integrate t sec^2 (2t) dt

F = t ⇨ df = dt

dg = sec² 2t dt ⇨ g = (1/2) tan 2t

⇔

integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt

u = 2t ⇨ du = 2 dt

As integral of tan u = - ln (cos (u)), you get:

integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant

integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant

integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer