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Unformatted text preview: CALCULUS 133: TECHNIQUES OF INTEGRATION The purpose of these techniques is the following: you are given the problem of finding an antiderivative of a complicated function, and these techniques allow you to reduce it to finding an antiderivative of a simpler function. Eventually you reduce the problem to finding an antiderivative of a “standard function”, for instance sin or cos or e x . A list of standard functions, for which we know the antiderivatives is given on page 383. I expect you to know 1,2,3,5,6,11, 13 and 14. 1. Integration by substitution This technique involves introducing a new variable u for x , which will hopefully simplify the integral. We let u = f ( x ) be some function of x and then du = f ( x ) dx . Then substitute in u and du in your integral. You have to be clever to make sure all the x ’s cancel. Be careful when you integrate definite integrals to change the bounds. Example To find R e x 1+ e x dx , let u = 1 + e x so that du = e x dx . Then Z e x 1 + e x dx = Z 1 u du = ln | u | + C = ln(1 + e x ) + C. Example To find R π/ 2 cos x 1+sin 2 x dx , let u = sin x so that du = cos x dx . Then when x = 0, u = 0 and when x = π/ 2, u = 1. Therefore, Z π/ 2 cos x 1 + sin 2 x dx = Z 1 1 1 + u 2 du = tan- 1 ( u ) 1 = tan- 1 (1)- tan- 1 (0) = √ 2 2- 0 = √ 2 2 . Example To find R tan x dx = R sin x cos x dx , let u = cos x , then du =- sin x dx . Therefore, Z sin x cos x dx =- Z 1 u du =- ln | u | + C =- ln | cos x | + C. 2. Trigonometric Integrals To compute R sin n x dx , R cos n x dx or R sin m x cos n x dx , where m and n are positive integers, the following ideas are handy. If sinpositive integers, the following ideas are handy....
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