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

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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.**