Question Number 21595 by Isse last updated on 29/Sep/17
$$\mathrm{3sec}^{\mathrm{2}} \mathrm{3}{xtan}\mathrm{3}{x} \\ $$
Commented by mrW1 last updated on 29/Sep/17
$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{question}? \\ $$
Answered by $@ty@m last updated on 29/Sep/17
$${Let}\:\mathrm{tan3}{x}={u}\: \\ $$$$\Rightarrow\mathrm{3sec}^{\mathrm{2}} \:\mathrm{3}{xdx}={du}\: \\ $$$$\therefore\int\mathrm{3sec}^{\mathrm{2}} \mathrm{3}{xtan}\mathrm{3}{xdx} \\ $$$$=\int{udu} \\ $$$$=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{\mathrm{2}} \mathrm{3}{x}+\mathrm{C}\: \\ $$