Question Number 62266 by aliesam last updated on 18/Jun/19 | ||
$$\int\frac{\mathrm{2}{sin}\left({x}\right)+\mathrm{3}{cos}\left({x}\right)}{\mathrm{3}{sin}\left({x}\right)+\mathrm{4}{cos}\left({x}\right)}{dx} \\ $$ | ||
Commented by maxmathsup by imad last updated on 19/Jun/19 | ||
$${let}\:{A}\:=\int\:\:\:\frac{\mathrm{2}{sinx}\:+\mathrm{3}{cosx}}{\mathrm{3}{sinx}\:+\mathrm{4}{cosx}}\:{dx}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}\:=\int\:\:\:\frac{\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{6}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{4}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int\:\:\frac{\mathrm{4}{t}+\mathrm{3}−\mathrm{3}{t}^{\mathrm{2}} }{\left(\mathrm{6}{t}+\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\frac{−\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{4}{t}+\mathrm{3}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(−\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{3}{t}+\mathrm{2}\right)}\:{dt}\:=\:\int\:\:\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{2}\right)}{dt}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{2}\right)} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}\:−\mathrm{2}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{9}−\mathrm{4}\left(−\mathrm{4}\right)\:=\mathrm{9}+\mathrm{16}\:=\mathrm{25}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{5}}{\mathrm{4}}\:=\mathrm{2}\:\:,\:{t}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{5}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}}{\left({t}−\mathrm{2}\right)\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\frac{\mathrm{2}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}\right)}{\left({t}−\mathrm{2}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{a}}{{t}−\mathrm{2}}\:+\frac{{b}}{\mathrm{2}{t}+\mathrm{1}}\:+\frac{{ct}+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow\mathrm{2}} \:\:\:\left({t}−\mathrm{2}\right){F}\left({t}\right)\:=\frac{\mathrm{2}\left(\mathrm{12}−\mathrm{8}−\mathrm{3}\right)}{\mathrm{5}.\mathrm{5}}\:=\frac{\mathrm{2}}{\mathrm{25}} \\ $$$${b}\:={lim}_{{t}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\left(\mathrm{2}{t}+\mathrm{1}\right){F}\left({t}\right)\:=\frac{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{4}}\:+\mathrm{2}−\mathrm{3}\right)}{−\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{5}}{\mathrm{4}}}\:=\frac{\mathrm{2}\left(\mathrm{3}+\mathrm{8}−\mathrm{12}\right)}{\mathrm{4}\left(−\frac{\mathrm{5}}{\mathrm{2}}\right).\frac{\mathrm{5}}{\mathrm{4}}}\:=\frac{\mathrm{4}}{\mathrm{25}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{2}}{\mathrm{25}\left({t}−\mathrm{2}\right)}\:+\frac{\mathrm{4}}{\mathrm{25}\left(\mathrm{2}{t}+\mathrm{1}\right)}\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${lim}_{{t}\rightarrow+\infty} \:{tF}\left({t}\right)\:=\mathrm{0}\:=\:{a}+\frac{{b}}{\mathrm{2}}\:+{c}\:\:\Rightarrow{c}\:=−{a}−\frac{{b}}{\mathrm{2}}\:=−\frac{\mathrm{2}}{\mathrm{25}}\:−\frac{\mathrm{2}}{\mathrm{25}}\:=−\frac{\mathrm{4}}{\mathrm{25}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{2}}{\mathrm{25}\left({t}−\mathrm{2}\right)}\:+\frac{\mathrm{4}}{\mathrm{25}\left(\mathrm{2}{t}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{4}}{\mathrm{25}}{t}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\frac{\mathrm{3}}{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{25}}\:+\frac{\mathrm{4}}{\mathrm{25}}\:+{d}\:=\frac{\mathrm{3}}{\mathrm{25}}\:+{d}\:\Rightarrow{d}\:=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{25}}\:=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{25}}\right)=\mathrm{3}.