MAT137Y1 Lecture : MAT 137Y 2007-08Winter Session, Self Generated Solutions to Problem Set 12.pdf
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MAT137Y1 Full Course Notes
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Mat 137y 2007-08 winter session, solutions to problem set 12. 1 (she 8. 1) (cid:90: let u = 3 2cos . Then: let u = x2, then du = 2xdx, so sin . 2: for parts (a) and (b), the results are fairly obvious by applying the appropriate substitution. Despite the fact that the answers are different, it turns out that they differ by a constant, since. 2 tan2 x +c1 = (sec2 x 1) +c1 = 2 sec2 x + so c1 and c2 differ by 1. 2 (she 8. 2: we integrate by parts and let u = x2 and dv = xe x2 dx. Then du = 2x dx and v = 1. 2: we multiply the expression and integrate by parts, giving us. 2 (cid:90) (2x + x2)2 dx = e x2 + (cid:90) x2e x2 1 e x2 +c. 4x (ln2)2 + (cid:90) 3 (cid:90) e3x cos2x dx = (cid:90) 4 2x (cid:21) (ln2)2 dx.
