MATH128 Lecture Notes - Lecture 2: Trigonometric Substitution

• School: University of Waterloo
• Department: Mathematics
• Course: MATH128
• Professor: Robert Garbary

Document Summary

Problem solve the following integral using a trigonometric substitution: dx. Solution let x = 4 sin with . Z 16 x2 x2 dx = z p16 42 sin2 . = cot + c. We want the solution in terms of x so we draw the triangle. Therefore our nal solution is cot = Problem 47 make a substitution to express the integrand as a rational function and then evaluate the integral: Z e2x e2x + 3ex + 2 dx. Z u du = z u2 + 3u + 2 (u + 2)(u + 1) u du. Simplifying we get and and hence u (u + 2)(u + 1) B u + 1 u = au + a + bu + 2b. A + b = 1 and 2b + a = 0. 2b + b = 1 = b = 1. 2 ln|u + 2| ln|u + 1| + c. 2 ln|ex + 2| ln|ex + 1| + c.