Question Number 38130 by gunawan last updated on 22/Jun/18
$$\mathrm{1}.\:\int\mathrm{tan}^{\mathrm{3}} \left(\mathrm{2}{x}\right)\mathrm{sec}^{\mathrm{5}} \left(\mathrm{2}{x}\right)\:{dx}\: \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{tan}^{\mathrm{5}} \left({x}\right)\mathrm{sec}^{\mathrm{6}} \left({x}\right)\:{dx}\: \\ $$$$\mathrm{3}.\:\int\mathrm{tan}^{\mathrm{6}} \left({ay}\right)\:{dy}\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
$$\left.\mathrm{1}\right)\int\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right){sin}\left(\mathrm{2}{x}\right)}{{cos}^{\mathrm{8}} \left(\mathrm{2}{x}\right)}{dx} \\ $$$${t}={cos}\mathrm{2}{x}\:\:\:{dt}=−\mathrm{2}{sin}\mathrm{2}{xdx} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{8}} }×{dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left\{\frac{{t}^{−\mathrm{7}} }{−\mathrm{7}}−\frac{{t}^{−\mathrm{5}} }{\left(−\mathrm{5}\right)}\right\}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{14}}×\frac{\mathrm{1}}{{cos}^{\mathrm{7}} \left(\mathrm{2}{x}\right)}−\frac{\mathrm{1}}{\mathrm{10}}×\frac{\mathrm{1}}{{cos}^{\mathrm{5}} \left(\mathrm{2}{x}\right)}+{c} \\ $$
Answered by Joel579 last updated on 22/Jun/18
$$\left(\mathrm{3}\right) \\ $$$${I}\:=\:\int\:\mathrm{tan}^{\mathrm{6}} \:\left({ay}\right)\:{dy}\:\:\:\:\left({u}\:=\:{ay}\:\:\rightarrow\:\:{du}\:=\:{a}\:{dy}\right) \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\:\int\:\left(\mathrm{sec}^{\mathrm{2}} \:{u}\:−\:\mathrm{1}\right)^{\mathrm{3}} \:{du} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\:\int\:\left(\mathrm{sec}^{\mathrm{6}} \:{u}\:−\:\mathrm{3sec}^{\mathrm{4}} \:{u}\:+\:\mathrm{3sec}^{\mathrm{2}} \:{u}\:−\:\mathrm{1}\right)\:{du} \\ $$
Commented by Joel579 last updated on 22/Jun/18
Answered by Joel579 last updated on 22/Jun/18
$$\left(\mathrm{3}\right) \\ $$$${I}\:=\:\int\:\mathrm{tan}^{\mathrm{6}} \:\left({ay}\right)\:{dy}\:\:\:\left({u}\:=\:{ay}\:\:\rightarrow\:\:{du}\:=\:{a}\:{dy}\right) \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\:\int\:\mathrm{tan}^{\mathrm{2}} \:{u}\:.\:\mathrm{tan}^{\mathrm{4}} \:{u}\:{du} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\:\int\:\left(\mathrm{sec}^{\mathrm{2}} \:{u}\:.\:\mathrm{tan}^{\mathrm{4}} \:{u}\:−\:\mathrm{tan}^{\mathrm{4}} \:{u}\right)\:{du} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\:\int\:\left[\mathrm{sec}^{\mathrm{2}} \:{u}\:.\:\mathrm{tan}^{\mathrm{4}} \:{u}\:−\:\left(\mathrm{sec}^{\mathrm{2}} \:{u}\:−\:\mathrm{1}\right)\mathrm{tan}^{\mathrm{2}} \:{u}\right]\:{du} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\:\int\:\left[\mathrm{sec}^{\mathrm{2}} \:{u}\:.\:\mathrm{tan}^{\mathrm{4}} \:{u}\:−\:\mathrm{sec}^{\mathrm{2}} \:{u}\:.\:\mathrm{tan}^{\mathrm{2}} \:{u}\right]\:{du}\:+\:\int\:\mathrm{tan}^{\mathrm{2}} \:{u}\:{du} \\ $$$${t}\:=\:\mathrm{tan}\:{u}\:\:\rightarrow\:\:{dt}\:=\:\mathrm{sec}^{\mathrm{2}} \:{u}\:{du} \\ $$$${I}\:=\:\frac{\mathrm{1}}{{a}}\:\int\:{t}^{\mathrm{4}} \:−\:{t}^{\mathrm{2}} \:{dt}\:+\:\mathrm{tan}\:{u}\:−\:{u} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\left(\frac{{t}^{\mathrm{5}} }{\mathrm{5}}\:−\:\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\:+\:\mathrm{tan}\:{u}\:−\:{u}\right)\:+\:{C} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{3}}} \frac{{sin}^{\mathrm{5}} {x}}{{cos}^{\mathrm{5}} {x}}×\frac{\mathrm{1}}{{cos}^{\mathrm{6}} {x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{3}}} \frac{{sinx}×\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} {dx}}{{cos}^{\mathrm{11}} {x}} \\ $$$$=\frac{\left(−\mathrm{1}\right)}{}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{3}}} \frac{\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} {x}+{cos}^{\mathrm{4}} {x}}{{cos}^{\mathrm{11}} {x}}×{d}\left({cosx}\right) \\ $$$$=\left(−\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{3}}} \left\{{cos}^{−\mathrm{11}} {x}−\mathrm{2}{cos}^{−\mathrm{9}} {x}+{cos}^{−\mathrm{7}} \left({x}\right)\right\}\left({dcosx}\right) \\ $$$$=\mid\frac{−\mathrm{1}}{−\mathrm{10}}×{cos}^{−\mathrm{10}} {x}+\mathrm{2}\frac{{cos}^{−\mathrm{8}} {x}}{−\mathrm{8}}−\frac{\mathrm{1}}{−\mathrm{6}}×{cos}^{−\mathrm{6}} {x}\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{3}}} \\ $$$$ \\ $$
Commented by gunawan last updated on 22/Jun/18
$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir} \\ $$