Evaluate the indefinite integral
integral sec(t/2) dt=
a)ln |sec t +tan t| +C
b)ln |sec (t/2) +tan (t/2)| +C
c)2tan^2 (t/2)+C
d)2ln cos(t/2) +C
e)2ln |sec (t/2)+tan (t/2)| +C
integral sec(t/2) dt =
2*integral sec(t/2) d(t/2)=
2*ln[sec (t/2) + tan (t/2)] +C
Thanks
To evaluate the indefinite integral ∫ sec(t/2) dt, we can use integration techniques, specifically substitution.
Let's start by making a substitution. We'll let u = t/2, therefore du = (1/2) dt.
Now we can rewrite the integral in terms of u:
∫ sec(t/2) dt = ∫ sec(u) (2 du)
Using the identity sec(u) = 1/cos(u), we can rewrite the integral further:
∫ sec(t/2) dt = ∫ (1/cos(u)) (2 du)
Now the integral can be simplified:
2 ∫ (1/cos(u)) du = 2 ∫ sec(u) du
The integral of sec(u) can be found using the natural logarithm function. The integral of sec(u) is ln |sec(u) + tan(u)| + C.
Therefore, substituting back, we have:
∫ sec(t/2) dt = 2 ∫ sec(u) du = 2 ln |sec(u) + tan(u)| + C
Substituting u = t/2 back into the equation, we get:
∫ sec(t/2) dt = 2 ln |sec(t/2) + tan(t/2)| + C
So, the correct option is (e) 2 ln |sec(t/2) + tan(t/2)| + C.