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Question Number 126749 by bemath last updated on 24/Dec/20

Commented by MJS_new last updated on 24/Dec/20

t=tan (x/2) leads to 2∫(((t+1)^2 )/(t^4 +10t^2 +1))dt now find the 2 square factors of the denominator t^4 +10t^2 +1=(t^2 +5−2(√6))(t^2 +5+2(√6)) and decompose...

$${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{2}\int\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{now}\:\mathrm{find}\:\mathrm{the}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{denominator} \\ $$$${t}^{\mathrm{4}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{1}=\left({t}^{\mathrm{2}} +\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right)\left({t}^{\mathrm{2}} +\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right) \\ $$$$\mathrm{and}\:\mathrm{decompose}... \\ $$

Answered by liberty last updated on 24/Dec/20

I=∫_0 ^π ((1+sin x)/(2−(2cos^2 x−1))) dx =∫_0 ^π ((1+sin x)/(3−2cos^2 x)) dx I_1 =∫_0 ^π (1/(3−2cos^2 x)) dx I_2 =∫_0 ^π ((sin x)/(3−2cos^2 x)) dx = −∫_0 ^π ((d(cos x))/(3−2cos^2 x)) I_2 =∫_(−1) ^1 (dt/(3−2t^2 )) = 2∫_0 ^1 (dt/(((√3)−t(√2))((√3)+t(√2)))) I=I_1 +I_2

$${I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{2}−\left(\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}\right)}\:{dx}\:=\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}}\:{dx} \\ $$$${I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{1}}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}}\:{dx}\: \\ $$$${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:{x}}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}}\:{dx}\:=\:−\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}} \\ $$$${I}_{\mathrm{2}} =\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\mathrm{3}−\mathrm{2}{t}^{\mathrm{2}} }\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dt}}{\left(\sqrt{\mathrm{3}}−{t}\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{3}}+{t}\sqrt{\mathrm{2}}\right)} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$

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