Using the identities:

the integral reduces to:

Perform a substitution to get

The substitution would probably have given a quicker answer, but the first thing that came to mind was another trig sub:

Giving:

Noting that for all real and so for the bounds of this integral, we may simplify to get

Note: and can be evaluated by using an appropriate right angled triangle.

Expanding the square and rationalising the denominator gives

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