Thomas' Calculus: Early Transcendentals

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MAT 104 Quiz 1, due Feb 21, 2003 on simple substitutions, integration by parts and partial fractions 1. (10 points) Find e 1 /x x 3 dx . We make the substitution t = 1 /x and then use integration by parts. If t = 1 /x , then dt = ( - 1 /x 2 ) dx so dt = - t 2 du . Thus e 1 /x x 3 dx = - t 3 e t dt t 2 = - te t dt = - te t + e t dt = - te t + e t + C = - e 1 /x x + e 1 /x + C. Here we have used integration by parts with u = t and dv = e t dt so du = dt and v = e t . 2. (10 points) Find π/ 2 0 cos x 2 - cos 2 x dx . (Hint: Use the identity sin 2 x + cos 2 x = 1.) The denominator 2 - cos 2 x is the same as 2 - (1 - sin 2 x ) = 1 + sin 2 x . The integral becomes π/ 2 0 cos x 1 + sin 2 x dx = sin π/ 2 sin 0 du 1 + u 2 from the subsitution u = sin x and du = cos x dx. So we get arctan(1) - arctan 0 = π 4 - 0 = π 4 . 3. (10 points) Find 3 x 2 arctan x 3 dx . Note: You may be used to calling the inverse tangent function tan - 1 instead of arctan. Both notations are standard. So arctan x 3 means exactly the same as tan - 1 ( x 3 ). If t = x 3 then dt = 3 x 2 dx and we get arctan( t ) dt . Now use integration by parts with u = arctan t and dv = dt . Thus du = dt/ (1 +
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