da sin z 12 COS 12 Now substitute and simplify to get a trig integral in terms of t The trig integral should no longer involve a square root Tz cos 1 dt 1 14 dt 1 144m da Part 2 Evaluating the Trig Integral From previous integration techniques we know that cos t dt Use this to evaluate the trig integral from Part 1 and then substitute back to get the final result in terms of a To substitute back in terms of a you would use sin t cos t The final result would then be sin t cos t 2