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Question Number 147346 by mnjuly1970 last updated on 20/Jul/21

	Answered by qaz last updated on 20/Jul/21
	$a_n =2∫_0 ^(π/2) (1−sin x)^n sin xd(sin x) ⇒a_n =2∫_0 ^1 (1−u)^n udu Σ_(n=1) ^∞ (a_n /n)=Σ_(n=1) ^∞ (2/n)∫_0 ^1 (1−u)^n udu =2∫_0 ^1 uΣ_(n=1) ^∞ (((1−u)^n )/n)du =−2∫_0 ^1 uln(1−(1−u))du =−2∫_0 ^1 ulnudu =−2{(1/2)u^2 lnu∣_0 ^1 −(1/2)∫_0 ^1 udu} =(1/2)$
	$a_{n} = 2 \int_{0}^{π / 2} {(1 - \sin x)}^{n} \sin xd (\sin x)$ $\Rightarrow a_{n} = 2 \int_{0}^{1} {(1 - u)}^{n} udu$ $\underset{n = 1}{\sum^{\infty}} \frac{a_{n}}{n} = \underset{n = 1}{\sum^{\infty}} \frac{2}{n} \int_{0}^{1} {(1 - u)}^{n} udu$ $= 2 \int_{0}^{1} u \underset{n = 1}{\sum^{\infty}} \frac{{(1 - u)}^{n}}{n} du$ $= - 2 \int_{0}^{1} uln (1 - (1 - u)) du$ $= - 2 \int_{0}^{1} ulnudu$ $= - 2 {\frac{1}{2} u^{2} lnu ∣_{0}^{1} - \frac{1}{2} \int_{0}^{1} udu}$ $= \frac{1}{2}$
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