This is simply an alternate form of an integral that we’ve solved already. We merely need to apply a couple of little tricks to condition the integrand and then substitute in the solution. Begin by completing the square under the radical,

(Eq’n 1)

Make the substitution y=x-a or y=a-x. Given that a>x, we must have the integral as a positive number, so for the differential we take y=x-a so that dy=dx. Under the radical which version of y we use makes no difference in the integral, but because we still want a positive number later we choose y=a-x. We thus have

(Eq’n 2)

Example Application:

Component of Another Integral

See .

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