271s1 - Solutions to MAT 271 Test #1 (1) (15 points)...

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(1) (15 points) Evaluate the integral Z t 2 sin(2 t ) dt . Solution: Use integration by parts twice: Z t 2 sin(2 t ) dt = t 2 · - cos(2 t ) 2 - Z - 2 t cos(2 t ) 2 dt u = t 2 v 0 = sin(2 t ) u 0 = 2 t v = - cos(2 t ) 2 = - t 2 cos(2 t ) 2 + Z t cos(2 t ) dt = - t 2 cos(2 t ) 2 + t · sin(2 t ) 2 - Z 1 · sin(2 t ) 2 dt u = t v 0 = cos(2 t ) u 0 = 1 v = sin(2 t ) 2 = - t 2 cos(2 t ) 2 + t sin(2 t ) 2 + cos(2 t ) 4 + C. Grading: +5 points for each integration by parts, +5 points for the final answer. Grading for common mistakes: +7 points (total) for choosing u and v 0 the wrong way and continuing; - 1 point for no + C . (2) (15 points) Evaluate the integral Z sin 3 x cos 2 xdx . Solution: This is a trigonometric integral, and the power of sin x is odd, so we will keep one of these sin x ’s as a factor and convert the rest to cosines; then we let u = cos x . Z sin 3 x cos 2 xdx = Z sin x (sin 2 x )cos 2 xdx = Z sin x (1 - cos 2 x )cos 2 xdx = Z - (1 - u 2 ) u 2 du u = cos x du/dx = - sin x du = - sin xdx = Z u 4 - u 2 du = u 5 5 - u 3 3 + C = cos 5 x 5 - cos 3 x 3 + C. Grading: +5 points for recognizing the integral was a trig integral, +5 points for the substitution u = cos x , +5 points for integrating the polynomial and sub- stituting. Grading for common mistakes:
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271s1 - Solutions to MAT 271 Test #1 (1) (15 points)...

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