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Assignment 13 Assigned Mon Oct 4 We refer to the integral table in the back of the book Section 7 5 Problem 3 I don t see this one in the table in the back of the book But it s a very easy substitution x dx u 2 du x u 2 dx du u x 2 2 u du u 2 3 2 u 4u1 2 C 3 2 x 2 3 2 4 x 2 1 2 C 3 Section 7 5 Problem 9 Use Integral 51 with a 2 to get x 2 2x 6 4x x2 8 x 2 1 2 x 4x x dx sin C 6 2 2 x 2 x 2 x 3 4x x2 4 sin 1 C 3 2 Section 7 5 Problem 15 Use Integral 108 with a 3 and b 3 to get e3x e3t cos 3t dt 2 3 cos 3x 3 sin 3x C 3 32 40 e3x cos 3x sin 3x C 6 Section 7 5 Problem 19 Use Integral 101 with n 2 and a 1 to get x3 1 x3 2 1 1 x tan x dx tan x dx 3 3 1 x2 So now we have to do a new integral Dividing x3 by 1 x2 gives x3 x so x3 dx 1 x2 x x 1 x2 x 1 1 dx x2 ln 1 x2 C 1 x2 2 2 Substituting this in above gives x3 1 1 x2 tan 1 x dx tan 1 x x2 ln 1 x2 C 3 6 6 Section 7 5 Problem 25 Use Integral 62 c with a cos 1 3 and b 1 4 and x to get sin 13 14 sin 31 14 cos d C 3 4 2 31 14 2 13 14 12 12 6 sin sin C 12 7 7 41 Section 7 5 Problem 29 First make a substitution 1 sin x dx 2 u sin 1 u du x u2 dx 2u du Now use Integral 99 with n 1 and a 1 and x u to get 1 u2 u2 sin 1 u u sin 1 u du du 2 2 1 u2 This last integral can be evaluated using Integral 33 with a 1 and x u 1 1 u2 du sin 1 u u 1 u2 C 2 2 1 u2 Putting it all together gives u2 1 1 u sin 1 u du sin 1 u sin 1 u u 1 u2 C 2 4 4 Then we substitute back with u x and remember there was an extra 2 in the original formula to get 1 1 sin 1 x dx x sin 1 x sin 1 x x 1 x C 2 2 1 1 x sin 1 x x x2 C 2 2 Section 7 5 Problem 31 First make the subsitution x u2 to get x u u2 dx 2u du 2 du 1 x 1 u2 1



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