MAT137Y1 Midterm: 2009 Test 3 solution
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MAT137Y1 Full Course Notes
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Mat137y 2009 2010 winter session, solutions to term test 3 (cid:90) du (cid:90) x dx, thereby giving us u 5/2 du = 2. 3 (lnx) 3/2 +c: evaluate the following integrals. dx x(lnx)5/2 . Then du = 1 (cid:90) (cid:90) 1 (8%) (i) (8%) (ii) arctanxdx. (cid:90) 1 arctanxdx = 0 (cid:90) dx (2 + x2)2 . (10%) (iii) 4 ln2. ln2 0 = (cid:20) xarctanx 1. We apply a trig substitution: let x = Then dx = (cid:90) (2 + 2tan2 )2 d = We draw a right triangle with x being the length of the opposite side and. 2 being the length of the adjacent side. By the pythagorean theorem the length of the hypoteneuse is. Hence, x2+2 and cos = x2 + 2. 8 arctan arctan (cid:19) (cid:19) (cid:18) x (cid:18) x . 4 x x2 + 2 (cid:18) x (cid:19) x2 + 2. +c (cid:90) 1 x + 2x2 x3 x(x2 + 1)2 (10%) (iv)
