let-f-x-0-pi-4-dt-x-tant-with-x-real-1-find-aexplicit-form-of-f-x-2-find-also-g-x-0-pi-4-dt-x-tant-2-3-give-f-n-x-at-form-of-integral-4-calculate-0-pi-4-dt-2

Question Number 66060 by mathmax by abdo last updated on 08/Aug/19
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{{x}+{tant}}\:\:{with}\:{x}\:{real} \\$$$$\left.\mathrm{1}\right)\:{find}\:{aexplicit}\:{form}\:{of}\:{f}\left({x}\right) \\$$$$\left.\mathrm{2}\right){find}\:{also}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\left({x}+{tant}\right)^{\mathrm{2}} } \\$$$$\left.\mathrm{3}\right){give}\:{f}^{\left({n}\right)} \left({x}\right){at}\:{form}\:{of}\:{integral} \\$$$$\left.\mathrm{4}\right){calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\mathrm{2}+{tant}}\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\left(\mathrm{2}+{tant}\right)^{\mathrm{2}} } \\$$
Commented by mathmax by abdo last updated on 10/Aug/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{dt}}{{x}+{tant}}\:\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give} \\$$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{{x}+\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({x}−{xu}^{\mathrm{2}} \:+\mathrm{2}{u}\right)}{du} \\$$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({xu}^{\mathrm{2}} −\mathrm{2}{u}−{x}\right)}{du}\:{let}\:{decompose} \\$$$${F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({xu}^{\mathrm{2}} −\mathrm{2}{u}−{x}\right)} \\$$$${xu}^{\mathrm{2}} −\mathrm{2}{u}−{x}=\mathrm{0}\:\rightarrow\Delta^{'} \:=\mathrm{1}+{x}^{\mathrm{2}} \:\Rightarrow{u}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\:\:{and}\:{u}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}} \\$$$$\left(\:{we}\:{suppose}\:{x}\neq\mathrm{0}\right)\:\Rightarrow{F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{{x}\left({u}−{u}_{\mathrm{1}} \right)\left({x}−{u}_{\mathrm{2}} \right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\$$$$=\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\$$$${a}\:={lim}_{{u}\rightarrow{u}_{\mathrm{1}} } \left({u}−{u}_{\mathrm{1}} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}}{{x}\left(\mathrm{2}\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\right)\left({u}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{u}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left({u}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)} \\$$$${b}\:={lim}_{{u}\rightarrow{u}_{\mathrm{2}} } \left({u}−{u}_{\mathrm{2}} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}}{{x}\left(−\mathrm{2}\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\right)\left({u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{u}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}}{−\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left({u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)} \\$$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:+{c}\:\Rightarrow{c}\:=−\left({a}+{b}\right) \\$$$${F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{{x}}\:=−\frac{{a}}{{u}_{\mathrm{1}} }−\frac{{b}}{{u}_{\mathrm{2}} }\:+{d}\:\Rightarrow{d}\:=\frac{\mathrm{1}}{{x}}+\frac{{a}}{{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}_{\mathrm{2}} }\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:{F}\left({u}\right){du}\:=\left[{aln}\mid{u}−{u}_{\mathrm{1}} \mid+{bln}\mid{u}−{u}_{\mathrm{2}} \mid\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:+\left[\frac{{c}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)+{darctan}\left({u}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\$$$$\left.={aln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{1}} }{\mathrm{1}−{u}_{\mathrm{1}} }\mid+{bln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{2}} }{\mathrm{1}−{u}_{\mathrm{2}} }\mid+\frac{{c}}{\mathrm{2}}{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right)+{d}\:{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right\} \\$$$$\Rightarrow{f}\left({x}\right)=\mathrm{2}{aln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{1}} }{\mathrm{1}−{u}_{\mathrm{1}} }\mid+\mathrm{2}{bln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{2}} }{\mathrm{1}−{u}_{\mathrm{2}} }\mid+{cln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)+\frac{{d}\pi}{\mathrm{4}} \\$$$${a}\:=\frac{\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\right)^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)} \\$$$$=\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{1}+{x}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)}\:=\frac{\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{2}{x}^{\mathrm{2}} \right)} \\$$$$=\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left({x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{1}\right)}\:\:{rest}\:{to}\:{simplify}\:{other}\:{coefficie}\mathrm{7}{ts}… \\$$$$\\$$
Commented by mathmax by abdo last updated on 10/Aug/19
