Use C for the constant

integral of t sec² 2t dt : Just use an integration by parts : 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

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