Question Number 107496 by mathdave last updated on 11/Aug/20
Answered by bemath last updated on 11/Aug/20
![∦BeMath∦ =∫ (1+tan^2 x)^2 sec^2 x dx =∫ (1+u^2 )^2 du [ u = tan x ] =∫ (1+2u^2 +u^4 ) du =u+(2/3)u^3 +(1/5)u^5 + C =tan x(1+(2/3)tan^2 x+(1/5)tan^4 x)+C](https://www.tinkutara.com/question/Q107497.png)
$$\:\:\:\nparallel\mathcal{B}{e}\mathcal{M}{ath}\nparallel \\ $$$$=\int\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} \:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}\: \\ $$$$=\int\:\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \:{du}\:\:\left[\:{u}\:=\:\mathrm{tan}\:{x}\:\right] \\ $$$$=\int\:\left(\mathrm{1}+\mathrm{2}{u}^{\mathrm{2}} +{u}^{\mathrm{4}} \right)\:{du}\: \\ $$$$={u}+\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{5}}{u}^{\mathrm{5}} \:+\:{C} \\ $$$$=\mathrm{tan}\:{x}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{2}} {x}+\frac{\mathrm{1}}{\mathrm{5}}\mathrm{tan}\:^{\mathrm{4}} {x}\right)+{C} \\ $$