The discussion focuses on evaluating the indefinite integral ∫ sin(x²) dx using the Maclaurin series expansion. The Maclaurin series for sin(x) is given by the formula ∑ (-1)ⁿx²ⁿ⁺¹/(2n+1)!, and for sin(x²), it is represented as ∑ (-1)x⁴ⁿ⁺²/(2n+1)!. Participants concluded that to find the indefinite integral, one should integrate the Maclaurin series term by term and apply a summation sign in front of the result.
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KTiaam said:Homework Statement
Evaluate the indefinite integral as an infinite series ∫ sin(x2) dx
Homework Equations
The Macluarin series of sin x =
∞
Ʃ (-1)nx2n+1/(2n+1)!
n=0
The Macluarin series for sin(x2) =
∞
Ʃ (-1)x4n+2/(2n+1)!
n=0
The Attempt at a Solution
Do i evaluate the integral of sin(x2)
from 0 to 1? then add the first couple of numbers to get a number or find a pattern?