Question

Integrate t sec^2 (2t) dt

Answer 1

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 

Answer 2

Answer:

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

Where C is Constant of Proportionality.