Integration Questions

Question Number 123552 by Lordose last updated on 26/Nov/20

$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{n}} \left(\mathrm{x}\right)\mathrm{dx} \\$$

Answered by TANMAY PANACEA last updated on 26/Nov/20

$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{n}−\mathrm{2}} {x}\left({sec}^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{n}−\mathrm{2}} {x}×{d}\left({tanx}\right)−{I}_{{n}−\mathrm{2}} \\$$ $$=\mid\frac{{tan}^{{n}−\mathrm{1}} {x}}{{n}−\mathrm{1}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −{I}_{{n}−\mathrm{2}} \\$$ $$=\frac{\mathrm{1}}{{n}−\mathrm{1}}−{I}_{{n}−\mathrm{2}} \\$$ $${I}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}−{I}_{{n}−\mathrm{2}} \\$$

Commented bypeter frank last updated on 26/Nov/20

$$\mathrm{thank}\:\mathrm{you} \\$$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

$${If}\:{it}\:{was}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}^{{n}} {x}\:{dx} \\$$ $${then}\:{we}\:{can}\:{get}\:{a}\:{closed}\:{form} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {x}\:{cos}^{−{n}} {x}\:{dx} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{t}\right)^{\frac{−{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} {dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}^{\mathrm{2}} {x}={t} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{n}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}\right)}=\frac{\pi}{\mathrm{2}{sin}\left(\frac{{n}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{p}} {x}\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{p}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{p}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}+{p}} {dt}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{\mathrm{2}{n}+{p}+\mathrm{1}} \\$$

Answered by mnjuly1970 last updated on 26/Nov/20

$$\Omega\overset{{tan}\left({x}\right)={y}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{y}^{{n}} {dy}}{\mathrm{1}+{y}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{y}^{{n}} −{y}^{{n}+\mathrm{2}} }{\mathrm{1}−{y}^{\mathrm{4}} }{dy} \\$$ $$\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{t}^{\frac{{n}−\mathrm{3}}{\mathrm{4}}} \:−{t}^{\frac{{n}−\mathrm{1}}{\mathrm{4}}} }{\mathrm{1}−{t}}{dt} \\$$ $$\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{H}_{\frac{{n}−\mathrm{1}}{\mathrm{4}}} −\mathrm{H}_{\frac{{n}−\mathrm{3}}{\mathrm{4}}} \right) \\$$ $$\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{{n}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right)\:\checkmark \\$$ $$\\$$ $$\\$$ $$\: \\$$ $$\:\: \\$$ $$\: \\$$

Commented byDwaipayan Shikari last updated on 26/Nov/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{\mathrm{4}} {x}\:{dx}=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)+\frac{\mathrm{4}}{\mathrm{3}}−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{4}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\pi{cot}\left(\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{8}}{\mathrm{3}}\right)=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{3}} \\$$

Commented byLordose last updated on 02/Dec/20

$$\mathrm{Exactly} \\$$