Integration Questions

Question Number 34992 by MJS last updated on 14/May/18

$$\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\mathrm{cos}\left({x}\right)}{\mathrm{1}+\mathrm{2sin}\left(\mathrm{2}{x}\right)}{dx} \\$$

Commented by math khazana by abdo last updated on 14/May/18

$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{cosx}}{\mathrm{1}+\mathrm{2}{sin}\left(\mathrm{2}{x}\right)}{dx} \\$$$${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{cosx}}{\mathrm{1}+\mathrm{4}{cosx}\:{sinx}}{dx}\:\:.{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\$$$${give}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{4}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$$=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\:\:\mathrm{1}+\frac{\mathrm{8}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)}{dt} \\$$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:\:+\mathrm{8}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt} \\$$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\:+\mathrm{8}{t}\:−\mathrm{8}{t}^{\mathrm{3}} }{dt} \\$$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:−\mathrm{8}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{8}{t}\:+\mathrm{1}}{dt}\:\:{let}\:{drcompose} \\$$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:−\mathrm{8}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{8}{t}\:+\mathrm{1}} \\$$$${the}\:{roots}\:{of}\:{p}\left({t}\right)={t}^{\mathrm{4}} \:−\mathrm{8}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{8}{t}\:+\mathrm{1}\:{are} \\$$$${t}_{\mathrm{1}} \:\sim\mathrm{7},\mathrm{59}\:\:\:\:\:\:{t}_{\mathrm{2}} \sim\mathrm{1},\mathrm{3}\:\:\:\:\:{t}_{\mathrm{3}} \sim−\mathrm{0},\mathrm{76}\:\:\:{t}_{\mathrm{4}} \sim−\mathrm{0},\mathrm{13}\:\Rightarrow \\$$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}\:−{t}_{\mathrm{1}} \right)\left({t}\:−{t}_{\mathrm{2}} \right)\left({t}\:−{t}_{\mathrm{3}} \right)\left({t}\:−{t}_{\mathrm{4}} \right)} \\$$$$=\:\frac{{a}}{{t}−{t}_{\mathrm{1}} }\:+\frac{{b}}{{t}−{t}_{\mathrm{2}} }\:+\frac{{c}}{{t}−{t}_{\mathrm{3}} }\:+\frac{{d}}{{t}\:−{t}_{\mathrm{4}} } \\$$$${a}={lim}_{{t}\rightarrow{t}_{\mathrm{1}} } \left({t}−{t}_{\mathrm{1}} \right){F}\left({t}\right)=\:\frac{\mathrm{1}−{t}_{\mathrm{1}} ^{\mathrm{2}} }{\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)\left({t}_{\mathrm{1}} \:−{t}_{\mathrm{3}} \right)\left({t}_{\mathrm{1}} \:−{t}_{\mathrm{4}} \right)} \\$$$$=\xi\left({t}_{\mathrm{1}} \right)\:\left(\mathrm{1}−{t}_{\mathrm{1}} ^{\mathrm{2}} \right)\:\:{with}\:\xi\left({t}_{\mathrm{1}} \right)=\:\frac{\mathrm{1}}{\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{3}} \right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{4}} \right)} \\$$$${b}\:=\xi\left({t}_{\mathrm{2}} \right)\left(\mathrm{1}−{t}_{\mathrm{2}} ^{\mathrm{2}} \right)\:{with}\:\xi\left({t}_{\mathrm{2}} \right)=\:\frac{\mathrm{1}}{\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{3}} \right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{4}} \right)} \\$$$${c}\:=\xi\left({t}_{\mathrm{3}} \right)\left(\mathrm{1}−{t}_{\mathrm{3}} ^{\mathrm{2}} \right) \\$$$${d}=\xi\left({t}_{\mathrm{4}} \right)\left(\mathrm{1}−{t}_{\mathrm{4}} ^{\mathrm{2}} \right)\:\Rightarrow \\$$$${I}\:\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} {F}\left({t}\right){dt} \\$$$$\left.\:=\:\mathrm{2}\:\left[\:{aln}\mid{t}−{t}_{\mathrm{1}} \mid\:+{b}\:{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+{c}\:{ln}\mid{t}−{t}_{\mathrm{3}} \right)\:+{d}\:{ln}\mid{t}−{t}_{\mathrm{4}} \mid\right]_{\mathrm{0}} ^{+\infty} \\$$$$=\mathrm{2}\left[\:{ln}\mid\frac{\left({t}−{t}_{\mathrm{1}} \right){a}}{\left({t}\:−{t}_{\mathrm{2}} \right)^{{b}} }\:\mid\:+\:{ln}\mid\:\frac{\left({t}−{t}_{\mathrm{3}} \right)^{{c}} }{\left({t}−{t}_{\mathrm{4}} \right)^{{d}} }\mid\right]_{\mathrm{0}} ^{+\infty} \\$$$$=\mathrm{2}\left\{−{ln}\mid\frac{\left(−{t}_{\mathrm{1}} \right)^{{a}} }{\left(−{t}_{\mathrm{2}} \right)^{{b}} }\:\mid−{ln}\mid\frac{\left(−{t}_{\mathrm{3}} \right)^{{c}} }{\left(−{t}_{\mathrm{4}} \right)^{{d}} }\mid\right\} \\$$$$=\mathrm{2}\left\{\:{bln}\mid{t}_{\mathrm{2}} \mid\:−{aln}\mid{t}_{\mathrm{1}} \mid\:\:+{dln}\mid{t}_{\mathrm{4}} \mid\:−{c}\:{ln}\mid{t}_{\mathrm{3}} \mid\right\}... \\$$$${the}\:{approximate}\:{value}\:{of}\:{I}\:{isdetermined}... \\$$$$\\$$

