Question Number 106120 by Her_Majesty last updated on 02/Aug/20 | ||
$${question}\:\mathrm{106075}\:{again} \\ $$$$\int\frac{\mathrm{1}+{cosx}}{\left(\mathrm{99}{cosx}−\mathrm{70}{sinx}+\mathrm{210}\right){cosx}−\mathrm{66}{sinx}+\mathrm{110}}{dx}=? \\ $$$${using}\:{t}={tan}\left({x}/\mathrm{2}\right)\:{I}\:{get} \\ $$$$−\mathrm{4}\int\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{3}} −\mathrm{22}{t}^{\mathrm{2}} +\mathrm{272}{t}−\mathrm{419}} \\ $$$${can}\:{someone}\:{factorize}\:{the}\:{denominator}? \\ $$ | ||
Commented by Sarah85 last updated on 03/Aug/20 | ||
$$\mathrm{typo}:\:{t}^{\mathrm{4}} −\mathrm{8}{t}^{\mathrm{3}} −\mathrm{22}{t}^{\mathrm{2}} +\mathrm{272}{t}−\mathrm{419} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${t}=\begin{cases}{\mathrm{2}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}−\sqrt{\mathrm{15}}}\\{\mathrm{2}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}−\sqrt{\mathrm{15}}}\\{\mathrm{2}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}+\sqrt{\mathrm{15}}}\\{\mathrm{2}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}+\sqrt{\mathrm{15}}}\end{cases} \\ $$ | ||
Commented by Her_Majesty last updated on 03/Aug/20 | ||
$${thanks}\:{but}\:{how}\:{you}\:{got}\:{it}? \\ $$ | ||
Commented by 1549442205PVT last updated on 03/Aug/20 | ||
$$\mathrm{t}^{\mathrm{4}} −\mathrm{8t}^{\mathrm{3}} −\mathrm{22t}^{\mathrm{2}} +\mathrm{272t}−\mathrm{419}=\mathrm{0} \\ $$$$\Leftrightarrow\left[\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{15}}\right)\boldsymbol{\mathrm{t}}+\mathrm{11}−\mathrm{6}\sqrt{\mathrm{15}}\:\right]× \\ $$$$\left[\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{15}}\right)\boldsymbol{\mathrm{t}}+\mathrm{11}+\mathrm{6}\sqrt{\mathrm{15}}\right]=\mathrm{0} \\ $$$$\Leftrightarrow\left[\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}\right)\boldsymbol{\mathrm{t}}−\left(\mathrm{9}+\mathrm{10}\sqrt{\mathrm{5}}\right)\right]× \\ $$$$\left[\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}\right)\boldsymbol{\mathrm{t}}−\left(\mathrm{9}−\mathrm{10}\sqrt{\mathrm{5}}\right)\right]=\mathrm{0} \\ $$ | ||
Answered by 1549442205PVT last updated on 03/Aug/20 | ||
$${t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{3}} −\mathrm{22}{t}^{\mathrm{2}} +\mathrm{272}{t}−\mathrm{419} \\ $$$$=\left(\mathrm{t}^{\mathrm{2}} +\mathrm{10}.\mathrm{374065851t}−\mathrm{19}.\mathrm{19728528}\right)× \\ $$$$\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}.\mathrm{374065851}+\mathrm{21}.\mathrm{82600076}\right) \\ $$$$\mathrm{two}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{be} \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{1}.\mathrm{602856312},\mathrm{x}_{\mathrm{2}} =−\mathrm{11}.\mathrm{97692216} \\ $$ | ||