656 7313 let u ln t 2 and dv dt du 2 ln t t dt and

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7.3.13: Let u = (ln t ) 2 and dv = dt : du = 2 ln t t dt and choose v = t . Then (ln t ) 2 dt = t (ln t ) 2 2 ln t dt. Next let u = ln t and dv = dt : du = 1 t dt and choose v = t . Thus (ln t ) 2 dt = t (ln t ) 2 2 t ln t 1 dt = t (ln t ) 2 2 t ln t + 2 t + C. 7.3.14: Let u = (ln t ) 2 and dv = t dt . Then du = 2 ln t t dt ; choose v = 1 2 t 2 . Thus t (ln t ) 2 dt = 1 2 ( t ln t ) 2 t ln t dt = 1 2 ( t ln t ) 2 1 2 t 2 ln t + 1 4 t 2 + C. (The last equality follows from the result in Problem 6.) 7.3.15: Let u = x and dv = ( x + 3) 1 / 2 dx : du = dx and choose v = 2 3 ( x + 3) 3 / 2 . Then x ( x + 3) 1 / 2 dx = 2 3 x ( x + 3) 3 / 2 2 3 ( x + 3) 3 / 2 dx = 2 3 x ( x + 3) 3 / 2 4 15 ( x + 3) 5 / 3 + C = ( x + 3) 3 / 2 2 3 x 4 15 x 4 5 + C = ( x + 3) 3 / 2 6 x 12 15 + C = 2 5 ( x 2)( x + 3) 3 / 2 + C = 2 5 ( x 2 + x 6) x + 3 + C. 7.3.16: Let u = x 2 and dv = x (1 x 2 ) 1 / 2 : du = 2 x dx ; choose v = 1 3 (1 x 2 ) 3 / 2 . Then x 3 (1 x 2 ) 1 / 2 dx = 1 3 x 2 (1 x 2 ) 3 / 2 + 2 3 x (1 x 2 ) 3 / 2 dx = 1 3 x 2 (1 x 2 ) 3 / 2 2 15 (1 x 2 ) 5 / 2 + C = (1 x 2 ) 3 / 2 1 3 x 2 + 2 15 (1 x 2 ) + C = (1 x 2 ) 3 / 2 3 x 2 + 2 15 + C = 1 15 (3 x 4 x 2 2) 1 x 2 + C. 7.3.17: Let u = x 3 and dv = x 2 ( x 3 + 1) 1 / 2 dx : du = 3 x 2 dx and choose v = 2 9 ( x 3 + 1) 3 / 2 . Then x 5 ( x 3 + 1) 1 / 2 dx = 2 9 x 3 ( x 3 + 1) 3 / 2 2 3 x 2 ( x 3 + 1) 3 / 2 dx = 2 9 x 3 ( x 3 + 1) 3 / 2 4 45 ( x 3 + 1) 5 / 2 + C = 1 45 ( x 3 + 1) 3 / 2 10 x 3 4( x 3 + 1) + C = 1 45 ( x 3 + 1) 3 / 2 (6 x 3 4) + C = 2 45 ( x 3 + 1) 3 / 2 (3 x 3 2) + C = 2 45 ( x 3 + 1) 1 / 2 (3 x 6 + x 3 2) + C. 657
7.3.18: Let u = sin θ and dv = sin θ dθ : du = cos θ dθ and choose v = cos θ . Then sin 2 θ dθ = sin θ cos θ + cos 2 θ dθ = sin θ cos θ + (1 cos 2 θ ) = sin θ cos θ + θ sin 2 θ dθ ; 2 sin 2 θ dθ = θ sin θ cos θ + 2 C ; sin 2 θ dθ = 1 2 ( θ sin θ cos θ ) + C. 7.3.19: Let u = csc θ and dv = csc 2 θ dθ : du = csc θ cot θ and choose v = cot θ . Then csc 3 θ dθ = csc θ cot θ csc θ cot 2 θ dθ = csc θ cot θ (csc θ )(csc 2 θ 1) = csc θ cot θ csc 3 θ dθ + csc θ dθ ; 2 csc 3 θ dθ = csc θ cot θ + ln | csc θ cot θ | + 2 C ; csc 3 θ dθ = 1 2 csc θ cot θ + 1 2 ln | csc θ cot θ | + C. Mathematica returns the antiderivative in the form C 1 2 cot θ csc θ 1 2 ln cos θ 2 + 1 2 ln sin θ 2 , whereas Maple yields an answer that is essentially the same as the one we obtained “by hand.” 7.3.20: Let u = sin(ln t ) and dv = dt : du = 1 t cos(ln t ) and choose v = t . Then sin(ln t ) dt = t sin(ln t ) cos(ln t ) dt. Now let u = cos(ln t ) and dv = dt : du = 1 t sin(ln t ) dt , and choose v = t . Thus sin(ln t ) dt = t sin(ln t ) t cos(ln t ) sin(ln t ) dt ; sin(ln t ) dt = 1 2 t sin(ln t ) 1 2 t cos(ln t ) + C. 7.3.21: Let u = arctan x and dv = x 2 dx : du = 1 1 + x 2 dx and choose v = 1 3 x 3 . Then x 2 arctan x dx = 1 3 x 3 arctan x 1 3 x 3 x 2 + 1 dx = 1 3 x 3 arctan x 1 3 x x x 2 + 1 + C = 1 3 x 2 arctan x 1 6 x 2 + 1 6 ln( x 2 + 1) + C. 658
7.3.22: Let u = ln(1 + x 2 ) and dv = dx : du = 2 x 1 + x 2 dx and choose v = x . Then ln(1 + x 2 ) dx = x ln(1 + x 2 ) 2 x 2 1 + x 2 dx = x ln(1 + x 2 ) 2 2 1 + x 2 dx = x ln(1 + x 2 ) 2 x + 2 arctan x + C.

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