calculate Ω = ∫_0 ^( (π/2)) (( x^( 2) )/(( sin(x)+cos(x))^( 2) ))dx=?


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Question Number 174838 by mnjuly1970 last updated on 12/Aug/22

$c a l c u l a t e$ $Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{(s i n (x) + c o s (x))}^{2}} d x = ?$
	Answered by princeDera last updated on 12/Aug/22
	$calculate Ω = ∫_0 ^( (π/2)) (( x^( 2) )/(( sin(x)+cos(x))^( 2) ))dx=? Ω = ∫_0 ^(π/2) (x^2 /(1 + sin (2x)))dx sin (2x) = ((1 − e^(−4ix) )/(2ie^(−2ix) )) Ω = 2i∫_0 ^(π/2) ((x^2 e^(−2ix) )/(2ie^(−2ix) + 1 − e^(−4ix) ))dx let z = e^(−2ix) dz =−2ie^(−2ix) dx ⇒ −(dz/(2iz))=dx and x = −(1/(2i))ln (z) Ω = −(1/(4i))∫_C ((ln^2 (z) )/(2iz+1−z^2 ))dz where the contour C is in the first quadrant Ω = (1/(4i))∫_C ((ln^2 (z))/((z−i)^2 ))dz double poles appear at z=i and they are in the contour Ω = (1/(4i))×2πi[(d/dz)ln^2 (z)]_(z=i) = (π/2){((2ln (i))/i)} ln (i) = i(π/2) Ω = (π/2)×((2iπ)/(2i)) = (π^2 /2) Ω = (π^2 /2)$
	$c a l c u l a t e$ $Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{(s i n (x) + c o s (x))}^{2}} d x = ?$ $Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{1 + \sin (2 x)} d x$ $\sin (2 x) = \frac{1 - e^{- 4 i x}}{2 {i e}^{- 2 i x}}$ $Ω = 2 i \int_{0}^{\frac{π}{2}} \frac{x^{2} e^{- 2 i x}}{2 {i e}^{- 2 i x} + 1 - e^{- 4 i x}} d x$ $l e t z = e^{- 2 i x}$ $d z = - 2 {i e}^{- 2 i x} d x \Rightarrow - \frac{d z}{2 i z} = d x a n d x = - \frac{1}{2 i} \ln (z)$ $Ω = - \frac{1}{4 i} \int_{C} \frac{\ln^{2} (z)}{2 i z + 1 - z^{2}} d z w h e r e t h e c o n t o u r C i s i n t h e f i r s t q u a d r a n t$ $Ω = \frac{1}{4 i} \int_{C} \frac{\ln^{2} (z)}{{(z - i)}^{2}} d z$ $d o u b l e p o l e s a p p e a r a t z = i a n d t h e y a r e i n t h e c o n t o u r$ $Ω = \frac{1}{4 i} \times 2 π i {[\frac{d}{d z} \ln^{2} (z)]}_{z = i} = \frac{π}{2} {\frac{2 \ln (i)}{i}}$ $\ln (i) = i \frac{π}{2}$ $Ω = \frac{π}{2} \times \frac{2 i π}{2 i} = \frac{π^{2}}{2}$ $Ω = \frac{π^{2}}{2}$
	Commented by Tawa11 last updated on 12/Aug/22

	$Great sir$
	Commented by Frix last updated on 12/Aug/22

	$something went wrong . . .$
	Commented by Frix last updated on 12/Aug/22

	$after substituting$ $z = e^{- 2 i x}$ $I get$ $Ω = - \frac{1}{4} \int \frac{(z^{2} + 2 i z - 1) \ln^{2} z}{{(z^{2} + 1)}^{2}} d z$ $I cannot solve this but your result is$ $Ω \approx 4.935$ $and approximating I get$ $Ω = \underset{0}{\int^{π / 2}} \frac{x^{2}}{{(\sin x + \cos x)}^{2}} d x \approx 0.862$
	Commented by princeDera last updated on 12/Aug/22