$$\left.\mathrm{2}\right){we}\:{have}\:{f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\left({x}+{tant}\right)^{\mathrm{2}} }\:=−{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)=−{f}^{'} \left({x}\right) \\$$$${rest}\:{to}\:{calculate}\:{f}^{'} \left({x}\right) \\$$$$\left.\mathrm{3}\right){we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\left({x}+{tant}\right)}\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+{tant}\right)^{{n}+\mathrm{1}} }{dt} \\$$
Commented by mathmax by abdo last updated on 10/Aug/19
$$\left.\mathrm{4}\right)\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give}\: \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\left(\mathrm{2}+{tant}\right)}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}−\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}{u}\right)} \\$$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{u}−\mathrm{2}\right)}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} −{u}−\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)}{du} \\$$$${let}\:{decompose}\:{F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} −{u}−\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\$$$${u}^{\mathrm{2}} −{u}−\mathrm{1}=\mathrm{0}\rightarrow\Delta\:=\mathrm{1}+\mathrm{4}\:=\mathrm{5}\:\Rightarrow{u}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:{and}\:{u}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$$${F}\left({u}\right)=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\$$$${a}\:=\frac{{u}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{5}}\left({u}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}}{\:\sqrt{\mathrm{5}}\left(\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}\right)}\:=\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}}{\:\sqrt{\mathrm{5}}\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{4}\right)}\:=\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\left(\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}\right)} \\$$$$=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\left(\mathrm{5}+\sqrt{\mathrm{5}}\right)}\:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{5}\sqrt{\mathrm{5}}+\mathrm{5}}\:=\frac{\mathrm{1}}{\mathrm{5}} \\$$$${b}\:=\frac{{u}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}}{−\sqrt{\mathrm{5}}\left({u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}}{\left(−\sqrt{\mathrm{5}}\right)\left(\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}\right)}\:=\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}}{\left(−\sqrt{\mathrm{5}}\right)\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{4}\right)} \\$$$$=\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{5}}}{\left(−\sqrt{\mathrm{5}}\right)\left(\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}\right)}\:=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\left(−\sqrt{\mathrm{5}}\right)\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)}\:=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{5}}−\mathrm{5}}\:=\frac{\mathrm{1}}{\mathrm{5}} \\$$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)=\mathrm{0}\:={a}+{b}+{c}\:\Rightarrow{c}\:=−\frac{\mathrm{2}}{\mathrm{5}} \\$$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=−\frac{{a}}{{u}_{\mathrm{1}} }−\frac{{b}}{{u}_{\mathrm{2}} }\:+{d}\:\Rightarrow{d}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}{u}_{\mathrm{1}} }\:+\frac{\mathrm{1}}{\mathrm{5}{u}_{\mathrm{2}} } \\$$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}\left\{\frac{{u}_{\mathrm{1}} +{u}_{\mathrm{2}} }{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }\right\}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}\:\frac{\mathrm{1}}{−\mathrm{1}}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\:=\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow \\$$$${F}\left({u}\right)=\:\frac{\mathrm{1}}{\mathrm{5}\left({u}−{u}_{\mathrm{1}} \right)}\:+\frac{\mathrm{1}}{\mathrm{5}\left({u}−{u}_{\mathrm{2}} \right)}\:+\frac{\frac{−\mathrm{2}}{\mathrm{5}}{u}\:+\frac{\mathrm{4}}{\mathrm{5}}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\$$$$\int\:{F}\left({u}\right){du}\:=\frac{\mathrm{1}}{\mathrm{5}}{ln}\mid{u}−{u}_{\mathrm{1}} \mid+\frac{\mathrm{1}}{\mathrm{5}}{ln}\mid{u}−{u}_{\mathrm{2}} \mid−\frac{\mathrm{1}}{\mathrm{5}}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)+\frac{\mathrm{4}}{\mathrm{5}}\:{arctanu}\:+{c} \\$$$$\Rightarrow\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:{F}\left({u}\right){du}= \\$$$$\left[\frac{\mathrm{1}}{\mathrm{5}}{ln}\mid{u}−{u}_{\mathrm{1}} \mid+\frac{\mathrm{1}}{\mathrm{5}}{ln}\mid{u}−{u}_{\mathrm{2}} \mid−\frac{\mathrm{1}}{\mathrm{5}}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)+\frac{\mathrm{4}}{\mathrm{5}}{arctan}\left({u}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\$$$$…{rest}\:{to}\:{finish}\:{the}\:{calculus}… \\$$