ((2k−n)/(x^2 +(2k−n)^2 ))$
	${s i n}^{n} x = {(\frac{e^{i x} - e^{- i x}}{2 i})}^{n}$ $= \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} C_{n}^{k} {(e^{i x})}^{k} {(- e^{- i x})}^{n - k}$ $= \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{n - k} e^{i k x - i (n - k) x}$ $= \frac{{(- 1)}^{n}}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} e^{i (2 k - n) x}$ $= \frac{{(- 1)}^{n}}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} {c o s (2 k - n) x + i s i n (2 k - n) x}$ $\Rightarrow L ({s i n}^{n} x)$ $= \frac{{(- 1)}^{n}}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} L (c o s (2 k - n) x$ $+ i \frac{{(- 1)}^{n}}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} L (s i n (2 k - n) x$ $L (c o s (2 k - n) x)$ $= \int_{0}^{\infty} c o s (2 k - n) t e^{- x t} d t$ $= R e (\int_{0}^{\infty} e^{- x t + i (2 k - n) t} d t)$ $a n d \int_{0}^{\infty} e^{(- x + i (2 k - n)) t} d t$ $= {[\frac{1}{- x + i (2 k - n)} e^{(- x + i (2 k - n)) t}]}_{0}^{\infty}$ $= - \frac{1}{- x + i (2 k - n)}$ $= \frac{1}{x - i (2 k - n)} = \frac{x + i (2 k - n)}{x^{2} + {(2 k - n)}^{2}}$ $\Rightarrow R e (. . . .) = \frac{x}{x^{2} + {(2 k - n)}^{2}} \Rightarrow$ $L ({s i n}^{n} x)$ $= \frac{{(- 1)}^{n}}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} \frac{x}{x^{2} + {(2 k - n)}^{2}}$ $+ i \frac{{(- 1)}^{n}}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} \frac{2 k - n}{x^{2} + {(2 k - n)}^{2}}$
	Commented by Gbenga last updated on 03/Aug/22

	$t h a n k s s k$
	Commented by Tawa11 last updated on 03/Aug/22

	$Great sir$
	Commented by Mathspace last updated on 03/Aug/22

	$t h a n k s$
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