∫_0 ^π Sin ^(100) xdx


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Question Number 172741 by Kalebwizeman last updated on 30/Jun/22

$\underset{0}{\int^{π}} S i n \overset{100}{} x d x$
	Answered by aleks041103 last updated on 01/Jul/22

	$I_{n} = 2 \int_{0}^{π / 2} {s i n}^{n} (x) d x$ ${s i n}^{n} x = (1 - {c o s}^{2} x) {s i n}^{n - 2} x$ $\Rightarrow \frac{1}{2} I_{n} = \int_{0}^{π / 2} {s i n}^{n - 2} x d x - \int_{0}^{π / 2} {c o s}^{2} x {s i n}^{n - 2} x d x =$ $= \frac{1}{2} I_{n - 2} - \int_{0}^{π / 2} (c o s x) (c o s x {s i n}^{n - 2} x d x) =$ $= \frac{1}{2} I_{n - 2} - \int_{0}^{π / 2} (c o s x) ({(s i n x)}^{n - 2} d (s i n x)) =$ $= \frac{1}{2} I_{n - 2} - \int_{0}^{π / 2} (c o s x) d (\frac{{s i n}^{n - 1} x}{n - 1}) =$ $= \frac{1}{2} I_{n - 2} - \frac{1}{n - 1} \int_{0}^{π / 2} (c o s x) d ({s i n}^{n - 1} x)$ $\int_{0}^{π / 2} (c o s x) d ({s i n}^{n - 1} x) =$ $= {[{c o s x s i n}^{n - 1} x]}_{0}^{π / 2} - \int_{0}^{π / 2} ({s i n}^{n - 1} x) d (c o s x) =$ $= \int_{0}^{π / 2} {s i n}^{n} x d x = \frac{1}{2} I_{n}$ $\Rightarrow I_{n} = I_{n - 2} - \frac{1}{n - 1} I_{n}$ $\Rightarrow (1 + \frac{1}{n - 1}) I_{n} = I_{n - 2}$ $\Rightarrow I_{n} = \frac{n - 1}{n} I_{n - 2}$ $\Rightarrow I_{2 m} = \frac{2 m - 1}{2 m} I_{2 (m - 1)}$ $\Rightarrow I_{2 m} = (\underset{s = 0}{\prod^{k - 1}} \frac{2 (m - s) - 1}{2 (m - s)}) I_{2 (m - k)}$ $k = m$ $\Rightarrow I_{2 m} = (\underset{s = 0}{\prod^{m - 1}} \frac{2 (m - s) - 1}{2 (m - s)}) I_{0}$ $I_{0} = 2 \int_{0}^{π / 2} {s i n}^{0} x d x = π$ $\Rightarrow I_{2 m} = π \underset{m - s = s^{'} = 1}{\prod^{m}} \frac{2 s^{'} - 1}{2 s^{'}}$ $I_{2 m} = \frac{1.3.5.7 . . . . . (2 m - 1)}{2.4.6.8 . . . . . (2 m)} π$ $\frac{1.3.5.7 . . . . . (2 m - 1)}{2.4.6.8 . . . . . (2 m)} = \frac{1.2.3 . . . . . (2 m - 1) (2 m)}{{(2.4.6.8 . . . . . (2 m))}^{2}} = \frac{(2 m)!}{{(2^{m} m!)}^{2}}$ $\Rightarrow \int_{0}^{π} {s i n}^{2 m} x d x = \frac{(2 m)! π}{4^{m} {(m!)}^{2}}$ $\Rightarrow \int_{0}^{π} {s i n}^{100} x d x = \frac{100! π}{4^{100} {(50!)}^{2}}$
	Commented by Tawa11 last updated on 01/Jul/22

	$Great sir$
	Commented by Kalebwizeman last updated on 02/Jul/22

	$t h a n k y o u s o m u c h s i r$
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