Question Number 126749 by bemath last updated on 24/Dec/20
Commented by MJS_new last updated on 24/Dec/20
$${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{2}\int\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{now}\:\mathrm{find}\:\mathrm{the}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{denominator} \\ $$$${t}^{\mathrm{4}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{1}=\left({t}^{\mathrm{2}} +\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right)\left({t}^{\mathrm{2}} +\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right) \\ $$$$\mathrm{and}\:\mathrm{decompose}… \\ $$
Answered by liberty last updated on 24/Dec/20
$${I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{2}−\left(\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}\right)}\:{dx}\:=\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}}\:{dx} \\ $$$${I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{1}}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}}\:{dx}\: \\ $$$${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:{x}}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}}\:{dx}\:=\:−\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} {x}} \\ $$$${I}_{\mathrm{2}} =\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\mathrm{3}−\mathrm{2}{t}^{\mathrm{2}} }\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dt}}{\left(\sqrt{\mathrm{3}}−{t}\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{3}}+{t}\sqrt{\mathrm{2}}\right)} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$