MATH 181 Study Guide - Final Guide: Hypotenuse, Trigonometric Substitution, Partial Fraction Decomposition
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1 / 25: evaluate z x ln(x) dx. Letting u = ln(x) and dv = x dx yields du = 1. The integration by parts formula is x dx and v = 1. 2 x2 ln(x) z 1 x2 ln(x) x2 ln(x) x2 . 2 / 25: evaluate the integral z e2x cos(x) dx. Letting u = e2x and dv = cos(x) dx yields du = 2e2x dx and v = sin(x). Z u dv = uv z v du. Z e2x cos(x) dx = e2x sin(x) z sin(x) 2e2x dx. Letting u = e2x and dv = sin(x) dx yields du = 2e2x dx and v = cos(x). Z e2x cos(x) dx = e2x sin(x) 2z e2x sin(x) dx. Z e2x cos(x) dx = e2x sin(x) 2(cid:20) e2x cos(x) + 2z e2x cos(x) dx(cid:21) Z e2x cos(x) dx = e2x sin(x) + 2e2x cos(x) 4z e2x cos(x) dx.
