Integration Questions

Question Number 35949 by ajfour last updated on 26/May/18

$$\int_{\mathrm{0}} ^{\:\:\alpha} \frac{\mathrm{tan}\:\theta}{\sqrt{{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta−{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}}\:{d}\theta\:=\:? \\$$

Commented by abdo mathsup 649 cc last updated on 26/May/18

$${let}\:{put}\:{I}_{\alpha} \:=\:\int_{\mathrm{0}} ^{\alpha} \:\:\:\frac{{tan}\theta}{\sqrt{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}}{d}\theta\: \\$$$${we}\:{hsve}\:{the}\:{formula}\:\:\mathrm{1}+{tan}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}\:\Rightarrow \\$$$${cos}^{\mathrm{2}} \theta\:=\:\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\:{and}\:\:{sin}^{\mathrm{2}} \theta\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\Rightarrow \\$$$${a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\:−{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\$$$$=\:\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:−{b}^{\mathrm{2}} \:\:\frac{{tan}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:=\:\frac{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} \:{tan}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\Rightarrow \\$$$${I}_{\alpha} \:=\:\int_{\mathrm{0}} ^{\alpha} \:\frac{{tan}\theta\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}{\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta}}\:{d}\theta\:\:{changement}\:{tan}\theta\:={t} \\$$$${give}\:{I}_{\alpha} \:=\:\:\int_{\mathrm{0}} ^{{tan}\alpha} \:\:\:\:\frac{{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} }}\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$$=\:\int_{\mathrm{0}} ^{{tan}\left(\alpha\right)} \:\:\:\:\frac{{t}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\:\frac{{dt}}{\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} }}\:\:{by}\:{parts} \\$$$${u}^{'} \:=\:\frac{{t}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:{and}\:\:{v}\:=\:\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} }}\:\: \\$$$${I}_{\alpha} \:\:=\:\left[\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} }}\right]_{\mathrm{0}} ^{{tan}\left(\alpha\right)} −\int_{\mathrm{0}} ^{{tan}\left(\alpha\right)} \:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}{b}^{\mathrm{2}} {t}\right)\left({a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} {dt} \\$$$$=\frac{\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}}{\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} \:{tan}^{\mathrm{2}} \alpha}}\:−\:\frac{\mathrm{1}}{\mid{a}\mid}\:\:−{b}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{{tan}\left(\alpha\right)} \:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\left(\:{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} {dt} \\$$$${let}\:{J}\:=\:\int_{\mathrm{0}} ^{{tan}\left(\alpha\right)} \:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\left({a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} {dt} \\$$$$={c}\:\int_{\mathrm{0}} ^{{tan}\left(\alpha\right)} \:\:\:\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\left({a}^{\mathrm{2}} \:\:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} {t}^{\mathrm{2}} }}{dt}\:\:...{be}\:{continued}\:... \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18

$$\int_{\mathrm{0}} ^{\alpha} \frac{{tan}\theta.{sec}\theta}{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta}}{d}\theta \\$$$$\int_{\mathrm{0}} ^{\alpha} \frac{{tan}\theta{sec}\theta}{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {sec}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} }}{d}\theta \\$$$$\frac{\mathrm{1}}{{b}}\int\frac{{d}\left({sec}\theta\right)}{\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−{sec}^{\mathrm{2}} \theta}} \\$$$${now}\:{use}\:{formula}\:\int\frac{{dx}}{\sqrt{{A}^{\mathrm{2}} −{X}^{\mathrm{2}} }} \\$$$$\frac{\mathrm{1}}{{b}}×\mid{sin}^{−\mathrm{1}} \left(\frac{{sec}\theta}{\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}}\right)\mid_{\mathrm{0}} ^{\alpha} \\$$$$\frac{\mathrm{1}}{{b}}×\left\{{sin}^{−\mathrm{1}} \left(\frac{{sec}\alpha}{\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}}\right)−{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }}\right)\right\} \\$$

Commented by ajfour last updated on 26/May/18

$${Thanks}. \\$$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18

$${its}\:{ok}... \\$$

Commented by ajfour last updated on 26/May/18

$${i}\:{applied}\:{your}\:{technique}\:{again} \\$$$${and}\:{corrected}\:{my}\:{answer}\:{to} \\$$$${Q}.\mathrm{35940}\:.\:{Thanks}\:{again}. \\$$