int_parts_sol - MATH 112-SPRING 2008-SECTION 5 Achilleas...

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MATH 112-SPRING 2008-SECTION 5 Achilleas Sinefakopoulos asin@math.cornell.edu 1. Find the error in the proof of the following claim. Claim : 0 = 1. Proof . By integration by parts (with dv = dx and u = 1 x ), we get Z 1 x dx = x · 1 x - Z x ± - 1 x 2 dx = 1 + Z 1 x dx. So by cancelling Z 1 x dx from both sides, we get 0 = 1. Solution . Cancelling Z 1 x dx is not allowed because it denotes a“family” of functions whose derivatives is 1 x and not just one function. That’s why we include “ + C ” in an indefinite integral. 2. Find the following indefinite integrals. (i) Z x ln xdx = x 2 2 ln x - x 2 4 + C Hint : Apply integration by parts with u = ln x and dv = xdx . (ii) Z x sin - 1 xdx = x 2 2 sin - 1 x - 1 4 sin - 1 x + x 4 1 - x 2 + C Hint : Apply integration by parts with u = sin - 1 x and dv = xdx . (iii) Z (ln x ) 2 dx = x (ln x ) 2 - 2 x ln x + 2 x + C Hint : Apply integration by parts with u = (ln x ) 2 and dv = dx . Then apply integration by parts again to find Z ln xdx . (iv) Z sin 3 xdx = - 1 3 sin 2 x cos x - 2 3 cos x + C (v) Z ln( x + 1 + x 2 ) dx = x ln( x + 1 + x 2 ) - 1 + x 2
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This note was uploaded on 02/28/2010 for the course MATH 1120 taught by Professor Gross during the Spring '06 term at Cornell University (Engineering School).

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