Question Number 174838 by mnjuly1970 last updated on 12/Aug/22
$$ \\ $$$$\:\:\:\:\:{calculate} \\ $$$$ \\ $$$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{2}} }{\left(\:{sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\:\mathrm{2}} }{dx}=? \\ $$$$ \\ $$
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)](https://www.tinkutara.com/question/Q174839.png)
$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:{calculate} \\ $$$$ \\ $$$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{2}} }{\left(\:{sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\:\mathrm{2}} }{dx}=? \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}\:+\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)}{dx} \\ $$$$\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:=\:\frac{\mathrm{1}\:−\:{e}^{−\mathrm{4}{ix}} \:}{\mathrm{2}{ie}^{−\mathrm{2}{ix}} } \\ $$$$\Omega\:=\:\mathrm{2}{i}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{2}} {e}^{−\mathrm{2}{ix}} }{\mathrm{2}{ie}^{−\mathrm{2}{ix}} \:+\:\mathrm{1}\:−\:{e}^{−\mathrm{4}{ix}} }{dx} \\ $$$${let}\:{z}\:=\:{e}^{−\mathrm{2}{ix}} \\ $$$${dz}\:=−\mathrm{2}{ie}^{−\mathrm{2}{ix}} {dx}\:\Rightarrow\:−\frac{{dz}}{\mathrm{2}{iz}}={dx}\:{and}\:{x}\:=\:−\frac{\mathrm{1}}{\mathrm{2}{i}}\mathrm{ln}\:\left({z}\right) \\ $$$$\Omega\:=\:−\frac{\mathrm{1}}{\mathrm{4}{i}}\int_{{C}} \frac{\mathrm{ln}^{\mathrm{2}} \left({z}\right)\:}{\mathrm{2}{iz}+\mathrm{1}−{z}^{\mathrm{2}} }{dz}\:{where}\:{the}\:{contour}\:{C}\:{is}\:{in}\:{the}\:{first}\:{quadrant} \\ $$$$ \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{4}{i}}\int_{{C}} \frac{\mathrm{ln}\:^{\mathrm{2}} \left({z}\right)}{\left({z}−{i}\right)^{\mathrm{2}} }{dz}\: \\ $$$${double}\:{poles}\:{appear}\:{at}\:{z}={i}\:{and}\:{they}\:{are}\:{in}\:{the}\:{contour} \\ $$$$ \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{4}{i}}×\mathrm{2}\pi{i}\left[\frac{{d}}{{dz}}\mathrm{ln}\:^{\mathrm{2}} \left({z}\right)\right]_{{z}={i}} \:=\:\frac{\pi}{\mathrm{2}}\left\{\frac{\mathrm{2ln}\:\left({i}\right)}{{i}}\right\} \\ $$$$\mathrm{ln}\:\left({i}\right)\:=\:{i}\frac{\pi}{\mathrm{2}} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{2}}×\frac{\mathrm{2}{i}\pi}{\mathrm{2}{i}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$$\Omega\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 12/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Frix last updated on 12/Aug/22
$$\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}… \\ $$
Commented by Frix last updated on 12/Aug/22
$$\mathrm{after}\:\mathrm{substituting} \\ $$$${z}=\mathrm{e}^{−\mathrm{2i}{x}} \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$$\Omega=−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\left({z}^{\mathrm{2}} +\mathrm{2i}{z}−\mathrm{1}\right)\mathrm{ln}^{\mathrm{2}} \:{z}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dz} \\ $$$$\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{but}\:\mathrm{your}\:\mathrm{result}\:\mathrm{is} \\ $$$$\Omega\approx\mathrm{4}.\mathrm{935} \\ $$$$\mathrm{and}\:\mathrm{approximating}\:\mathrm{I}\:\mathrm{get} \\ $$$$\Omega=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{x}^{\mathrm{2}} }{\left(\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}\approx\mathrm{0}.\mathrm{862} \\ $$
Commented by princeDera last updated on 12/Aug/22
$$ \\ $$$${how}\:{did}\:{u}\:{get}\:{that}\:{first}\:{integral} \\ $$$$ \\ $$
Commented by Frix last updated on 13/Aug/22
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}}{dx}= \\ $$$$\mathrm{with}\:\mathrm{sin}\:\mathrm{2}{x}\:=\frac{\mathrm{e}^{\mathrm{4i}{x}} −\mathrm{1}}{\mathrm{2ie}^{\mathrm{2i}{x}} }\:\left(=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{4i}{x}} }{\mathrm{2ie}^{−\mathrm{2i}{x}} }\right) \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{2i}{x}} \left(\mathrm{2e}^{\mathrm{2i}{x}} +\left(\mathrm{e}^{\mathrm{4i}{x}} −\mathrm{1}\right)\mathrm{i}\right)}{\left(\mathrm{e}^{\mathrm{4i}{x}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}= \\ $$$$\mathrm{with}\:{z}=\mathrm{e}^{\mathrm{2i}{x}} \:\Leftrightarrow\:{x}=−\mathrm{i}\frac{\mathrm{ln}\:{z}}{\mathrm{2}}\:\Leftrightarrow\:{dx}=−\mathrm{i}\frac{{dz}}{\mathrm{2}{z}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{1}} {\overset{−\mathrm{1}} {\int}}\frac{\left({z}^{\mathrm{2}} −\mathrm{2i}{z}−\mathrm{1}\right)\mathrm{ln}^{\mathrm{2}} \:{z}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dz} \\ $$
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)](https://www.tinkutara.com/question/Q174873.png)
$$\:\:\Omega=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{2}} }{{sin}^{\:\mathrm{2}} \left({x}+\frac{\pi}{\mathrm{4}}\right)}{dx} \\ $$$$\:\:\:\:\overset{{i}.{b}.{p}} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\left[−{cot}\left({x}+\frac{\pi}{\mathrm{4}}\right){x}^{\:\mathrm{2}} \right]_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} +\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {xcot}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx}\right. \\ $$$$\:\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{8}}\:+\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}.{cot}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx} \\ $$$$\:\:\overset{{i}.{b}.{p}} {=}\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{8}}\:+\:\left\{\left[{xlnsin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {lnsin}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx}\right\} \\ $$$$\:\:=\frac{\pi^{\:\mathrm{2}} }{\mathrm{8}}\:+\frac{\pi}{\mathrm{2}}\:{ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:−\int_{−\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{4}}} {lncos}\left({x}\right){dx} \\ $$$$\:\:\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{8}}\:−\frac{\pi}{\mathrm{4}}\:{ln}\left(\mathrm{2}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {lncos}\left({x}\right){dx} \\ $$$$\:\:\underset{\boldsymbol{{ln}}\left(\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\right)^{\:\ast\ast} } {\overset{\boldsymbol{{fourier}}\:\boldsymbol{{series}}} {=}}\frac{\pi^{\:\mathrm{2}} }{\mathrm{8}}\:−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\mathrm{2}\left\{\frac{{G}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\right\} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{8}}\:+\frac{\pi}{\mathrm{4}}\:{ln}\left(\mathrm{2}\right)−{G} \\ $$$$\:\:\:\ast\ast\::\:\:{ln}\left({cos}\left({x}\right)\right)\overset{?} {=}−{ln}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left.\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {cos}\left(\mathrm{2}{nx}\right)\right)}{{n}}\: \\ $$