Thomas' Calculus: Early Transcendentals

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MAT 104 Midterm March 2003 Problem Possible score Average score 1 12 9.5 2 12 4.9 3 12 8.9 4 12 5.4 5 12 7.5 6 12 8.1 7 12 6.8 8 6 4.8 9 10 4.3 Total 100 60.2 1. (12 points) Find sin 3 (ln x ) cos 2 (ln x ) x dx . Let u = ln x . Then the integral becomes sin 3 u cos 2 u du = sin 2 u cos 2 u sin u du = (1 - cos 2 u ) cos 2 u sin u du. Substitute t = cos u to get - (1 - t 2 ) t 2 dt = ( t 4 - t 2 ) dt = t 5 5 - t 3 3 + C = cos 5 (ln x ) 5 - cos 3 (ln x ) 3 + C. 2. (12 points) Find ln( x 2 + x + 1) x 2 dx . Do integration by parts with u = ln( x 2 + x + 1) and dv = dx/x 2 . Then du = (2 x + 1) dx/ ( x 2 + x + 1) and v = - 1 /x . So ln( x 2 + x + 1) x 2 dx = - 1 x ln( x 2 + x + 1) + 2 x + 1 x ( x 2 + x + 1) Now use partial fractions to compute the integral on the right. 2 x + 1 x ( x 2 + x + 1) = A x + Bx + C x 2 + x + 1 = Ax 2 + Ax + A + Bx 2 + Cx x ( x 2 + x + 1) This leads to 0 = A + B , 2 = A + C , 1 = A and we conclude that B = - 1 and C = 1. So now we have 2 x + 1 x ( x 2 + x + 1) dx = dx x + 1 - x x 2 + x + 1 dx = ln | x | + 1 - x ( x + 1 / 2) 2 + 3 / 4 dx.
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In this integral we make the substitution u = x + 1 / 2 and then - u + 3 / 2 = - x + 1. So we get ln | x | + 1 - x ( x + 1 / 2) 2 + 3 / 4 dx = ln | x | - 1 2 2 u u 2 + 3 / 4 du + 3 2 du u 2 + 3 / 4 = ln | x | - 1 2 ln( u 2 + 3 / 4) + 2 3 3 2 arctan 2 u 3 + C Returning to the original variable x and combining all the pieces gives - 1 x ln( x 2 + x + 1) + ln | x | - 1 2 ln( x 2 + x + 1) + 3 arctan 2 x + 1 3 + C 3. (12 points) Find e 3 x arctan( e x ) dx . Start with a substitution w = e x . Then dw = e x dx . We get w 3 arctan( w ) dw w = w 2 arctan( w ) dw. Now do integration by parts, with u = arctan w and dv = w 2 dw . So du = dw/ (1+ w 2 ) and v = w 3 / 3.
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