Answered by MJS last updated on 14/May/18

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\mathrm{cos}\left({x}\right)}{\mathrm{1}+\mathrm{2sin}\left(\mathrm{2}{x}\right)}{dx}=\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{sin}\left({x}\right)}{\mathrm{2sin}\left(\mathrm{2}{x}\right)−\mathrm{1}}{dx}= \\$$$$=\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{sin}\left({x}\right)}{\mathrm{4sin}\left({x}\right)\mathrm{cos}\left({x}\right)−\mathrm{1}}{dx}= \\$$$$\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{{t}=\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}}\\{\mathrm{sin}\left({x}\right)=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}};\:\mathrm{cos}\left({x}\right)=−\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}}\end{bmatrix} \\$$$$=−\mathrm{4}\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{t}}{{t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{1}}{dt}= \\$$$$\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{{t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{1}=\mathrm{0}}\\{{t}_{\mathrm{1}} =−\mathrm{2}−\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}\\{{t}_{\mathrm{2}} =−\mathrm{2}−\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}\\{{t}_{\mathrm{3}} =−\mathrm{2}+\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}\\{{t}_{\mathrm{4}} =−\mathrm{2}+\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}\end{bmatrix} \\$$$$=−\mathrm{4}\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\frac{\mathcal{A}}{{t}−{t}_{\mathrm{1}} }+\frac{\mathcal{B}}{{t}−{t}_{\mathrm{2}} }+\frac{\mathcal{C}}{{t}−{t}_{\mathrm{3}} }+\frac{\mathcal{D}}{{t}−{t}_{\mathrm{4}} }\right){dt}= \\$$$$=−\mathrm{4}\left[\mathcal{A}\mathrm{ln}\mid{t}−{t}_{\mathrm{1}} \mid+\mathcal{B}\mathrm{ln}\mid{t}−{t}_{\mathrm{2}} \mid+\mathcal{C}\mathrm{ln}\mid{t}−{t}_{\mathrm{3}} \mid+\mathcal{D}\mathrm{ln}\mid{t}−{t}_{\mathrm{4}} \mid\right]_{−\mathrm{1}} ^{\mathrm{1}} = \\$$$$=−\mathrm{4}\left[\mathcal{A}\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{1}} \mid+\mathcal{B}\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{2}} \mid+\mathcal{C}\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{3}} \mid+\mathcal{D}\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{4}} \mid\overset{\frac{\pi}{\mathrm{2}}} {\right]}_{−\frac{\pi}{\mathrm{2}}} = \\$$$$\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\mathcal{A}=\frac{{t}_{\mathrm{1}} }{\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{3}} \right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{4}} \right)}=\frac{−\sqrt{\mathrm{6}}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{96}}}\\{\mathcal{B}=\frac{{t}_{\mathrm{2}} }{\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{3}} \right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{4}} \right)}=\frac{−\sqrt{\mathrm{6}}−\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{96}}}\\{\mathcal{C}=\frac{{t}_{\mathrm{3}} }{\left({t}_{\mathrm{3}} −{t}_{\mathrm{1}} \right)\left({t}_{\mathrm{3}} −{t}_{\mathrm{2}} \right)\left({t}_{\mathrm{3}} −{t}_{\mathrm{4}} \right)}=\frac{\sqrt{\mathrm{6}}−\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{96}}}\\{\mathcal{D}=\frac{{t}_{\mathrm{4}} }{\left({t}_{\mathrm{4}} −{t}_{\mathrm{1}} \right)\left({t}_{\mathrm{4}} −{t}_{\mathrm{2}} \right)\left({t}_{\mathrm{4}} −{t}_{\mathrm{3}} \right)}=\frac{\sqrt{\mathrm{6}}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{96}}}\end{bmatrix} \\$$$$=−\frac{\mathrm{1}}{\mathrm{24}}\left[\left(−\sqrt{\mathrm{6}}+\mathrm{3}\sqrt{\mathrm{2}}\right)\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}+\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\mid+\left(−\sqrt{\mathrm{6}}−\mathrm{3}\sqrt{\mathrm{2}}\right)\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}+\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\mid+\left(\sqrt{\mathrm{6}}−\mathrm{3}\sqrt{\mathrm{2}}\right)\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}−\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\mid+\left(\sqrt{\mathrm{6}}+\mathrm{3}\sqrt{\mathrm{2}}\right)\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}−\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\mid\overset{\frac{\pi}{\mathrm{2}}} {\right]}_{−\frac{\pi}{\mathrm{2}}} \approx \\$$$$\approx\mathrm{1}.\mathrm{091166} \\$$

