Question

Integrate t sec^2 (2t) dt

Answer 1

`\int {t* \sec^2(2t)} \, dt \\\\ \text{Integrating by parts}\\\\=(t* \tan 2t)/(2)-\int{((d)/(dt)(t) * \int{{\ sec^(2)(2t)}}) dt}\\\\=(t* \tan 2t)/(2)-\int{(\tan 2t)/(2)} dt\\\\=(t* \tan 2t)/(2)-(\log sec 2t)/(4)+C`

Where C is Constant of Proportionality.

Answer 2

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