Integration Questions

Question Number 89511 by M±th+et£s last updated on 17/Apr/20

$$\int_{\mathrm{0}} ^{\mathrm{3}\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{4}} \frac{\theta}{\mathrm{3}}\:{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{3}}}\:{d}\theta \\$$

Commented bymr W last updated on 17/Apr/20

$${please}\:{go}\:{to}\:{Q}\mathrm{89482}\:{and}\:{report}\:{that} \\$$ $${you}\:{also}\:{have}\:{this}\:{problem}\:{since}\:{a} \\$$ $${long}\:{time}. \\$$

Commented bymr W last updated on 17/Apr/20

$${did}\:{you}\:{report}\:{once}\:{that}\:{you}\:{can}\:{not} \\$$ $${edit}\:{your}\:{existing}\:{posts}?\:{now}\:{i}\:{have} \\$$ $${the}\:{same}\:{problem}. \\$$

Commented byM±th+et£s last updated on 17/Apr/20

$${i}\:{have}\:{this}\:{problem}\:{from}\:{a}\:{long}\:{time} \\$$ $${and}\:{till}\:{now}\:{i}\:{cant}\:{delete}\:{or}\:{edit}\:{my}\:{posts} \\$$

Commented byM±th+et£s last updated on 17/Apr/20

$${ok}\:{sir}\: \\$$

Commented byabdomathmax last updated on 17/Apr/20

$${sir}\:{mrw}\:{if}\:{you}\:{have}\:{a}\:{problem}\:{with}\:{your} \\$$ $${phone}\:{change}\:{it}\:{with}\:{another}\:{or}\:{work}\:{with} \\$$ $${a}\:{tablet}\:{like}\:{me}....{its}\:{eazy}.... \\$$

Commented bymr W last updated on 18/Apr/20

$${thanks}\:{for}\:{the}\:{hint}\:{sir}!\:{i}'{ll}\:{consider}. \\$$ $${but}\:{now}\:{the}\:{problem}\:{is}\:{solved}\:{by}\:{the} \\$$ $${developer}. \\$$

Answered by ajfour last updated on 19/Apr/20

$${let}\:\frac{\theta}{\mathrm{3}}=\mathrm{2}\phi\:\:\Rightarrow\:\:{d}\theta=\mathrm{6}{d}\phi \\$$ $${I}=\mathrm{6}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{4}} \mathrm{2}\phi\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\phi}\:{d}\phi \\$$ $$\:\:=\mathrm{12}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{4}} \sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{4}} \mathrm{2}\phi\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\phi}\:{d}\phi \\$$ $$\:\:=\:\mathrm{12}\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \sqrt{\left(\mathrm{1}+{T}^{\mathrm{2}} \right)^{\mathrm{3}} −{T}^{\:\mathrm{4}} }×\frac{{d}\phi}{\left(\mathrm{1}+{T}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} } \\$$ $${T}\:=\:\mathrm{tan}\:\mathrm{2}\phi\:\:\Rightarrow\:\:{d}\phi=\frac{{dT}}{\mathrm{2}\left(\mathrm{1}+{T}^{\:\mathrm{2}} \right)} \\$$ $$\:=\:\mathrm{6}\int_{\mathrm{0}} ^{\:\infty} \sqrt{\frac{\left(\mathrm{1}+{T}^{\:\mathrm{2}} \right)^{\mathrm{3}} −{T}^{\:\mathrm{4}} }{\mathrm{1}+{T}^{\:\mathrm{2}} }}\:\frac{{dT}}{\left(\mathrm{1}+{T}^{\:\mathrm{2}} \right)^{\mathrm{2}} } \\$$ $${let}\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{T}^{\:\mathrm{2}} }=\:{t}\:\:\Rightarrow\:\frac{\mathrm{2}{TdT}}{\left(\mathrm{1}+{T}^{\:\mathrm{2}} \right)^{\mathrm{2}} }=\:−{dt} \\$$ $$\:\:=\mathrm{6}\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−{t}\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)^{\mathrm{2}} }\:{dt} \\$$ $$\:\:\:=\mathrm{6}\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{{t}}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:{dt}\:\: \\$$ $$\:\:=\:\mathrm{6}\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{\mathrm{1}−{t}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }}\:{dt} \\$$ $$\:\:\:=\mathrm{6}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\sqrt{{t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\mathrm{1}}}{{t}+\mathrm{1}}{dt} \\$$ $$......... \\$$