Evaluate the Integral

Problem

(∫_0^1)(cos(x2)*d(x))

Solution

1. Identify the integral as a Fresnel integral of the first kind, which does not have an elementary antiderivative.

2. Represent the integrand cos(x2) using its Maclaurin series expansion:

cos(u)=(∑_n=0^∞)(((−1)n*u(2*n))/((2*n)!))

3. Substitute u=x2 into the series to find the expansion for cos(x2)

cos(x2)=(∑_n=0^∞)(((−1)n*(x2)(2*n))/((2*n)!))

4. Simplify the power of x in the series:

cos(x2)=(∑_n=0^∞)(((−1)n*x(4*n))/((2*n)!))

5. Integrate the series term-by-term with respect to x from 0 to 1

(∫_0^1)((∑_n=0^∞)(((−1)n*x(4*n))/((2*n)!))*d(x))=(∑_n=0^∞)(((−1)n)/((2*n)!))*(∫_0^1)(x(4*n)*d(x))

6. Apply the power rule for integration:

(∫_0^1)(x(4*n)*d(x))=[(x(4*n+1))/(4*n+1)]10=1/(4*n+1)

7. Combine the results into a single infinite series:

(∫_0^1)(cos(x2)*d(x))=(∑_n=0^∞)(((−1)n)/((2*n)!(4*n+1)))

8. Calculate the first few terms of the series to provide a numerical approximation:

n=0⇒1/(1⋅1)=1

n=1⇒(−1)/(2⋅5)=−0.1

n=2⇒1/(24⋅9)≈0.00463

n=3⇒(−1)/(720⋅13)≈−0.000107

9. Sum the terms to find the approximate value:

1−0.1+0.00463−0.000107≈0.90452

Final Answer

(∫_0^1)(cos(x2)*d(x))=(∑_n=0^∞)(((−1)n)/((2*n)!(4*n+1)))≈0.90452

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