ex-int-indef-s - Solution&Inde¡nite Integrals Question...

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Unformatted text preview: Solution &Inde¡nite Integrals Question 1 (a) Z (3 & x 2 ) 3 dx = Z & 27 & 27 x 2 + 9 x 4 & x 6 ¡ dx = 27 x & 27 x 3 3 + 9 x 5 5 & x 7 7 + C = 27 x & 9 x 3 + 9 5 x 5 & 1 7 x 7 + C (b) Z ¢ 1 & 1 x 2 £ q x p xdx = Z ¢ 1 & 1 x 2 £ p xx 1 = 2 dx = Z ¢ 1 & 1 x 2 £ p x 3 = 2 dx = Z ¢ 1 & 1 x 2 £ x 3 = 4 dx = Z & x 3 = 4 & x & 5 = 4 ¡ dx = 4 7 x 7 = 4 + 4 x & 1 = 4 + C = 4( x 2 + 7) 7 4 p x + C (c) Z ¢ a x + a 2 x 2 + a 3 x 3 £ dx = a ln j x j + a 1 ( & 1) x & 2+1 + a 3 1 ( & 2) x & 3+1 + C = a ln j x j & a 2 x & a 3 2 x 2 + C (d) Z x + 1 p x dx = Z ¢ x p x & 1 p x £ dx = Z & x 1 = 2 & x & 1 = 2 ¡ dx = 1 1 2 x 1 = 2+1 & 1 & 1 2 x & 1 = 2+1 + C = 2 3 x p x + 2 p x + C (e) Z (1 & x )(1 & 2 x )(1 & 3 x ) dx = Z & 1 & 6 x + 11 x 2 & 6 x 3 ¡ dx = 1 0 + 1 x 0+1 & 6 1 + 1 x 1+1 + 11 2 + 1 x 2+1 & 6 3 + 1 x 3+1 + C = x & 3 x 2 + 11 3 x 3 & 3 2 x 4 + C (f) Let x = tan u; dx = sec 2 udu and note that sec 2 u = 1 + tan 2 u Z x 2 1 + x 2 dx = Z x 2 + 1 & 1 1 + x 2 dx = Z ¢ 1 & 1 1 + x 2 £ dx = x & Z dx 1 + x 2 = x & Z sec 2 udu 1 + tan 2 u = x & Z du = x & u + c = x & tan & 1 x + C (g) Z (1 + sin x + cos x ) dx = x & cos x + sin x + C 1 (h) Let u = 1 & x; du = & dx Z 5 p 1 & 2 x + x 2 1 & x dx = Z 5 q (1 & x ) 2 1 & x dx = Z (1 & x ) 2 = 5 1 & x dx = Z (1 & x ) & 3 = 5 dx = Z u & 3 = 5 ( & du ) = & 1 1 & 3 = 5 u & 3 = 5+1 + C = & 5 2 5 p (1 & x ) 2 + C (i) Z (2 x + 3 x ) 2 dx = Z & 2 2 x + 2 (2 x 3 x ) + 3 2 x ¡ dx = Z & 2 2 x + 2 (6 x ) + 3 2 x ¡ dx = 4 x ln 4 + 2 6 x ln 6 + 9 x ln 9 + C (j) Let u = 2 x & 3 ; du = 2 dx Z (2 x & 3) 10 dx = Z u 10 du 2 = 1 2 1 10 + 1 u 10+1 + C = 1 22 (2 x & 3) 11 + C (k) Let u = x= 2 ; du = dx= 2 Z dx 1 + cos x = Z dx 2 cos 2 ( x= 2) = Z sec 2 udu = tan u + C = tan x= 2 + C (l) Let u = 1 & x 2 ; du = & 2 xdx Z xdx p 1 & x 2 = Z x p u du & 2 x = & 1 2 Z u & 1 = 2 du = & 1 2 1 & 1 2 + 1 u & 1 = 2+1 + C = & p 1 & x 2 + C (m) Let u = 1 + x 2 ; du = 2 xdx Z xdx (1 + x 2 ) 2 = Z x u 2 du 2 x = 1 2 Z u & 2 du = 1 2 1 & 2 + 1 u & 2+1 + C = & 1 2(1 + x 2 ) + C (n) Let u = p x; du = 1 2 p x dx Z dx p x (1 + x ) = Z 2 p xdu p x (1 + u 2 ) = 2 Z du 1 + u 2 = 2 tan & 1 u + C = 2 arctan p x + C (o) Let u = p x; du = 1 2 p x dx Z dx p x (1 & x ) = Z 2 p xdu p x p 1 & u 2 = 2 Z du p 1 & u 2 = 2 sin & 1 u + C = 2 sin & 1 p x + C (p) Let u = & x 2 ; du = & 2 xdx Z xe & x 2 dx = Z xe u du & 2 x = & 1 2 Z e u du = & 1 2 e & x 2 + C (q) Let u = e & x ; du = & e & x dx and y = u + p 1 + u 2 ; dy...
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This note was uploaded on 10/19/2011 for the course MATHEMATIC 33 taught by Professor Qian during the Spring '99 term at Hong Kong Shue Yan.

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ex-int-indef-s - Solution&Inde¡nite Integrals Question...

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