MATA37H3 Lecture Notes - Lecture 9: Improper Integral
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Mata37 - lecture 9 - integration by parts and improper integrals. Integration by parts continued: example 1: evaluate ex sin(x)dx (cid:90) (cid:90) Choose u1 = sin(x), dv1 = exdx, du1 = cos(x)dx, v1 = ex. Choose u2 = cos(x), dv2 = exdx, du2 = sin(x), v2 = ex. = sin(x)ex (cos(x)ex ex( sin(x)dx)) ex cos(x)dx (cid:90) (cid:90) ex sin(x)dx. I = sin(x)ex cos(x)ex i. 2i = sin(x)ex cos(x)ex + c. 8 x 2 csc(x)dx = (cid:90) . 2 (type i) dx; x = 0 [ 1, 1] (type ii) arctan(x)dx (type i: e. g. 2 (cid:90) 3 (cid:90) (cid:90) 1 (cid:90) 0. 1 (3x + 1)2 dx (type i) (cid:90) a. 1 (3x + 1)2 dx by de nition of type. Let u = 3x + 1, du = 3dx, du. X = 1 = u = 3(1) + 1 = 4. X = a = u = 3a + 1. Our improper integral converges to 1.
