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
Can you possibly take a clearer picture? Maybe I can help you with that.
Answer:
Between 0 and 1
Step-by-step explanation:
The given expresion is

Recall that:

We apply this property of exponents to get;

We rewrite to obtain a positive index as:

Therefore the value of the expression is between 0 and 1.
A protractor can be used!