quiz01_s103_solns - u = t, du = dt and dv = e t dt, v = e t...

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Name: Math 1B Quiz 1 Solutions Section 103 September 4, 2009 1. (4 points) With u = ln(cos x ) , du = - sin x cos x dx and dv = sin xdx, v = - cos x , we have Z sin x ln(cos x ) dx = - ln(cos x ) cos x - Z cos x sin x cos x dx. Cancelling and integrating yields the result: Z sin x ln(cos x ) dx = - ln(cos x ) cos x + cos x + C. Alternate method: Use substitution with t = cos x, dt = - sin x to obtain - R ln( t ) dt , and then solve that integral by parts, or by remembering a correct integral for ln( t ). 2. (4 points) Substituting with t = x, dt = 1 2 x dx , we have dx = 2 xdt = 2 tdt , so Z e x dx = Z 2 te t dt. Next using integration by parts with
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Unformatted text preview: u = t, du = dt and dv = e t dt, v = e t , we obtain 2 Z te t dt = 2( te t-Z e t dt ) = 2( te t-e t ) + C. The last step is to reverse the substitution: Z e √ x dx = 2 e √ x ( √ x-1) + C. 3. (2 points) Use the identity cos 2 x = 1-sin 2 x to get Z cos 3 xdx = Z (1-sin 2 x ) cos xdx. Use substitution with u = sin x, du = cos xdx : Z (1-sin 2 x ) cos xdx = Z (1-u 2 ) du = u-u 3 3 + C = sin x-1 3 sin 3 x + C. 1...
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This note was uploaded on 03/25/2011 for the course MATH 1B taught by Professor Reshetiken during the Fall '08 term at University of California, Berkeley.

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