\frac{\mathrm{23}}{\mathrm{50}}\:=\frac{\mathrm{69}}{\mathrm{50}}\:\Rightarrow \\ $$$$\int\:{F}\left({t}\right){dt}\:=\:\frac{\mathrm{2}}{\mathrm{25}}\:\int\frac{{dt}}{{t}−\mathrm{2}}\:+\frac{\mathrm{4}}{\mathrm{25}}\:\int\:\frac{{dt}}{\mathrm{2}{t}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{50}}\:\int\:\frac{\mathrm{8}{t}−\mathrm{69}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{25}}{ln}\mid{t}−\mathrm{2}\mid+\frac{\mathrm{4}}{\mathrm{50}}{ln}\mid\mathrm{2}{t}+\mathrm{1}\mid\:−\frac{\mathrm{4}}{\mathrm{50}}\:\int\:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:+\frac{\mathrm{69}}{\mathrm{50}}\:\int\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{25}}{ln}\mid{t}−\mathrm{2}\mid+\frac{\mathrm{4}}{\mathrm{50}}{ln}\mid\mathrm{2}{t}+\mathrm{1}\mid−\frac{\mathrm{4}}{\mathrm{50}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:+\frac{\mathrm{69}}{\mathrm{50}}\:{arctan}\left({t}\right)+{c}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{2}}{\mathrm{25}}{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}\mid+\frac{\mathrm{4}}{\mathrm{50}}{ln}\mid\mathrm{2}{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}\mid−\frac{\mathrm{4}}{\mathrm{50}}{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\:+\frac{\mathrm{69}}{\mathrm{100}}\:{x}\:+{c}\:. \\ $$$$ \\ $$$$ \\ $$ | ||
Answered by tanmay last updated on 18/Jun/19 | ||
$${N}_{{r}} ={a}\left({D}_{{r}} \right)+{b}\left(\frac{{dD}_{{r}} }{{dx}}\right) \\ $$$$\int\frac{{a}\left({D}_{{r}} \right)+{b}\left(\frac{{dD}_{{r}} }{{dx}}\right)}{{D}_{{r}} }{dx} \\ $$$${a}\int{dx}+{b}\int\frac{{d}\left({D}_{{r}} \right)}{{D}_{{r}} } \\ $$$${ax}+{bln}\left({D}_{{r}} \right)+{c} \\ $$$${ax}+{bln}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right)+{c} \\ $$$${now}\:{finding}\:{value}\:{of}\:{a}\:{and}\:{b} \\ $$$$\mathrm{2}{sinx}+\mathrm{3}{cosx}={a}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right)+{b}\frac{{d}}{{dx}}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right) \\ $$$$\mathrm{2}{sinx}+\mathrm{3}{cosx}={a}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right)+{b}\left(\mathrm{3}{cosx}−\mathrm{4}{sinx}\right) \\ $$$$\mathrm{2}{sinx}+\mathrm{3}{cosx}=\left(\mathrm{3}{a}−\mathrm{4}{b}\right){sinx}+\left(\mathrm{4}{a}+\mathrm{3}{b}\right){cosx} \\ $$$$\mathrm{3}{a}−\mathrm{4}{b}=\mathrm{2}\:\:×\mathrm{3} \\ $$$$\mathrm{4}{a}+\mathrm{3}{b}=\mathrm{3}\:\:\:×\mathrm{4} \\ $$$$\mathrm{9}{a}−\mathrm{12}{b}=\mathrm{6} \\ $$$$\mathrm{16}{a}+\mathrm{12}{b}=\mathrm{12} \\ $$$$\mathrm{25}{a}=\mathrm{18}\rightarrow{a}=\frac{\mathrm{18}}{\mathrm{25}} \\ $$$$\mathrm{3}{b}=\mathrm{3}−\mathrm{4}{a} \\ $$$$\mathrm{3}{b}=\mathrm{3}−\frac{\mathrm{72}}{\mathrm{25}}\rightarrow{b}=\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\: \\ $$$$\frac{\mathrm{18}}{\mathrm{25}}\boldsymbol{{x}}+\frac{\mathrm{1}}{\mathrm{25}}\boldsymbol{{ln}}\left(\mathrm{3}\boldsymbol{{sinx}}+\mathrm{4}\boldsymbol{{cosx}}\right)+\boldsymbol{{c}} \\ $$ | ||
Commented by aliesam last updated on 18/Jun/19 | ||
$${thank}\:{you}\:{sir}\: \\ $$ | ||
Commented by tanmay last updated on 18/Jun/19 | ||
$${most}\:{welcome}\:{sir} \\ $$ | ||