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Question Number 205279 by gopikrishnan last updated on 14/Mar/24
∫^(π/2) _(-π/2) ((8(√2)cosx)/((1+e^(sinx) )(1+sin^4 x)))dx=aπ+blog(3+2(√2)) then find a+b
$$\underset{-\pi/\mathrm{2}} {\int}^{\pi/\mathrm{2}} \frac{\mathrm{8}\sqrt{\mathrm{2}}{cosx}}{\left(\mathrm{1}+\overset{{sinx}} {{e}}\right)\left(\mathrm{1}+{si}\overset{\mathrm{4}} {{n}x}\right)}{dx}={a}\pi+{blog}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:{then}\:{find}\:{a}+{b} \\ $$
Answered by Berbere last updated on 14/Mar/24
x→−x Ω=∫_(−(π/2)) ^(π/2) ((8(√2)cos(x))/((1+e^(sin(x)) )(1+sin^4 (x))))dx=∫_(π/2) ^(−(π/2)) −((8(√2)cos(−x))/((1+e^(−sin(x)) )(1+sin^4 (x))))dx =Ω 2Ω=∫_(−(π/2)) ^(π/2) ((8(√2)cos(x))/(1+sin^4 (x)));sin(x)=y 2Ω=∫_(−1) ^1 ((8(√2)dy)/(1+y^4 ))⇒Ω=∫_0 ^1 ((8(√2)dy)/((y^2 +1+y(√2))(y^2 +1−y(√2)))) ((8(√2))/((y^4 +1)))=((ay+b)/(y^2 +1+y(√2)))+((cy+d)/(y^2 +1−y(√2))) a+c=0,(−a(√2)+b+c(√2)+d)=0 b+d=8(√2);(a+c+d(√2)−b(√2))=0 a−c=8 d−b=0 d=b=4(√2);a=4,c=−4 ((4y+4(√2))/(y^2 +y(√2)+1))+((−4y+4(√2))/(y^2 +1−y(√2))) ((2(2y+(√2)))/(y^2 +y(√2)+1))+((2(√2))/((y+(1/( (√2))))^2 +(1/2)))+((−2(2y−(√2)))/(y^2 +1−y(√2)))+((2(√2))/((y−(1/2))^2 +(1/2))) ∫((8(√2))/(y^4 +1))dy=2ln(((y^2 +y(√2)+1)/(y^2 −y(√2)+1)))+4tan^(−1) (y(√2)+1)+4tan^(−1) (y(√2)−1)+c Ω=2ln(((2+(√2))/(2−(√2))))+4(tan^(−1) ((√2)+1)+tan^(−1) ((√2)−1)−tan^(−1) (1)−tan^(−1) (−1)) =2ln(((6+4(√2))/2))+4(tan^(−1) ((√2)+1)+tan^(−1) ((1/( (√2)+1)))) 2ln(3+2(√2))+4.(π/2)=2π+2ln(3+2(√(2)))
$${x}\rightarrow−{x} \\ $$$$\Omega=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{8}\sqrt{\mathrm{2}}{cos}\left({x}\right)}{\left(\mathrm{1}+{e}^{{sin}\left({x}\right)} \right)\left(\mathrm{1}+{sin}^{\mathrm{4}} \left({x}\right)\right)}{dx}=\int_{\frac{\pi}{\mathrm{2}}} ^{−\frac{\pi}{\mathrm{2}}} −\frac{\mathrm{8}\sqrt{\mathrm{2}}{cos}\left(−{x}\right)}{\left(\mathrm{1}+{e}^{−{sin}\left({x}\right)} \right)\left(\mathrm{1}+{sin}^{\mathrm{4}} \left({x}\right)\right)}{dx} \\ $$$$=\Omega \\ $$$$\mathrm{2}\Omega=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{8}\sqrt{\mathrm{2}}{cos}\left({x}\right)}{\mathrm{1}+{sin}^{\mathrm{4}} \left({x}\right)};{sin}\left({x}\right)={y} \\ $$$$\mathrm{2}\Omega=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{8}\sqrt{\mathrm{2}}{dy}}{\mathrm{1}+{y}^{\mathrm{4}} }\Rightarrow\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{8}\sqrt{\mathrm{2}}{dy}}{\left({y}^{\mathrm{2}} +\mathrm{1}+{y}\sqrt{\mathrm{2}}\right)\left({y}^{\mathrm{2}} +\mathrm{1}−{y}\sqrt{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\left({y}^{\mathrm{4}} +\mathrm{1}\right)}=\frac{{ay}+{b}}{{y}^{\mathrm{2}} +\mathrm{1}+{y}\sqrt{\mathrm{2}}}+\frac{{cy}+{d}}{{y}^{\mathrm{2}} +\mathrm{1}−{y}\sqrt{\mathrm{2}}} \\ $$$${a}+{c}=\mathrm{0},\left(−{a}\sqrt{\mathrm{2}}+{b}+{c}\sqrt{\mathrm{2}}+{d}\right)=\mathrm{0} \\ $$$${b}+{d}=\mathrm{8}\sqrt{\mathrm{2}};\left({a}+{c}+{d}\sqrt{\mathrm{2}}−{b}\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$${a}−{c}=\mathrm{8} \\ $$$${d}−{b}=\mathrm{0} \\ $$$${d}={b}=\mathrm{4}\sqrt{\mathrm{2}};{a}=\mathrm{4},{c}=−\mathrm{4} \\ $$$$\frac{\mathrm{4}{y}+\mathrm{4}\sqrt{\mathrm{2}}}{{y}^{\mathrm{2}} +{y}\sqrt{\mathrm{2}}+\mathrm{1}}+\frac{−\mathrm{4}{y}+\mathrm{4}\sqrt{\mathrm{2}}}{{y}^{\mathrm{2}} +\mathrm{1}−{y}\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}\left(\mathrm{2}{y}+\sqrt{\mathrm{2}}\right)}{{y}^{\mathrm{2}} +{y}\sqrt{\mathrm{2}}+\mathrm{1}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left({y}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}+\frac{−\mathrm{2}\left(\mathrm{2}{y}−\sqrt{\mathrm{2}}\right)}{{y}^{\mathrm{2}} +\mathrm{1}−{y}\sqrt{\mathrm{2}}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\int\frac{\mathrm{8}\sqrt{\mathrm{2}}}{{y}^{\mathrm{4}} +\mathrm{1}}{dy}=\mathrm{2}{ln}\left(\frac{{y}^{\mathrm{2}} +{y}\sqrt{\mathrm{2}}+\mathrm{1}}{{y}^{\mathrm{2}} −{y}\sqrt{\mathrm{2}}+\mathrm{1}}\right)+\mathrm{4tan}^{−\mathrm{1}} \left({y}\sqrt{\mathrm{2}}+\mathrm{1}\right)+\mathrm{4tan}^{−\mathrm{1}} \left({y}\sqrt{\mathrm{2}}−\mathrm{1}\right)+{c} \\ $$$$\Omega=\mathrm{2}{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)+\mathrm{4}\left(\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left(−\mathrm{1}\right)\right) \\ $$$$=\mathrm{2}{ln}\left(\frac{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)+\mathrm{4}\left(\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\right)\right) \\ $$$$\mathrm{2}{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)+\mathrm{4}.\frac{\pi}{\mathrm{2}}=\mathrm{2}\pi+\mathrm{2}{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right. \\ $$
Commented by gopikrishnan last updated on 16/Mar/24
thank u
$${thank}\:{u} \\ $$

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