I = ∫_0^πx log sin x dx ⋯ (i) = ∫_0^π(π x)log sin (π x)dx ⋯ (ii) By adding (i) and (ii), we get 2I = ∫_0^ππ log sin x dx ⇒ I=2 π/2∫_0^ ...

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Inequalities of Integrals

∫0πx log sin ...

Question

$\int_{0}^{π} x l o g s i n x d x =$

\frac{π}{2} l o g \frac{1}{2}

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\frac{π^{2}}{2} l o g \frac{1}{2}

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$π l o g \frac{1}{2}$

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$π^{2} l o g \frac{1}{2}$

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Solution

The correct option is B $\frac{π^{2}}{2} l o g \frac{1}{2}$
$I = \int_{0}^{π} x l o g s i n x d x \dots (i)$
$= \int_{0}^{π} (π - x) l o g s i n (π - x) d x \dots (i i)$
By adding (i) and (ii), we get
$2 I = \int_{0}^{π} π l o g s i n x d x \Rightarrow I = \frac{2 π}{2} \int_{0}^{\frac{π}{2}} l o g s i n x d x$
$= π (\frac{π}{2} l o g \frac{1}{2}) = \frac{π^{2}}{2} l o g \frac{1}{2}$

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