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midtermsol2 - CALCULUS 153: MIDTERM 2 Please answer all...

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Unformatted text preview: CALCULUS 153: MIDTERM 2 Please answer all questions in a blue book that’s provided to you. Don’t forget to write your name. There are two sides to this exam. Problem 1 (32 points) . Compute the following integrals (8 points each) (a) Z tan x sin 2 x dx (b) Z dx x 2- 4 x + 5 (c) Z arctan x dx (d) Z x + 1 x 2- 2 x + 1 dx. Solution (a) Z tan x sin 2 x dx = Z sin x cos x (1- cos 2 x ) dx Letting u = cos x we get Z sin x cos x (1- cos 2 x ) dx = Z ( 1 u- u )(- du ) =- ln | u | + u 2 2 + C =- ln | cos x | + cos 2 x 2 + C. (b) Z dx x 2- 4 x + 5 = Z dx ( x- 2) 2 + 1 = arctan( x- 2) + C. (c) We use IBP. Let u = arctan x and dv = dx so that du = (1+ x 2 )- 1 dx and v = x . Then Z arctan x dx = x arctan x- Z x 1 + x 2 = x arctan x- 1 2 ln(1 + x 2 ) + C. (d) Note that x 2- 2 x + 1 = ( x- 1) 2 . A partial fraction decomposition gives that x + 1 x 2- 2 x + 1 = 1 x- 1 + 2 ( x- 1) 2 . Therefore, Z x + 1 x 2- 2 x + 1 dx = Z dx x- 1 + 2 Z dx ( x- 1) 2 = ln | x- 1 | - 2 x- 1 + C....
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This test prep was uploaded on 04/07/2008 for the course MATH 153 taught by Professor Masson during the Fall '07 term at UChicago.

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midtermsol2 - CALCULUS 153: MIDTERM 2 Please answer all...

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