Commented by MJS last updated on 14/May/18

$$\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{get}\:\mathrm{this}\:\mathrm{out}\:\mathrm{of}\:\mathrm{my}\:\mathrm{head}... \\$$$$\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{solution},\:\mathrm{please}\:\mathrm{check}\:\mathrm{for} \\$$$$\mathrm{mistakes}\:\mathrm{of}\:\mathrm{method}. \\$$$$\mathrm{I}\:\mathrm{can}\:\mathrm{guarantee}\:\mathrm{the}\:\mathrm{correctness}\:\mathrm{of}\:\mathrm{the}\:\mathrm{zeros} \\$$$${t}_{\mathrm{1}} ,\:{t}_{\mathrm{2}} ,\:{t}_{\mathrm{3}} ,\:{t}_{\mathrm{4}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{coefficients}\:\mathcal{A},\:\mathcal{B},\:\mathcal{C},\:\mathcal{D}. \\$$

Commented by ajfour last updated on 14/May/18

$${for}\:{x}=\frac{\pi}{\mathrm{12}}\:,\:\:\frac{\mathrm{sin}\:{x}}{\mathrm{2sin}\:\mathrm{2}{x}−\mathrm{1}}\:{is}\:{not}\:{defined}\:. \\$$

Commented by MJS last updated on 14/May/18

$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{Cauchy}\:\mathrm{principal}\:\mathrm{value}: \\$$$$\int\frac{{p}}{{x}−{q}}{dx}={p}\mathrm{ln}\mid{x}−{q}\mid \\$$$${a}<{q}<{b};\:{h}>\mathrm{0} \\$$$$\underset{{a}} {\overset{{b}} {\int}}\frac{{p}}{{x}−{q}}{dx}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\underset{{a}} {\overset{{q}−{h}} {\int}}\frac{{p}}{{x}−{q}}{dx}+\underset{{q}+{h}} {\overset{{b}} {\int}}\frac{{p}}{{x}−{q}}{dx}\right)= \\$$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\left[{p}\mathrm{ln}\mid{x}−{q}\mid\right]_{{a}} ^{{q}−{h}} +\left[{p}\mathrm{ln}\mid{x}−{q}\mid\right]_{{q}+{h}} ^{{b}} \right)= \\$$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({p}\mathrm{ln}\mid{h}\mid−{p}\mathrm{ln}\mid{a}−{q}\mid+{p}\mathrm{ln}\mid{b}−{q}\mid−{p}\mathrm{ln}\mid{h}\mid\right)= \\$$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({p}\mathrm{ln}\mid{b}−{q}\mid−{p}\mathrm{ln}\mid{a}−{q}\mid\right)= \\$$$$={p}\mathrm{ln}\mid{b}−{q}\mid−{p}\mathrm{ln}\mid{a}−{q}\mid=\underset{{a}} {\overset{{b}} {\int}}\frac{{p}}{{x}−{q}}{dx} \\$$$${q}.{e}.{d}. \\$$

Commented by ajfour last updated on 14/May/18

$${not}\:{as}\:{sure},\:{you}\:{know}\:{better}. \\$$