∫_0 ^π x ln(sin x) dx = ?


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Question Number 82174 by jagoll last updated on 18/Feb/20

$\underset{0}{\int^{π}} x l n (\sin x) d x = ?$
Commented by mathmax by abdo last updated on 19/Feb/20
$let I =∫_0 ^π xln(sinx)dx ⇒I=∫_0 ^(π/2) x ln(sinx)dx +∫_(π/2) ^π xln(sinx)dx but ∫_(π/2) ^π x ln(sinx)dx =_(x=(π/2) +α) ∫_0 ^(π/2) ((π/2) +α)ln(cosα)dα =(π/2) ∫_0 ^(π/2) ln(cosα)dα +∫_0 ^(π/2) αln(cosα)dα ⇒ I =(π/2) ∫_0 ^(π/2) ln(cosα)dα +∫_0 ^(π/2) x(ln(sinx)+ln(cosx) dx =(π/2) ∫_0 ^(π/2) ln(cosα)dα +∫_0 ^(π/2) xln((1/2)sin(2x))dx =(π/2) ∫_0 ^(π/2) ln(cosα)dα −ln(2)∫_0 ^(π/2) x dx +∫_0 ^(π/2) xln(2x)dx_(→2x=t) =(π/2) ∫_0 ^(π/2) ln(cosα)dα −ln(2)[(x^2 /2)]_0 ^(π/2) +(1/2) ∫_0 ^π tln(t)(dt/2) =(π/2)∫_0 ^(π/2) ln(cosα)dα −ln(2){(π^2 /8)} +(1/4) I ⇒ (3/4) I =(π/2)(−(π/2)ln(2))−(π^2 /8)ln(2) =−(π^2 /4)ln(2)−(π^2 /8)ln(2) =−((3π^2 )/8)ln(2) ⇒ I =(4/3)(−((3π^2 )/8))ln(2) =−(π^2 /2)ln(2)$
$l e t I = \int_{0}^{π} x l n (s i n x) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} x l n (s i n x) d x + \int_{\frac{π}{2}}^{π} x l n (s i n x) d x$ $b u t \int_{\frac{π}{2}}^{π} x l n (s i n x) d x =_{x = \frac{π}{2} + α} \int_{0}^{\frac{π}{2}} (\frac{π}{2} + α) l n (c o s α) d α$ $= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α + \int_{0}^{\frac{π}{2}} α l n (c o s α) d α \Rightarrow$ $I = \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α + \int_{0}^{\frac{π}{2}} x (l n (s i n x) + l n (c o s x) d x$ $= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α + \int_{0}^{\frac{π}{2}} x l n (\frac{1}{2} s i n (2 x)) d x$ $= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α - l n (2) \int_{0}^{\frac{π}{2}} x d x + \int_{0}^{\frac{π}{2}} x l n (2 x) {d x}_{\to 2 x = t}$ $= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α - l n (2) {[\frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} + \frac{1}{2} \int_{0}^{π} t l n (t) \frac{d t}{2}$ $= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α - l n (2) {\frac{π^{2}}{8}} + \frac{1}{4} I \Rightarrow$ $\frac{3}{4} I = \frac{π}{2} (- \frac{π}{2} l n (2)) - \frac{π^{2}}{8} l n (2) = - \frac{π^{2}}{4} l n (2) - \frac{π^{2}}{8} l n (2)$ $= - \frac{3 π^{2}}{8} l n (2) \Rightarrow I = \frac{4}{3} (- \frac{3 π^{2}}{8}) l n (2) = - \frac{π^{2}}{2} l n (2)$
	Answered by Kunal12588 last updated on 19/Feb/20

	$I = \int_{0}^{π} x l o g (s i n x) d x$ $\Rightarrow I = \int_{0}^{π} (x - π) l o g (s i n x) d x$ $\Rightarrow I = \frac{π}{2} \int_{0}^{π} l o g (s i n x) d x$ $\Rightarrow I = π \int_{0}^{π / 2} l o g (s i n x) d x$ $[★ \int_{0}^{π / 2} l o g (s i n x) d x = \int_{0}^{π / 2} l o g (c o s x) d x = - \frac{π}{2} l o g (2)]$ $\Rightarrow I = π (\frac{- π}{2} l o g (2)) = \frac{- π^{2}}{2} l o g (2)$ $∴ \int_{0}^{π} x l o g (s i n x) d x = - \frac{π^{2}}{2} l o g (2)$ $[★ h e r e ‘ ‘ l o g a^{″} m e a n s ‘ ‘ {l o g}_{e} a^{″}]$
	Commented by Kunal12588 last updated on 19/Feb/20

	$★ T . P \int_{0}^{π / 2} l o g (s i n x) d x = \frac{- π}{2} l o g (2)$ $I = \int_{0}^{π / 2} l o g (s i n x) d x$ $\Rightarrow I = \int_{0}^{π / 2} l o g (c o s x) d x$ $\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n x c o s x) d x$ $\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n 2 x) d x - \frac{1}{2} \int_{0}^{π / 2} l o g 2 d x$ $\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n 2 x) d x - \frac{π}{4} l o g 2$ $l e t 2 x = t \Rightarrow d x = \frac{1}{2} d t$ $I = \frac{1}{4} \int_{0}^{π} l o g (s i n t) d t - \frac{π}{4} l o g 2$ $\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (c o s x) d x - \frac{π}{4} l o g 2$ $\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n x) d x - \frac{π}{4} l o g 2$ $\Rightarrow I = \frac{1}{2} I - \frac{π}{4} l o g 2$ $\Rightarrow \frac{1}{2} I = - \frac{π}{4} l o g 2$ $\Rightarrow I = - \frac{π}{2} l o g 2$ $\Rightarrow \int_{0}^{π / 2} l o g (s i n x) d x = \frac{- π}{2} l o g (2)$
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