APPM 1360 Final: appm1360fall2015examfinal_sol
Document Summary
Fall 2015: (a)(10 pts) evaluate the integral z. 4dx x2 x2 4 using trigonometric substitution. (b)(10 pts) solve for y: dy dx y x2 + 5x + 6 (for this part, assume that x > 0. ) Solution: (a)(10 pts) let x = 2 sec with 0 < < , then x2 4 = 2 tan , and dx = 2 sec tan . Note we could also use x = 2 csc( ) then z 8 csc cot d . 8 csc2 cot (b)(10 pts) separating variables yields dy y fractions on the right hand side, we have x2 + 5x + 6. =z sin( )d = cos( ) + c = Note that z dy/y = ln|y| + c1 and using partial x + 3(cid:19) + c2 x + 3(cid:19) dx = ln(x + 2) ln(x + 3) + c2 = ln(cid:18) x + 2 x. =z so dx (x + 2)(x + 3)
