One method: Expand both and in Taylor series; in powers of , and in powers of . We can save ourselves some effort if we note from the integral equation that ; and from repeatedly differentiating the integral equation, that (so the coefficient of the term in the expansion is 1), and that for (because we get the sine in the integrand for those terms). Thus, the Taylor series expansion for (about ) has the form .
The expansion for is .
Inserting these in the integral equation, we get
or after some expanding and rearrangement (and subtracting the term from both sides)
All of the integrals on the right side are readily evaluated. For example, retaining only terms through , we get
Equating coefficients for each power of on the left and right sides gives a system of equations for the coefficients Upon solving that system, we get . Repeating the calculation, this time including terms through , gives and leaves the lower-order coefficients essentially unchanged, evidence that the above approximation for is quite good for .