smidterm2sol

smidterm2sol - CALCULUS 153: SAMPLE MIDTERM 2 SOLUTIONS...

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CALCULUS 153: SAMPLE MIDTERM 2 SOLUTIONS Problem 1 (28 points) . Compute the following integrals. (1) Z sin 3 x cos 2 x dx (2) Z e - x sin x dx (3) Z x 2 1 + x 2 dx (4) Z arcsin x dx Solution (1) Z sin 3 x cos 2 x dx = Z (1 - cos 2 x ) cos 2 x sin x dx. Let u = cos x so that du = - sin x dx . Then Z (1 - cos 2 x ) cos 2 x sin x dx = - Z (1 - u 2 ) u 2 du = - u 3 3 + u 5 5 + C = cos 5 x 5 - cos 3 x 3 + C. (2) Let u = sin x and dv = e - x dx so that du = cos x dx and v = - e - x . Then by IBP, Z e - x sin x dx = - e - x sin x + Z e - x cos x dx. Performing IBP again ( u = cos x , dv = e - x dx ), one gets that Z e - x sin x dx = - e - x sin x - e - x cos x - Z e - x sin x dx so that Z e - x sin x dx = - e - x sin x + e - x cos x 2 + C. (3) Easy way: Z x 2 1 + x 2 dx = Z ± 1 - 1 1 + x 2 ² dx = x - arctan x + C. To do it the harder way, let x = tan u . You should get the same answer (this is good practice). 1
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2 CALCULUS 153: SAMPLE MIDTERM 2 SOLUTIONS (4) Use IBP. Let u = arcsin x , dv = dx so that du = 1 / 1 - x 2 dx , v
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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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smidterm2sol - CALCULUS 153: SAMPLE MIDTERM 2 SOLUTIONS...

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