find the first few terms of S

Question

The integral: is equal to 3pi.

(b) Now, let's evaluate the same integral using a power series. First, find the power series for the function . Then, integrate it from 0 to 2, and call the result S.

S should be an infinite series.

The first few terms of S are below:

a0 = 12

a1 = -4

a2 = 12/5

a3 = -12/7

a4 = 4/3

Please find:

a5 =

a6 =

a7 =

a8 =

a9 =

Answer 1

24/(4 + x²) = 6/(1 + x²/4) = 6(1 - x²/4 + x⁴/16 - x⁶/64 + ...+ (-1)ⁿx²ⁿ/4ⁿ + ...

∫₀²24/(x² + 4)dx = 6(x - x³/12 + x⁵/80 - x⁷/448 + .. + (-1)ⁿx²ⁿ⁺¹/[4ⁿ(2n + 1)] + ...)₀² =

= 12 - 4 + 12/5 - 12/7 + 4/3 - ... + (- 1)ⁿ·12/(2n + 1)

From here a_n = (-1)ⁿ·12/(2n + 1), n = 0, 1, 2, ...

n= 5; a₅ = - 12/11;

n = 6; a₆ = 12/13

n= 7; a₇ = - 4/5;

n = 8; a₈ = 12/17;

n = 9; a₉ = - 12/19