	$h o w d i d u g e t t h a t f i r s t i n t e g r a l$
	Commented by Frix last updated on 13/Aug/22

	$\underset{0}{\int^{π / 2}} \frac{x^{2}}{1 + \sin 2 x} d x =$ $with \sin 2 x = \frac{e^{4 i x} - 1}{{2 ie}^{2 i x}} (= \frac{1 - e^{4 i x}}{{2 ie}^{- 2 i x}})$ $= 2 \underset{0}{\int^{π / 2}} \frac{x^{2} e^{2 i x} ({2 e}^{2 i x} + (e^{4 i x} - 1) i)}{{(e^{4 i x} + 1)}^{2}} d x =$ $with z = e^{2 i x} \Leftrightarrow x = - i \frac{\ln z}{2} \Leftrightarrow d x = - i \frac{d z}{2 z}$ $= - \frac{1}{4} \underset{1}{\int^{- 1}} \frac{(z^{2} - 2 i z - 1) \ln^{2} z}{{(z^{2} + 1)}^{2}} d z$
	Answered by mnjuly1970 last updated on 13/Aug/22
	$Ω= (1/2) ∫_0 ^( (π/2)) (( x^( 2) )/(sin^( 2) (x+(π/4))))dx =^(i.b.p) (1/2) {[−cot(x+(π/4))x^( 2) ]_0 ^( (π/2)) +2∫_0 ^( (π/2)) xcot(x+(π/4))dx = (π^( 2) /8) +∫_0 ^( (π/2)) x.cot(x+(π/4))dx =^(i.b.p) (π^( 2) /8) + {[xlnsin(x+(π/4))]_0 ^(π/2) −∫_0 ^( (π/2)) lnsin(x+(π/4))dx} =(π^( 2) /8) +(π/2) ln(((√2)/2)) −∫_(−(π/4)) ^( (π/4)) lncos(x)dx = (π^( 2) /8) −(π/4) ln(2)−2∫_0 ^( (π/4)) lncos(x)dx =_(ln(cos(x))^( ∗∗) ) ^(fourier series) (π^( 2) /8) −(π/4)ln(2)−2{(G/2)−(π/4)ln(2)} = (π^( 2) /8) +(π/4) ln(2)−G ∗∗ : ln(cos(x))=^? −ln(2)−Σ_(n=1) ^∞ (((−1)^(n−1) cos(2nx)))/n)$
	$Ω = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{s i n}^{2} (x + \frac{π}{4})} d x$ $\overset{i . b . p}{=} \frac{1}{2} {{[- c o t (x + \frac{π}{4}) x^{2}]}_{0}^{\frac{π}{2}} + 2 \int_{0}^{\frac{π}{2}} x c o t (x + \frac{π}{4}) d x$ $= \frac{π^{2}}{8} + \int_{0}^{\frac{π}{2}} x . c o t (x + \frac{π}{4}) d x$ $\overset{i . b . p}{=} \frac{π^{2}}{8} + {{[x l n s i n (x + \frac{π}{4})]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} l n s i n (x + \frac{π}{4}) d x}$ $= \frac{π^{2}}{8} + \frac{π}{2} l n (\frac{\sqrt{2}}{2}) - \int_{- \frac{π}{4}}^{\frac{π}{4}} l n c o s (x) d x$ $= \frac{π^{2}}{8} - \frac{π}{4} l n (2) - 2 \int_{0}^{\frac{π}{4}} l n c o s (x) d x$ $\underset{l n {(c o s (x))}^{* }}{\overset{f o u r i e r s e r i e s}{=}} \frac{π^{2}}{8} - \frac{π}{4} l n (2) - 2 {\frac{G}{2} - \frac{π}{4} l n (2)}$ $= \frac{π^{2}}{8} + \frac{π}{4} l n (2) - G$ $ * : l n (c o s (x)) \overset{?}{=} - l n (2) - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} c o s (2 n x))}{n}$
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