7.1 Integration By Parts/13

$\int t\sec ^{2}\left(2t\right)dt$

$u=t\qquad dv=\sec ^{2}\left(2t\right)\qquad \int \sec ^{2}\left(2t\right)dt$

$du=dt\qquad v={\frac {1}{2}}\tan \left(2t\right)$

$\int t\sec ^{2}\left(2t\right)dt={\frac {1}{2}}\tan \left(2t\right)-{\frac {1}{2}}\int \tan \left(2t\right)dt={\frac {1}{2}}\tan \left(2t\right)-{\frac {1}{4}}\int \tan \left(u\right)du={\frac {1}{2}}\tan \left(2t\right)-{\frac {1}{4}}\ln |\sec 2t|+c$

${\begin{aligned}&u=2t\\[2ex]&du=2dt\\[2ex]&{\frac {du}{2}}=dt\\[2ex]\end{aligned}}$