Given an=integral of cosnx/(1+cosx), x=0 to x=pi/2 calculate a0 and a1.

To detemrine a0, we'll substitute n by
0:

a0 = Int cos0*x dx/(1+cosx)

But cos
0 = 1

a0 = Int dx/(1+cosx)

We'll apply
the half angle identity:

2[cos (x/2)]^2 = 1 + cos
x

We'll re-write a0:

a0 = Int dx/2[cos
(x/2)]^2

We notice that [tan (x/2)]' = 1/2[cos
(x/2)]^2

a0 = Int [tan (x/2)]' dx = tan
(x/2)

Now, we'll apply Leibniz Newton, to evaluate
a0:

a0 = F(pi/2) - F(0)

F(pi/2) = tan
(pi/4) = 1

F(0) = tan 0 = 0

a0 = tan
(pi/4) - tan 0 = 1

a0 = 1

Now, we'll
determine a1, substituting n by 1:

a1 = Int cos x dx/(1 + cos
x)

a1 = Int dx - Int dx/(1 + cos
x)

We'll recognize in the second term from the right side, the
original form of a0.

a1 = Int dx -
a0

Since a0 = 1, we'll get:

a1 = Int dx
- 1

Int dx = x

Now, we'll apply Leibniz
Newton, to evaluate Int dx:

Int dx = F(pi/2) - F(0) =
pi/2

a1 = pi/2 - 1

The
requested terms are: a0 = 1 and a1 = pi/2 - 1.