quiz01_s107_solns

quiz01_s107_solns - dt and dv = sin tdt v =-cos t we obtain...

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Name: Math 1B Quiz 1 Solutions Section 107 September 4, 2009 1. (4 points) With u = ln( x 2 ) , du = 2 x x 2 dx = 2 x dx and dv = xdx, v = x 2 2 , we have Z x ln( x 2 ) dx = x 2 2 ln( x 2 ) - Z xdx = x 2 2 (ln( x 2 ) - 1) + C. Alternate method: Use substitution with t = x 2 , dt = 2 xdx to obtain 1 2 R ln( t ) dt , and then solve that integral by parts, or by remembering a correct integral for ln( t ). Note: The answer may also be expressed as x 2 (ln x - 1 2 ) + C (or some variant) by using properties of log. 2. (4 points) Substituting with t = x, dt = 1 2 x dx , we have dx = 2 xdt = 2 tdt , so Z sin xdx = Z 2 t sin tdt. Next using integration by parts with u = t, du =
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Unformatted text preview: dt and dv = sin tdt, v =-cos t , we obtain 2 Z t sin tdt = 2(-t cos t + Z cos tdt ) = 2(-t cos t + sin t ) + C. The last step is to reverse the substitution: Z sin √ xdx = 2(-√ x cos √ x + sin √ x ) + C. 3. (2 points) Use the identity sin 2 x = 1-cos 2 x to get Z sin 3 xdx = Z (1-cos 2 x ) sin xdx. Use substitution with u = cos x, du =-sin xdx : Z (1-cos 2 x ) sin xdx =-Z (1-u 2 ) du = u 3 3-u + C = 1 3 cos 3 x-cos 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 Berkeley.

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